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Q `=Sheet1BSheet2YSheet3," AB  correct, well done!7A  this is correct since Cr is in Group 6  well done.E  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet.$B  incorrect as Rh is a 4d element.;A  this is incorrect since the complex is neutral overall.;D  this is incorrect since the complex is neutral overall.;E  this is incorrect since the complex is neutral overall.HA  incorrect as there are 2 F environments, hence must be 2 resonances.HC  incorrect as there are 2 F environments, hence must be 2 resonances.$C  incorrect as Tc is a 4d element.RA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5QRE  incorrect: Fe is a 3d element so OK there, but Fe is in Gp 8, so Fe(IV) is d4.PQBA  incorrect as a tetrahedral ion cannot have a c4 rotation axis.23$B  incorrect as there's no c3 axis.C  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axesOP.D  incorrect as there is no c3 rotation axis.oD  incorrect  you've spotted way too many isomers. Make sure sure you understand why B is the correct answer.oE  incorrect  you've spotted way too many isomers. Make sure sure you understand why B is the correct answer.A  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)..D  correct, as shown at the end of lecture 2.tE  incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2).A  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes.C  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around.B  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes.[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.[B  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.[C  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.No answer given: B is correct.4No answer given: A is correct since Cr is in Group 6No answer given:  E is correc as there are 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet.ED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3.CDWNo answer given: D is correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3.kNo answer given: E is correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.@No answer given: D is correct, as shown at the end of lecture 2.kNo answer given: A is correct as the ratio of spheres in the close packed array to octahedral holes is 1:1.uNo answer given: B is correct as this is a quarter of the way from one corner to the opposite corner across the cube.YNo answer given: A is correct as monoclinic cells have a `" b `" c, = = 90and `" 90.1352FDElianAdamsPhilipAdler1352C1BeckyAldus0F8981SarahAndersonNatalieAndrews1st nameemail
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Feedback 9Feedback 10Feedback 11Feedback 12Feedback 13Feedback 14Feedback 15MarkpD  incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states.%gC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states.B  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). No answer given: E is correct as CO is a sigma donor/pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states).5bE  incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane.&'No answer given: C is correct as 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes.A  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!B  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone.;C  incorrect as trans isomer does not have optical isomers No answer given: B is correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone.+ 05 8T W E  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states).vA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it.YB  incorrect. You have to choose the smallest cell possible within the highest symmetry.PC  correct. First choose the highest symmetry then the smallest cell within it.RD  incorrect. First choose the highest symmetry then the smallest cell within it.eE  incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry.cNo answer given: C is correct  first choose the highest symmetry then the smallest cell within it.@B  incorrect as cubic cells have a = b = c and = = = 90.GA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90.>C  incorrect as triclinic cells have a `" b `" c and `" `" .GD  incorrect as orthorhombic cells have a `" b `" c and = = = 90.IE  incorrect as hexagonal cells have a = b `" c, = = 90 and `" 90.cA  incorrect as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell).aB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell).cC  incorrect as there are 4 large spheres (81/8 + 6) and 8 small sp<heres (all inside the cell).D  incorrect as you have just counted the number of circles! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell).E  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell).sNo answer given: B is correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell).A  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face.aB  correct. This is a quarter of the way from one corner to the opposite corner across the cube.C  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge.D  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face.zE  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway.YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1.[B  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1.[C  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1.[D  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1.[E  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1.7A  incorrect. See calculation at the end of lecture 2.oB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2).7C  incorrect. See calculation at the end of lecture 2.^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.pNo answer given: D is correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.C  you got the charge right, but remember that Cl counts as 1 charge and O counts as 2 charge, hence Vanadium must have the oxidation state (IV).B  incorrect since although there are 2 F environments, there is only 1 F in one environment (trans NMe3) and 2 mutually trans F s this cannot give rise to 2 triplets.D  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings).B  incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons.C  incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons.D  incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons.E  incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons..doc]C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Trial\pathname13536F1352C40F88D60F897BB0F88DE1352F30F891E13536D0F896C0F89671352A01353ADC0F88870F896213524A1352F813FBD21352FED0F89891353BD1353BE0F88D31352C813529E1352C5E0F884513524D13FBDC0F88BC13537E13524B13532AF0F895D0F89320F88370F88BFG13FBD413FBD51352F7H13FBE013FBE51353BF0F88710F888F13524C0F8833Joanne0F89ABI13534E0F89AC0F899C0F88470F88491353BB13FC0D0F8994J13524F13FBE613FBD913FBDE13FC521353BC0F883BK0F897413FBD31353AB13FC531353C0
short namesurname@Excellent effort  you scored full marks. Keep up the good work!=Well done on scoring such a high mark. Keep up the good work!You've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!`You've passed, so well done, but make sure you think carefully about what you can do to improve.Not too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.You were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrongN/AOverall feedback_C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\JoeJoBloggsJosephjbloggs@soton.ac.uk
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!f@?@@@@@?@@??@@?@/ !D;BfB  correct, well done!/ D;Bf:7A  this is correct since Cr is in Group 6  well done./<
D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./ D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./d D;Bf1.D  incorrect as there is no c3 rotation axis./ D;#BfroE  incorrect  you've spotted way too many isomers. Make sure sure you understand why B is the correct answer./
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;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./ D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./ D;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./ D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./! D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./" !D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./# "D;HMBf:7C  incorrect. See calculation at the end of lecture 2./$ #D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./% $D;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y3('x )D&DDDD'nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA1.doc/)h (D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.~
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!f@?@@@@@@?@?@@@@"@/, D;BfB  correct, well done!/\ D;Bf:7A  this is correct since Cr is in Group 6  well done./ D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./h D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./ D;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/$ D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./D D
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./ D;*/BfheE  incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry./d D;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./ D;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./! !D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./" "D;BGBf^[D  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./#! #D;HMBf1.D  correct, as shown at the end of lecture 2./$" $D;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./%# %D;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y3( (D&DDDD'nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA2.doc/)$ )D;YhBfYou've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!

