Answers and marking scheme

Inorganic Chemistry I August 2010 Answers Answers and marking scheme 1I.1 Molecular Structure a) With the aid of molecular orbital energy level diagr...
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Inorganic Chemistry I August 2010 Answers

Answers and marking scheme 1I.1 Molecular Structure a) With the aid of molecular orbital energy level diagrams rationalise the bond strength values shown below. Include drawings of the relevant molecular orbitals in your answer. Molecule

B2

C2

N2

Bond strength (kJ mol-1)

297

607

945

ANSWER: Students should draw a general MO energy level diagram for diatomics where MO mixing is significant (i.e. the one that applies to these three bond (this was discussed at length during the lectures).

Marks given in the following way: correct orientation – i.e. if pz is used to form the sigma bond it means that the molecule is “sitting” on this axis (1 mark); correct order of the energy levels including the mixing (2 marks); drawings of MOs (4 marks). 1

Inorganic Chemistry I August 2010 Answers Once the electrons are placed on the right MOs, the following information regarding bond order (and number of electrons in bonding/antibonding MOs) can be obtained:

B2

C2

N2

Valence electrons

6

8

10

Electrons in bonding MO

4

6

8

Electron in antibonding MO

2

2

2

Bond Order

1

2

3

Clearly, the strongest bond should be for N2 (consistent with the experimental values) since is the only one in the series that has a BO of 3. This is followed by C2 (BO = 2) and finally B2 (BO = 1). This is consistent with the experimental values provided. (3 marks)

b) Answer ALL parts of this question: i) Using VSEPR, state the pseudostructure and structure of the following compounds. Include a drawing of the structures in your answer. SnCl2

[H3O]+

BBr3

ANSWER: The predicted structures from VSEPR are: SnCl2

BBr3

[H3O] +

Pseudostruct: Trigonal planar

Trigonal planar

Tetrahedral

Structure:

Bent

Trigonal planar

Trigonal pyramidal

(2 marks)

(2 marks)

(2 marks)

ii) Give a suitable hybridisation scheme for the central atom in each case. 2

Inorganic Chemistry I August 2010 Answers

ANSWER: SnCl2 sp2 BBr3 sp2 + [H3O] sp3 (3 marks) c) [PtCl4]2– and POCl3 belong to the D4h and C3v point groups respectively. Determine the structures of these two molecules and explain your reasoning. ANSWER: [PtCl4] 2– is square planar

(1 mark)

POCl3 is tetrahedral

(1 mark)

Since [PtCl4] 2– belongs to the D4h point group it should have a σh mirror plane. From the two possible geometries that this molecule could have (namely tetrahedral or square planar) only the latter could have a σh. In addition to this, a molecule belonging to a D4h point group should have a C4 and 4C2. This is consistent with a square planar geometry (the C4 is perpendicular to the plane of the molecule; two C2 along the Cl-PtCl bonds; 2 C2 bisecting the bonds and parallel to D4h point group). POCl3 belongs to the C3v point group, hence, it should have 3σv mirror planes but no σh mirror plane. In addition, it should have a C3 axis only. This is consistent with a tetrahedral molecule in which one of the substituents is different to the other three (the C3 axis is located along the O-P bond; three planes parallel to C3 each containing Cl-P-O and bisecting the ClP-Cl bonds). (4 marks)

d) Answer ALL parts of this question: i) Using VSEPR predict whether CO2 is linear or bent. ANSWER: CO2 is linear (there are no lone pairs around the carbon and all other electron are involved in bonding – i.e. an AX2 type moelcule):

O

C

O (2 Marks)

ii) What hybridization is appropriate for the C atom? ANSWER: The following hybridization will allow us to rationalize the structure:

3

Inorganic Chemistry I August 2010 Answers

The two remaining 2p orbitals remain as such. (2 marks) iii) Using VB, show a bonding scheme using the hybridization you have suggested. ANSWER: The two sp hybrid orbitals can form a  bond with each oxygen atom (which is sp2). The two remaining p orbitals on the carbon, can form  bonds with a p orbital from each oxygen. (2 marks)

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Inorganic Chemistry I August 2010 Answers

1I2. PERIODICITY AND INORGANIC REACTIVITY ANSWERS (a) (i) Boron (1s2 2s2 2p1) has three valence electrons and can therefore bond to three other atoms. The predicted shape for this molecule (VSEPR) is trigonal planar with a sp2 hybridised boron atom. This gives three B-Cl 2c-2e sigma bonds. Boron now has 6e around it and is therefore electron deficient (2e short of its octet). This is compensated for by back donation of a pair of electrons from each Cl (2pz) into the empty pz on the boron atom. [6 Marks] (ii)

BF3.NMe3 + BCl3  BCl3.NMe3 + BF3 [1 Mark]

BCl3.NMe3 + BF3  no reaction [1 Mark] The boron trihalides are Lewis acids and the NMe3 is a Lewis base [1]. BCl3 is more acidic than BF3 [1].This is because the pi-orbital overlap and hence relief of electron deficiency is best for B-F (2p-2p) rather than B-Cl (2p-3p) due to the size/energy mismatch and longer bond for B-Cl [1].