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!f@?@@@@@?@@??@@@"@/% D;BfB  correct, well done!/% D;Bf:7A  this is correct since Cr is in Group 6  well done./X& D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./' D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./( D;BfebE  incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane./H) D;#BfroE  incorrect  you've spotted way too many isomers. Make sure sure you understand why B is the correct answer./(* D
;$)BfspD  incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states./+ D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./+ D;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./ , D;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!P !D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./". "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./#@/ #D;HMBf1.D  correct, as shown at the end of lecture 2./$/ $D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./%`0 %D;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y3(] (D&DDDD'nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA3.doc/)1 )D;YhBfYou've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
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!f@?@@@@@@?@?@@@@"@/2 D;BfB  correct, well done!/,3 D;Bf:7A  this is correct since Cr is in Group 6  well done./3 D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./85 D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./5 D;Bf1.D  incorrect as there is no c3 rotation axis./$6 D;#BfroE  incorrect  you've spotted way too many isomers. Make sure sure you understand why B is the correct answer./7 D
;$)BfspD  incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states./7 D;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./8 D;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./ 9 D;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!9 !D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./"; "D;BGBf^[E  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./#; #D;HMBf1.D  correct, as shown at the end of lecture 2./$4< $D;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./%@= %D;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y3(2 %D&DDDD'nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA4.doc/)> )D;YhBfYou've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!~
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!f@??@@@@?@@??@@@ @/? %D;BfB  correct, well done!/@ D;Bf:7A  this is correct since Cr is in Group 6  well done./@ D ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./A D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./A D;Bf1.D  incorrect as there is no c3 rotation axis./B D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./ C D
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./C D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./D D;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./ pE D;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!4F D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./"pG !D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./#$H "D;HMBf1.D  correct, as shown at the end of lecture 2./$H #D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./%DI $D;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y3( $D&DDDD'nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA5.doc/)J )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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!f@?@@@@??@@@?@@@@/K D;BfB  correct, well done!/K D;Bf:7A  this is correct since Cr is in Group 6  well done./;A  this is incorrect since the complex is neutral overall./
L D
;Bf:7A  this is correct since Cr is in Group 6  well done./
D
;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./
P D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./
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;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/
D
;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./
, D
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./
D
;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./
@ D
;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./
У D
;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./
! !D
;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./
"X "D
;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./
# #D
;HMBf1.D  correct, as shown at the end of lecture 2./
$l $D
;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./
%, %D
;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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!f@?@@?@@@@@@@@@@@/4 D;BfB  correct, well done!/d D;Bf:7A  this is correct since Cr is in Group 6  well done./ԩ D ;BfKHC  incorrect as there are 2 F environments, hence must be 2 resonances./h D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./ D;BfEBA  incorrect as a tetrahedral ion cannot have a c4 rotation axis./ D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./ D
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./l D;*/BfheE  incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry./8 D;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./ Ȯ D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./! !D;<ABfC  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge./" "D;BGBf^[E  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./# #D;HMBf1.D  correct, as shown at the end of lecture 2./$ $D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./%ز %D;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y((((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsB6.doc/)` )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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!f@?@@@@@?@@@@@?@@/$ D;Bf>;D  this is incorrect since the complex is neutral overall./ D;Bf:7A  this is correct since Cr is in Group 6  well done./ D ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./ D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./ D;BfebE  incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane./` D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./ D
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./Ի D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./ļ D;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./ T D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!x !D;<ABf}zE  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway./"p "D;BGBf^[D  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./#( #D;HMBfwtE  incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2)./$ $D;NRBfA  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./%4 %D;SXBf^[B  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y((((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsB7.doc/) )D;YhBfNot too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
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!f@?@@@@@@@@@@@@?@/ D;BfC  you got the charge right, but remember that Cl counts as 1 charge and O counts as 2 charge, hence Vanadium must have the oxidation state (IV)./ D;Bf:7A  this is correct since Cr is in Group 6  well done./ D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./ D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./H D;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/x D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./ D
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./ D;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./ D;05BfIE  incorrect as hexagonal cells have a = b `" c, = = 90 and `" 90./ $ D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!H !D;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./" "D;BGBf^[B  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./# #D;HMBf:7C  incorrect. See calculation at the end of lecture 2./$4 $D;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./%@ %D;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y(X(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsB8.doc/) )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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'y((((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC1.doc/)l )D;YhBfN/A~
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!f@?@?@?@??@??@@?@/t D;BfB  correct, well done!/ D;Bf:7A  this is correct since Cr is in Group 6  well done./ D ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./ D
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/ D;Bf1.D  incorrect as there is no c3 rotation axis./ D;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/@ D
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./ D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./ D;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./ D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./! !D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./" "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./# #D;HMBfroB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2)./$ $D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./%D %D;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y((((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC2.doc/) )D;YhBfYou were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
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!f@?@@@@@??@@?@?@&@/8 D;BfB  correct, well done!/h D;Bf:7A  this is correct since Cr is in Group 6  well done./ D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./t D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./ D;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/0 D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./P D
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./ D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./ D;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./ @ D;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./! !D;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./" "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./# #D;HMBf:7C  incorrect. See calculation at the end of lecture 2./$ $D;NRBfA  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./% %D;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y((((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC3.doc/) )D;YhBf@=Well done on scoring such a high mark. Keep up the good work!
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!f@??@@@@@@@??@@? @/ D;BfB  correct, well done!/H D;Bf:7A  this is correct since Cr is in Group 6  well done./ D ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./L D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./ D;Bf1.D  incorrect as there is no c3 rotation axis./8 D;#Bf>;C  incorrect as trans isomer does not have optical isomers/ D
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./ D;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./$ D;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./ D;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!x !D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./" "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./#h #D;HMBf1.D  correct, as shown at the end of lecture 2./$ $D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./% %D;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y(@(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC4.doc/) )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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!f@?@@@@@@?@??@@@$@/ D;BfB  correct, well done!/ D;Bf:7A  this is correct since Cr is in Group 6  well done./ D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./ D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./ D;Bf1.D  incorrect as there is no c3 rotation axis./l D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./ D
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./ D;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./ D;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./ 0 D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!T !D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./" "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./#D #D;HMBfroB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2)./$$ $D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./% %D;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y((((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC5.doc/)p
)D;YhBfYou've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!~
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!f@?@@?@??????@@@@/ D;BfB  correct, well done!/ D;Bf:7A  this is correct since Cr is in Group 6  well done./ D ;BfKHC  incorrect as there are 2 F environments, hence must be 2 resonances./ D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./@
D;BfEBA  incorrect as a tetrahedral ion cannot have a c4 rotation axis./
D;#Bf>;C  incorrect as trans isomer does not have optical isomers/@ D
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./ D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./ D;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./ D;6;BffcA  incorrect as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./! !D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./" "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./# #D;HMBfroB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2)./$ $D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./%l %D;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y(/x(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC6.doc/) )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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;Bf/*{ D;Bf/* D;#Bf/*~ D
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'y(0x(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC7.doc/) )D;YhBfN/A
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!f@?@@@@?@@@@?@@@$@/ D;BfB  correct, well done!/ D;Bf:7A  this is correct since Cr is in Group 6  well done./@ D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./ D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./h D;Bf'$B  incorrect as there's no c3 axis./ D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./ D
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./( D;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./ D;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./ L D;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./! !D;<ABfC  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge./"$! "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./#! #D;HMBf1.D  correct, as shown at the end of lecture 2./$8" $D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./%" %D;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y(T1x(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC8.doc/)$ )D;YhBfYou've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
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!f@?@@@????@@?@@@ @/% D;BfB  correct, well done!/% D;Bf:7A  this is correct since Cr is in Group 6  well done./<& D ;BfWB  incorrect since although there are 2 F environments, there is only 1 F in one environment (trans NMe3) and 2 mutually trans F s this cannot give rise to 2 triplets./' D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./ ( D;Bf'$B  incorrect as there's no c3 axis./l( D;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/) D
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./* D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./+ D;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./ d, D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./! !D;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./"L. "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./#/ #D;HMBf1.D  correct, as shown at the end of lecture 2./$`/ $D;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./% 0 %D;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y(,2x(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD1.doc/)1 )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.~
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'y(3x(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD2.doc/)X3 )D;YhBfN/A
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!f@?@@@@?@?@?@@@@ @/`3 D;BfB  correct, well done!/3 D;Bf:7A  this is correct since Cr is in Group 6  well done./4 D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./5 D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./(6 D;Bf1.D  incorrect as there is no c3 rotation axis./6 D;#Bf>;C  incorrect as trans isomer does not have optical isomers/7 D
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./T8 D;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./8 D;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./ 9 D;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!L: !D;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./"; "D;BGBf^[D  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./#@< #D;HMBfwtE  incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2)./$,= $D;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./%8> %D;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y(>(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD3.doc/)? )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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!f@?@@?@?@@@@@@@?@/@ D;BfB  correct, well done!/@ D;Bf:7A  this is correct since Cr is in Group 6  well done./(A D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./B D
;Bf'$C  incorrect as Tc is a 4d element./C D;BfEBA  incorrect as a tetrahedral ion cannot have a c4 rotation axis./C D;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./D D
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./F D;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./F D;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./ 0G D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!TH !D;<ABfC  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge./"hI "D;BGBf^[C  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./# J #D;HMBf1.D  correct, as shown at the end of lecture 2./$J $D;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./%K %D;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y(3x(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD4.doc/)M )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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!f@@@?@?@?@@@?@@@ @/N D;BfB  correct, well done!/N D;BfB  incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons./P D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./Q D
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/XR D;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/S D;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/T D
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./V D;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./W D;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./ W D;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./!X !D;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./"Y "D;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./#8Z #D;HMBf1.D  correct, as shown at the end of lecture 2./$Z $D;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./%[ %D;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'y(X\(((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD6.doc/)0] )D;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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!f @?@@@@?@@@@??@@$@/ ^ !D ;BfB  correct, well done!/ _ !D ;Bf:7A  this is correct since Cr is in Group 6  well done./ t_ !D ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./ a !D
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./ a !D ;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/ b !D ;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./ c !D
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./ @e !D ;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./ e !D ;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./ df ! D ;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./ !(g !!D ;<ABfD  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face./ "Xh !"D ;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./ #i !#D ;HMBf:7A  incorrect. See calculation at the end of lecture 2./ $i !$D ;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./ %C  incorrect as triclinic cells have a `" b `" c and `" `" ./" Ѐ # D";6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./"! #!D";<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./""Ђ #"D";BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./"# ##D";HMBf1.D  correct, as shown at the end of lecture 2./"$ #$D";NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./"% #%D";SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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#!f#@?@?@@@?@@@@@@@"@/# $D#;BfB  correct, well done!/#ć $D#;Bf:7A  this is correct since Cr is in Group 6  well done./#4 $D# ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./#Љ $D#