(b)

(i) H2O expected to be bent and based on tetrahedral with two lone pairs [1] hence X would be sp3 hybridised [1] H2O slightly off ideal sp3 angle due to lp-lp repulsions > bp-bp repulsions [1] H2S close to 90 deg since S not hybridised much at all [1]– little or no contribution from s orbital –[1] therefore H 1s overlaps with S 3p orbitals [1] which are orientated at 90 deg to each other [1]. (ii) SF6 < SeF6 < TeF6. [2] Each are octahedral compounds [1] SF6 is kinetically inert. The octahedral arrangement of the 6F around S makes it very difficult for reagents to attack the S atom [2]. TeF6 and SeF6 are octahedral molecules like SF6 but the group 16 elements are descended they get larger and therefore more accessible to attack [2]. (iii) M + H2O -> M(OH)2 + H2 [1] Expect reactivity with water to increase down the group (as with Group 1) [1]. Mg is passivated and kinetically inert at ambient temperature due to ready formation of protective oxide coating on the metal surface [2] 2Mg + O2 -> 2MgO [1] Be is even less reactive with water than Mg [1] again because of its protective oxide coating which it readily forms in air due to the very strong Be-O bonds [1]

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Inorganic Chemistry I August 2010 Answers

1I3.

a)

Coordination Chemistry

Discuss any TWO of the following: (i)

Explain the origin of colour from charge-transfer processes between metals and ligands and how they differ from d-d transitions. Give one example each of MLCT and LMCT complexes.

Charge-transfer occurs when there is an electronic transition which results in a displacement of charge from one atom to another. This can occur in transition metal complexes between metals and ligands. (1 mark). Unlike d-d transitions, they are not symmetry-forbidden and are often 10,000 times stronger (1 mark) MLCT: electron transfer from filled metal orbital to empty ligand’s empty orbital (usually *). Most commonly encountered in compounds containing pi-acceptor ligands, such as CO, CN- and bipy (any reasonable example will be accepted). LMCT: is the opposite, when the ligand donated electrons into empty metal orbitals, found in intensely coloured d0 metal complexes, e.g. K[MnO4], [CrO4] 2-, etc. (3 marks) (ii)

Give the formal name for the complex K2Mn(SO4)2.6H2O, according to the IUPAC nomenclature. Given that the complex has a magnetic moment of 5.9 B.M., determined the electronic configuration of the complex. Illustrate how this complex undergoes ionisation isomerism. Potassium disulfatomanganate(II) [2 marks], d5, high-spin.[1 mark] Ionisation isomerism: K2[Mn(SO4)2].6H2O ↔ [Mn(H2O)2(SO4)] .4H2O + K2SO4 [2 marks] could also displace both sulphates. (iii) Sigma donor ligands have only one pair of electrons to donate to the metal centre (1 mark), while pi-acceptor ligands have (additionally) available empty pi* orbitals to accept electrons from filled d-orbitals from the metal (1 mark). Any reasonable examples (1 mark).

(2 x 1 mark)

b)

Answer ALL parts of this question, with reference to the following: 6

Inorganic Chemistry I August 2010 Answers Values of ligand field splitting parameter (o, in cm-1) for octahedral iron complexes are given in the following Table. ion Fe2+ Fe3+ (i)

6 H2O 10400 14300

Ligand set 6 CN– 32800 35000

Comment on the difference in o observed between [Fe(H2O)6]2+ and [Fe(H2O)6]3+, and that between [Fe(H2O)6]2+ and [Fe(CN)6]4–.

The greater the charge on the metal, the greater is o. If the charge is large, then this should cause the ligands to be attracted towards the metal more closely, and so interact more strongly with the d-orbitals. [1 mark] Increase in oxidation state from+2 to +3 normally cases an increase in o of roughly 50%, as is observed for [Fe(H2O)6] 2+ and [Fe(H2O)6] 3+ [1 mark]. o is also dependent on the position of the ligand in the Spectrochemical Series. [1 mark] In this case, CN- is a strong field ligand (pi-acceptor), while H2O is a weak field ligand, thus o of [Fe(H2O)6] 2+ is smaller than [Fe(CN)6] 4–.[1 mark] (ii)

Draw the crystal field splitting diagrams for [Fe(H2O)6]2+ and [Fe(H2O)6]3+ in their ground state electronic configuration (both high spin). Calculate the crystal field stabilisation energies (in kJ mol-1) for both of these complexes, use P to denote Pairing energy.

d6 (t2g4eg2) - 1 mark. CFSE (Fe2+) = 4 x –0.4o + 2 x 0.6o + P = –0.4o + P [1 mark] = –49.8 + P kJ mol-1 [1 mark] (10,000 cm-1 = 119.7 kJ mol-1) [plus 1 mark for working] (iii) Predict the magnetic moment you will expect for complex [Fe(CN)6]3–. This is expected to be low-spin due to large o (1 mark), so expect d5 (t2g5eg0) = 1unpaired electron (1 mark), therefore SO = √n(n+2) = 1.73 B.M. (1 mark). (iv)

Illustrate stepwise and overall stability constants for the following reaction: [Fe(H2O)6]3+ ↔ [Fe(CN)6]3–

[Fe(H2O)6]3+ + CN– ↔ [Fe(CN)(H2O)5]2+ [Fe(CN)(H2O)5]2+ + CN– ↔ [Fe(CN)2(H2O)4]+ [Fe(CN)2(H2O)4] + + CN– ↔ [Fe(CN)3(H2O)3] [Fe(CN)3(H2O)3] + CN– ↔ [Fe(CN)4(H2O)2]– [Fe(CN)4(H2O)2]– + CN– ↔ [Fe(CN)5(H2O)]2– [Fe(CN)5(H2O)]2– + CN– ↔ [Fe(CN)6]3– illustrate: 1 mark]

1 2 3 4 5 6

[any to

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Inorganic Chemistry I August 2010 Answers

Stepwise, Kn = [Fe(CN)n(H2O)6-n]/{ [Fe(CN)n-1(H2O)7-n] [ CN–]} Overall, n = [Fe(CN)n(H2O)6-n]/ {[Fe (H2O)6] [CN–]n} [1 mark] therefore6 = K1 x K2 x….K6 [1 mark]

[1 mark]

half marks if general formula is given, instead of the example given, i.e. [M] and [L].

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