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/#x $D#;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/# $D#;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./#Ȍ $D#
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./# $D#;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./# $D#;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./# $ D#;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./#!P $!D#;<ABfC  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge./#"d $"D#;BGBf^[C  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./## $#D#;HMBf1.D  correct, as shown at the end of lecture 2./#$ $$D#;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./#%< $%D#;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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$!f$@?@@@@@@?@@?@@@&@/$ܕ %D$;BfB  correct, well done!/$ %D$;Bf:7A  this is correct since Cr is in Group 6  well done./$ %D$ ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./$ %D$
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./$ %D$;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/$8 %D$;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./$X %D$
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./$Ȝ %D$;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./$l %D$;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./$ % D$;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./$! %!D$;<ABfC  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge./$"ԟ %"D$;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./$# %#D$;HMBfwtE  incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2)./$$t %$D$;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./$%4 %%D$;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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%!f%@?@@@@?@????@@@"@/%@ &D%;BfB  correct, well done!/%p &D%;Bf:7A  this is correct since Cr is in Group 6  well done./% &D% ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./% &D%
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./%l &D%;BfebE  incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane./%4 &D%;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./%T &D%
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./% &D%;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./%L &D%;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./% ܫ & D%;6;BffcA  incorrect as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./%! &!D%;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./%" &"D%;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./%# &#D%;HMBfwtE  incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2)./%$ &$D%;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./%%@ &%D%;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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&!f&@?@@@@@?@@??@??@/&ز 'D&;BfB  correct, well done!/& 'D&;Bf:7A  this is correct since Cr is in Group 6  well done./&x 'D& ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./&x 'D&
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./& 'D&;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/&4 'D&;#Bf>;C  incorrect as trans isomer does not have optical isomers/& 'D&
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./& 'D&;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./& 'D&;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./& ' D&;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./&! '!D&;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./&" '"D&;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./&# '#D&;HMBfroB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2)./&$ '$D&;NRBfA  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./&% '%D&;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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'!f'@?@@@@@@@@??@@@$@/' (D';BfB  correct, well done!/'( (D';Bf:7A  this is correct since Cr is in Group 6  well done./' (D' ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./'4 (D'
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./' (D';BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/' (D';#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./' (D'
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./'d (D';*/BfheE  incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry./'0 (D';05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./' ( D';6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./'! (!D';<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./'" ("D';BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./'#t (#D';HMBf1.D  correct, as shown at the end of lecture 2./'$ ($D';NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./'% (%D';SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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(!f(@?@@?@??@@??@@@ @/(4 )D(;BfB  correct, well done!/(d )D(;Bf:7A  this is correct since Cr is in Group 6  well done./( )D( ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./(p )D(
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./( )D(;BfEBA  incorrect as a tetrahedral ion cannot have a c4 rotation axis./( )D(;#Bf>;C  incorrect as trans isomer does not have optical isomers/( )D(
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./(P )D(;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./(@ )D(;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./( ) D(;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./(! )!D(;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./(" )"D(;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./(#t )#D(;HMBf1.D  correct, as shown at the end of lecture 2./($ )$D(;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./(% )%D(;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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)!f)@?@@@@@@@@@@@?@ @/) *D);BfB  correct, well done!/) *D);Bf:7A  this is correct since Cr is in Group 6  well done./) *D) ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./) *D)
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./) *D);BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/)H *D);#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./)h *D)
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./)8 *D);*/Bf\YB  incorrect. You have to choose the smallest cell possible within the highest symmetry./) *D);05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./) l * D);6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./)!0 *!D);<ABfD  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face./)"` *"D);BGBf^[B  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./)# *#D);HMBf1.D  correct, as shown at the end of lecture 2./)$x *$D);NRBfA  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./)% *%D);SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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*!f*@?@@@@@@?@@?@@@(@/* +D*;BfB  correct, well done!/* +D*;Bf:7A  this is correct since Cr is in Group 6  well done./* +D* ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./* +D*
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./* +D*;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/*D +D*;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./*d +D*
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./*4 +D*;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./* +D*;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./* h + D*;6;BffcC  incorrect as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./*!0 +!D*;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./*" +"D*;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./*# +#D*;HMBf1.D  correct, as shown at the end of lecture 2./*$ +$D*;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./*% +%D*;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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+!f+@?@@@@@@?@@?@@@$@/+ ,D+;BfB  correct, well done!/+ ,D+;Bf:7A  this is correct since Cr is in Group 6  well done./+p ,D+ ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./+ ,D+
;BfURE  incorrect: Fe is a 3d element so OK there, but Fe is in Gp 8, so Fe(IV) is d4./+ ,D+;Bf'$B  incorrect as there's no c3 axis./+ ,D+;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./+ ,D+
;$)BfspD  incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states./+ ,D+;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./+ ,D+;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./+ 8 , D+;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./+! ,!D+;<ABf}zE  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway./+"t ,"D+;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./+#t ,#D+;HMBf1.D  correct, as shown at the end of lecture 2./+$ t ,$D+;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./+%t ,%D+;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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,!f,@??@@@??@??@@@@@/,t D,;BfB  correct, well done!/,t D,;Bf:7A  this is correct since Cr is in Group 6  well done./,0t D, ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./,t D,
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./,Pt D,;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/,t D,;#BfroD  incorrect  you've spotted way too many isomers. Make sure sure you understand why B is the correct answer./,`t D,
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./, t D,;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./,
t D,;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./, 4t  D,;6;BffcA  incorrect as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./,!t !D,;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./,"8
t "D,;BGBf^[D  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./,#
t #D,;HMBf1.D  correct, as shown at the end of lecture 2./,$Pt $D,;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./,%\t %D,;SXBf^[C  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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,'y,(L;C  incorrect as trans isomer does not have optical isomers/.0%t /D.
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./.&t /D.;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./.&t /D.;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./. 4't / D.;6;BffcA  incorrect as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./.!'t /!D.;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./."8)t /"D.;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./.#)t /#D.;HMBf1.D  correct, as shown at the end of lecture 2./.$L*t /$D.;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./.%+t /%D.;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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/!f/@??@@@@@@@?@@?@@//`t 0D/;BfB  correct, well done!//t 0D/;Bf:7A  this is correct since Cr is in Group 6  well done.//.t 0D/ ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances.//.t 0D/
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3.// /t 0D/;Bf1.D  incorrect as there is no c3 rotation axis.///t 0D/;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone.//0t 0D/
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states.//p1t 0D/;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it.//2t 0D/;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90.// 2t 0 D/;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell).//!h3t 0!D/;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face.//"4t 0"D/;BGBf^[C  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1.//#\5t 0#D/;HMBf1.D  correct, as shown at the end of lecture 2.//$5t 0$D/;NRBfA  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes.//%6t 0%D/;SXBf^[B  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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/'y/(>x(0((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsF7.doc//)h8t 0)D/;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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0!f0??@?@?@@@@??@@@@/0,9t 1D0;Bf>;A  this is incorrect since the complex is neutral overall./09t 1D0;Bf:7A  this is correct since Cr is in Group 6  well done./0:t 1D0 ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./0t 1D0
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./0?t 1D0;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./08@t 1D0;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./0 @t 1 D0;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./0!At 1!D0;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./0"Bt 1"D0;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./0#Ct 1#D0;HMBf1.D  correct, as shown at the end of lecture 2./0$Ct 1$D0;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./0%Dt 1%D0;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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0'y0(+t(1((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsF8.doc/0)(Ft 1)D0;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.~
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1!f1@?@@@??@?@@?@@@&@/1Ft 2D1;BfB  correct, well done!/1Gt 2D1;Bf:7A  this is correct since Cr is in Group 6  well done./1Gt 2D1 ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./1It 2D1
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./1Jt 2D1;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/1HKt 2D1;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/1lLt 2D1
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./1Mt 2D1;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./1dNt 2D1;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./1 Nt 2 D1;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./1!Ot 2!D1;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./1"Pt 2"D1;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./1#0Qt 2#D1;HMBf1.D  correct, as shown at the end of lecture 2./1$Qt 2$D1;NRBfB  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./1%Rt 2%D1;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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1'y1(PEt2((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG1.doc/1)8Tt 2)D1;YhBf@=Well done on scoring such a high mark. Keep up the good work!~
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2!f2@?@?@?@?@@??@@@@/2Tt 3D2;Bf>;E  this is incorrect since the complex is neutral overall./2,Ut 3D2;Bf:7A  this is correct since Cr is in Group 6  well done./2Ut 3D2 ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./2Wt 3D2
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/2DXt 3D2;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/2tYt 3D2;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/2Zt 3D2
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./2[t 3D2;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./2\t 3D2;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./2 l]t 3 D2;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./2!^t 3!D2;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./2"_t 3"D2;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./2#`t 3#D2;HMBf1.D  correct, as shown at the end of lecture 2./2$`t 3$D2;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./2%at 3%D2;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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2'y2(t3((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG2.doc/2)0ct 3)D2;YhBfYou were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
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4'y4(dt5((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG4.doc/4)Pft 5)D4;YhBfN/A~
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5!f5@??@@?@@@@?@@@@@/5Xft 6D5;BfB  correct, well done!/5ft 6D5;Bf:7A  this is correct since Cr is in Group 6  well done./5ft 6D5 ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./5gt 6D5
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./5ht 6D5;Bf1.D  incorrect as there is no c3 rotation axis./5xht 6D5;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/5it 6D5
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./5ljt 6D5;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./5kt 6D5;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./5 kt 6 D5;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./5!lt 6!D5;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./5"nt 6"D5;BGBf^[B  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./5#nt 6#D5;HMBfroB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2)./5$ot 6$D5;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./5%pt 6%D5;SXBf^[C  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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5'y5(tet6((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG5.doc/5)8rt 6)D5;YhBfNot too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
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6!f6@?@?@@@@?@??@@@$@/6st 7D6;BfB  correct, well done!/6(tt 7D6;Bf:7A  this is correct since Cr is in Group 6  well done./6tt 7D6 ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./64vt 7D6
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/6vt 7D6;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/6xt 7D6;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./6,yt 7D6
;$)BfspD  incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states./6zt 7D6;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./6zt 7D6;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./6 D{t 7 D6;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./6!ht 7!D6;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./6"}t 7"D6;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./6#X~t 7#D6;HMBf1.D  correct, as shown at the end of lecture 2./6$~t 7$D6;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./6%t 7%D6;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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6'y6(\qt7((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG6.doc/6)Tt 7)D6;YhBfYou've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!~
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7!f7@?@@@@???@@?@@@*@/7dt 8D7;BfB  correct, well done!/7t 8D7;Bf:7A  this is correct since Cr is in Group 6  well done./7t 8D7 ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./7t 8D7
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./7,t 8D7;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/7\t 8D7;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./7t 8D7
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./7Јt 8D7;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./7t 8D7;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./7 Pt 8 D7;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./7!t 8!D7;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./7"؋t 8"D7;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./7#t 8#D7;HMBf1.D  correct, as shown at the end of lecture 2./7$t 8$D7;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./7%t 8%D7;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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7'y7(`t8((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG7.doc/7)8t 8)D7;YhBf@=Well done on scoring such a high mark. Keep up the good work!
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8!f8??@?@@?@@@??@?@@/8t 9D8;Bf>;A  this is incorrect since the complex is neutral overall./8,t 9D8;Bf:7A  this is correct since Cr is in Group 6  well done./8t 9D8 ;BfWB  incorrect since although there are 2 F environments, there is only 1 F in one environment (trans NMe3) and 2 mutually trans F s this cannot give rise to 2 triplets./8t 9D8
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/8t 9D8;BfebE  incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane./8dt 9D8;#BfroE  incorrect  you've spotted way too many isomers. Make sure sure you understand why B is the correct answer./8Dt 9D8
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./8t 9D8;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./8 =D<;05Bf/< * = D<;6;Bf/<!*( =!D<;<ABf/<"*U ="D<;BGBf/<#*q =#D<;HMBf/<$* =$D<;NRBf/<%* =%D<;SXBf
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=!f=@??@@?@@?@??@@@"@/=D=;BfB  correct, well done!/=lt >D=;Bf:7A  this is correct since Cr is in Group 6  well done./=t >D= ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./=pt >D=
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./=t >D=;BfebE  incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane./=t >D=;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/=t >D=
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./=t >D=;*/BfURD  incorrect. First choose the highest symmetry then the smallest cell within it./=`t >D=;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./= t > D=;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./=!t >!D=;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./="t >"D=;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./=#t >#D=;HMBf1.D  correct, as shown at the end of lecture 2./=$t >$D=;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./=%t >%D=;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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='y=(t>((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsH5.doc/=)Tt >)D=;YhBfYou've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
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>!f>@?@?@@@@?@@@@@? @/>dt ?D>;BfB  correct, well done!/>t ?D>;Bf:7A  this is correct since Cr is in Group 6  well done./>t ?D> ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./>t ?D>
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/>Ht ?D>;Bf1.D  incorrect as there is no c3 rotation axis./>t ?D>;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./>t ?D>
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./>t ?D>;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./>t ?D>;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./> Pt ? D>;6;BffcC  incorrect as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./>!t ?!D>;<ABfD  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face./>"Ht ?"D>;BGBf^[D  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./>#t ?#D>;HMBf1.D  correct, as shown at the end of lecture 2./>$`t ?$D>;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./>% t ?%D>;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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>'y>(xt?((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsH6.doc/>)t ?)D>;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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?!f?@?@@@@@@@@@?@@@&@/?t @D?;BfB  correct, well done!/?t @D?;Bf:7A  this is correct since Cr is in Group 6  well done./?t @D? ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./?t @D?
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./?t @D?;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/?t @D?;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./?t @D?
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./?ht @D?;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./?t @D?;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./? t @ D?;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./?!`t @!D?;<ABfD  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face./?"t @"D?;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./?#Dt @#D?;HMBfroB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2)./?$$t @$D?;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./?%t @%D?;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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?'y?(t@((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsH7.doc/?)tt @)D?;YhBf@=Well done on scoring such a high mark. Keep up the good work!D
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@!f@@?@@@@@??@??@@@&@/@t AD@;BfB  correct, well done!/@ t AD@;Bf:7A  this is correct since Cr is in Group 6  well done./@t AD@ ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./@,t AD@
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./@t AD@;Bf'$B  incorrect as there's no c3 axis./@t AD@;#Bf>;C  incorrect as trans isomer does not have optical isomers/@t AD@
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./@t AD@;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./@t AD@;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./@ lt A D@;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./@!0t A!D@;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./@"lt A"D@;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./@# t A#D@;HMBf1.D  correct, as shown at the end of lecture 2./@$t A$D@;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./@%@t A%D@;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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@'y@(tA((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsH8.doc/@)t A)D@;YhBf@=Well done on scoring such a high mark. Keep up the good work!
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A!fA@?@?@@@@@@?@@@?@/ALt BDA;BfB  correct, well done!/At BDA;Bf:7A  this is correct since Cr is in Group 6  well done./At BDA ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./At BDA
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/A0t BDA;Bf1.D  incorrect as there is no c3 rotation axis./Aw BDA;#Bf>;C  incorrect as trans isomer does not have optical isomers/Aw BDA
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./Aw BDA;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./Aw BDA;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./A w B DA;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./A!w B!DA;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./A"w B"DA;BGBf^[C  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./A#w B#DA;HMBf1.D  correct, as shown at the end of lecture 2./A$w B$DA;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./A%w B%DA;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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A'yA(tB((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI1.doc/A)lw B)DA;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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B!fB@?@@@@@?@@@?@@@&@/B0 w CDB;BfB  correct, well done!/B` w CDB;Bf:7A  this is correct since Cr is in Group 6  well done./B w CDB ;BfWB  incorrect since although there are 2 F environments, there is only 1 F in one environment (trans NMe3) and 2 mutually trans F s this cannot give rise to 2 triplets./B(w CDB
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./Bw CDB;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/Bw CDB;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./Bw CDB
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./Bw CDB;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./Bw CDB;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./B Tw C DB;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./B!w C!DB;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./B"w C"DB;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./B#w C#DB;HMBf1.D  correct, as shown at the end of lecture 2./B$w C$DB;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./B%w C%DB;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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B'yB(dwC((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI2.doc/B)<w C)DB;YhBf@=Well done on scoring such a high mark. Keep up the good work!
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C!fC@??@@??@?@??@@@&@/Cw DDC;BfB  correct, well done!/Cw DDC;Bf:7A  this is correct since Cr is in Group 6  well done./CXw DDC ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./Cw DDC
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./Cxw DDC;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/Cw DDC;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/Cw DDC
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./C w DDC;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./Cw DDC;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./C Tw D DC;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./C!w D!DC;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./C"Tw D"DC;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./C#w D#DC;HMBf1.D  correct, as shown at the end of lecture 2./C$hw D$DC;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./C%( w D%DC;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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C'yC( wD((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI3.doc/C)!w D)DC;YhBf@=Well done on scoring such a high mark. Keep up the good work!
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D!fD@??@@@???@@?@@@(@/D0"w EDD;BfB  correct, well done!/D`"w EDD;Bf:7A  this is correct since Cr is in Group 6  well done./D"w EDD ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./Dd#w EDD
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./D#w EDD;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/D %w EDD;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./D@&w EDD
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./D'w EDD;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./D(w EDD;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./D )w E DD;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./D!)w E!DD;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./D"*w E"DD;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./D#P+w E#DD;HMBf1.D  correct, as shown at the end of lecture 2./D$+w E$DD;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./D%p,w E%DD;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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D'yD(wE((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI4.doc/D)w E)DD;YhBf@=Well done on scoring such a high mark. Keep up the good work!
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E!fE@?@@@?@@@@?@@@? @/Ex.w FDE;BfB  correct, well done!/E.w FDE;Bf:7A  this is correct since Cr is in Group 6  well done./E/w FDE ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./E0w FDE
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./E@1w FDE;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/Ep2w FDE;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/E3w FDE
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./Ed4w FDE;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./E5w FDE;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./E 5w F DE;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./E!\6w F!DE;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./E"7w F"DE;BGBf^[B  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./E#P8w F#DE;HMBf1.D  correct, as shown at the end of lecture 2./E$8w F$DE;NRBfB  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./E%9w F%DE;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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E'yE($wF((nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI5.doc/E)`;w F)DE;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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F/F*q{ GDF;Bf/F*{ GDF;Bf/F*{ GDF ;Bf/F* GDF
;Bf/F*I GDF;Bf/F*ŋ GDF;#Bf/F* GDF
;$)Bf/F*o GDF;*/Bf/F* GDF;05Bf/F *ȥ G DF;6;Bf/F!*5 G!DF;<ABf/F"*\ G"DF;BGBf/F#*u G#DF;HMBf/F$* G$DF;NRBf/F%* G%DF;SXBf
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F'yF($;C  incorrect as trans isomer does not have optical isomers/GtAw HDG
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./GBw HDG;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./GCw HDG;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./G HDw H DG;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./G!Ew H!DG;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./G"Ew H"DG;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./G#Fw H#DG;HMBf1.D  correct, as shown at the end of lecture 2./G$Fw H$DG;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./G%Gw H%DG;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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G'yG(\HwH(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI7.doc/G)4Iw H)DG;YhBf@=Well done on scoring such a high mark. Keep up the good work!
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H!fH@@??@???@@??@@?@/HIw IDH;Bf>;E  this is incorrect since the complex is neutral overall./H(Jw IDH;BfB  incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons./HPKw IDH ;BfKHA  incorrect as there are 2 F environments, hence must be 2 resonances./HKw IDH
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/HLw IDH;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/HMw IDH;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/HNw IDH
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./H4Pw IDH;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./H$Qw IDH;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./H Qw I DH;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./H!Rw I!DH;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./H"Tw I"DH;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./H#Tw I#DH;HMBfroB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2)./H$Uw I$DH;NRBfB  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./H%Vw I%DH;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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H'yH(WwI(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI8.doc/H)XXw I)DH;YhBfNot too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
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I!fI@?@?@?@@@@??@@@@/IZw JDI;BfB  correct, well done!/IHZw JDI;Bf:7A  this is correct since Cr is in Group 6  well done./IZw JDI ;BfKHC  incorrect as there are 2 F environments, hence must be 2 resonances./IL[w JDI
;BfURA  incorrect: Co is a 3d element, so OK there, but Co is in Gp 9, so Co(IV) is d5/I[w JDI;Bf1.D  incorrect as there is no c3 rotation axis./IT\w JDI;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/Ix]w JDI
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./IH^w JDI;*/BfheE  incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry./I_w JDI;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./I _w J DI;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./I!h`w J!DI;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./I"aw J"DI;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./I#Xbw J#DI;HMBf:7C  incorrect. See calculation at the end of lecture 2./I$bw J$DI;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./I%cw J%DI;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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I'yI(dwJ(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsJ1.doc/I)dew J)DI;YhBfNot too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
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J!fJ@?@@@@@@?@@@@@@&@/J$gw KDJ;BfB  correct, well done!/JTgw KDJ;Bf:7A  this is correct since Cr is in Group 6  well done./Jgw KDJ ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./J`iw KDJ
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./Jiw KDJ;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/Jkw KDJ;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./J;C  incorrect as trans isomer does not have optical isomers/Kvw LDK
;$)BfB  incorrect as Cl is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./K4xw LDK;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./K$yw LDK;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./K yw L DK;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./K!zw L!DK;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./K"w L"DK;BGBf^[C  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./K#w L#DK;HMBf1.D  correct, as shown at the end of lecture 2./K$,}w L$DK;NRBfB  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./K%L~w L%DK;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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K'yK(:wL(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsJ3.doc/K)w L)DK;YhBfNot too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
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L!fL@?@@@?@@@@?@@@?@/Lw MDL;BfB  correct, well done!/LЁw MDL;Bf:7A  this is correct since Cr is in Group 6  well done./L@w MDL ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./L܃w MDL
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./Lhw MDL;Bf1.D  incorrect as there is no c3 rotation axis./LȄw MDL;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/Lw MDL
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./Lw MDL;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./L`w MDL;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./L w M DL;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./L!w M!DL;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./L"Pw M"DL;BGBf^[B  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./L#w M#DL;HMBf:7C  incorrect. See calculation at the end of lecture 2./L$xw M$DL;NRBfB  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./L%w M%DL;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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L'yL(wM(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsJ4.doc/L)4w M)DL;YhBfYou were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
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M!fM@?@@@@@@?@@@?@@"@/Mw NDM;BfB  correct, well done!/Mȏw NDM;Bf:7A  this is correct since Cr is in Group 6  well done./M8w NDM ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./Mԑw NDM
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./M`w NDM;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/Mw NDM;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./Mw NDM
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./M w NDM;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./MĖw NDM;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./M Tw N DM;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./M!xw N!DM;<ABfC  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge./M"w N"DM;BGBf^[D  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./M#Dw N#DM;HMBf:7A  incorrect. See calculation at the end of lecture 2./M$w N$DM;NRBfC  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. Here the labels are swapped around./M%w N%DM;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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M'yM(PwN(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsJ5.doc/M)Pw N)DM;YhBfYou've done well  this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
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N!fN@?@@?@@@?@??@@@$@/N`w ODN;BfB  correct, well done!/Nw ODN;Bf:7A  this is correct since Cr is in Group 6  well done./Nw ODN ;BfKHC  incorrect as there are 2 F environments, hence must be 2 resonances./Nw ODN
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./N w ODN;BfEBA  incorrect as a tetrahedral ion cannot have a c4 rotation axis./Nw ODN;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./Nȡw ODN
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./Nw ODN;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./NC  incorrect as triclinic cells have a `" b `" c and `" `" ./P w Q DP;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./P!w Q!DP;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./P"w Q"DP;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./P#w Q#DP;HMBf1.D  correct, as shown at the end of lecture 2./P$(w Q$DP;NRBfB  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./P%Hw Q%DP;SXBf^[A  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
P&
P'yP(wQ(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsJ8.doc/P)w Q)DP;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.~
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Q!fQ@?@@@@?@?@@?@@@,@/Qw RDQ;BfB  correct, well done!/Qw RDQ;Bf:7A  this is correct since Cr is in Group 6  well done./Q;C  incorrect as trans isomer does not have optical isomers/Sw TDS
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./S\w TDS;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./Sw TDS;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./S w T DS;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./S!w T!DS;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./S"w T"DS;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./S#w T#DS;HMBf1.D  correct, as shown at the end of lecture 2./S$w T$DS;NRBfA  incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. and are the heights of octahedral holes./S%w T%DS;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
S&
S'yS(LT(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsK3.doc/S)w T)DS;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.~
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T!fT@?@@@@@@?@@?@@@.@/Tdw UDT;BfB  correct, well done!/Tw UDT;Bf:7A  this is correct since Cr is in Group 6  well done./Tw UDT ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./Tw UDT
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./T,w UDT;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/T\w UDT;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./Tw UDT
;$)BfqE  correct as CO is a sigma donor/ pi acceptor the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states)./Tw UDT;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./Tw UDT;05BfGA  correct as monoclinic cells have a `" b `" c, = = 90and `" 90./T w U DT;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./T!w U!DT;<ABfdaB  correct. This is a quarter of the way from one corner to the opposite corner across the cube./T"w U"DT;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./T#\w U#DT;HMBf1.D  correct, as shown at the end of lecture 2./T$w U$DT;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./T%w U%DT;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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T'yT(XU(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsK4.doc/T)w U)DT;YhBfC@Excellent effort  you scored full marks. Keep up the good work!
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U!fU??@@@@@?@@??@@@ @/Uw VDU;Bf>;A  this is incorrect since the complex is neutral overall./Uw VDU;Bf:7A  this is correct since Cr is in Group 6  well done./Upw VDU ;BfE  correct: 2 F environments the unique F couples to the other 2 mutually trans F s to give a triplet, while the 2 mutually trans F s resonate together but couple to the one unique F to give a doublet./Uw VDU
;Bf'$C  incorrect as Tc is a 4d element./Ux VDU;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/U<x VDU;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./U\x VDU
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./U,x VDU;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./Ux VDU;05BfGD  incorrect as orthorhombic cells have a `" b `" c and = = = 90./U x V DU;6;BfdaB  correct as there are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./U!px V!DU;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./U"x V"DU;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./U#`x V#DU;HMBfwtE  incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2)./U$Lx V$DU;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./U% x V%DU;SXBf\YE  correct as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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V!fV@?@@@@@@@@?@@@@@/V`x WDV;Bf>;E  this is incorrect since the complex is neutral overall./Vx WDV;Bf:7A  this is correct since Cr is in Group 6  well done./VHx WDV ;BfD  incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei one in one environment and two in another environment ie sequential couplings)./VHx WDV
;BfHED  correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3./Vx WDV;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/Vx WDV;#BfB  correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone./V$x WDV
;$)BfjgC  incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states./Vx WDV;*/BfSPC  correct. First choose the highest symmetry then the smallest cell within it./Vx WDV;05Bf@B  incorrect as cubic cells have a = b = c and = = = 90./V x W DV;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./V!@x W!DV;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./V"x W"DV;BGBf^[D  incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1./V#4x W#DV;HMBf1.D  correct, as shown at the end of lecture 2./V$x W$DV;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./V%Tx W%DV;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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V'yV( xW(F(nkC:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsK6.doc/V)x W)DV;YhBfc`You've passed, so well done, but make sure you think carefully about what you can do to improve.
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W!fW@?@@@???@@??@@@@/Wx XDW;BfB  correct, well done!/Wx XDW;Bf:7A  this is correct since Cr is in Group 6  well done./WDx XDW ;BfKHC  incorrect as there are 2 F environments, hence must be 2 resonances./Wx XDW
;Bf'$C  incorrect as Tc is a 4d element./W$x XDW;BfC  correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the ML axes/WTx XDW;#BfA  incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both!/Wxx XDW
;$)BfA  incorrect as O2 is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion)./Wx XDW;*/BfyvA  incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it./Wx XDW;05Bf>C  incorrect as triclinic cells have a `" b `" c and `" `" ./W < x X DW;6;BfE  incorrect as you have just counted the number of spheres! There are 4 large spheres (81/8 + 6) and 8 small spheres (all inside the cell)./W!`!x X!DW;<ABfA  incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face./W""x X"DW;BGBf\YA  correct as the ratio of spheres in the close packed array to octahedral holes is 1:1./W#P#x X#DW;HMBfroB  incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2)./W$0$x X$DW;NRBfa^D  correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom./W%$x X%DW;SXBf^[D  incorrect as CsCl is formed from Cs occupying cubic holes in a primitive cubic lattice.
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