In addition to questions like the ones on the assignments, students can expect to be asked questions of the following types:

define various terms (CFSE, high-spin versus low-spin complexes, magnetic moment, Gouy balance, chelating ligand, fac and mer isomers, ambidentate ligand, octahedral site preference energy, linkage isomers, ionization isomers, Lewis acid, Lewis base, metal complex, and other terms important in this course)

nomenclature; drawing structures of compounds given their names [e.g. cis-diamminedi-S-thiocyanatoplatinum(II); fac-triamminetrichlorocobalt(III)]; drawing all the isomers possible for a given complex

draw the d-orbitals and label them properly

explain or interpret data

1. Question: Use Lewis structures to explain why the Pt-N-C bond is linear in [Pt(NCS)₄]^2- but the Pt-S-C bond is bent in [Pt(SCN)₄]^2-. What are the hybridizations of N and S?

(a) Good student answer, full marks

The Lewis structure is shown above with hybridizations. When forming a bond with platinum through the nitrogen, the lone pair of electrons on the nitrogen is oriented 180° from the N-C bond. The Pt forms a coordinate covalent bond with the nitrogen through its lone pair of electrons which gives the Pt-N-C bond a 180^o bond angle (linear). The three lone pairs on the sulfur are in a tetrahedral arrangement with the C-S bond. The bond angle between the lone pairs and the C-S bond is therefore about 109.5^o. This means that when S is bonding to the platinum through one of its lone pairs, the Pt-S-C bond angle will have to be about 109.5^o (bent).

• Failure to connect the Pt-N-C or Pt-S-C angles to the likely hybridization at N or S.
• Failure to recognize that thiocyanate is an electron pair donor to the Lewis acid, Pt²⁺.
• Forgot to apply VSEPR theory to predicting angles.

 

 
 
 
 
 
 
 

(a) Good student answer, full marks.

SO₃^2- has lone pairs so it can bond through either atom. NO₃^- only has lone pairs of electrons on its oxygen atoms, so can bond only through oxygen. Lewis structures:

(b) Why students lost marks.

• Failure to distinguish between ambidentate ligands and chelating ligands
• Incorrect Lewis structures

(a) Good student answer with a diagram

Cu(II) is a d9 ion, and its electronic configuration looks like this: ---> Since there is space available in an eg orbital, color can result when absorption of radiation necessary to promote an electron from t2g to eg occurs.
Cu(I) is a d10 ion, and its electron configuration looks like this: ---> Since there is no space available in eg, t2g electrons cannot be promoted to eg and so no color results.

(b) Good student answer without a diagram

The Cu(II) cation has the configuration [Ar]3d⁹. The electron transfers within the split d-orbitals result in changes in energy and the molecule absorbs light in the visible spectrum. Therefore Cu(II) salts are colored. Cu(I) has 10 d-electrons, having the configuration [Ar]3d¹⁰. Since the d-orbital is completely filled, no electron transitions occur within the d-orbitals. Since it is electron transitions within the d-orbitals that correspond to the absorption of light in the visible part of the spectrum, an inability for these transitions to occur results in colorless salts.

• Failure to give an appropriate level of depth to an explanation. It is not enough to state that "The reason that Cu(II) is colored is that it is a d⁹ ion but the reason that Cu(I) is colorless is because it is a d¹⁰ ion."
• Failure to connect color with the energy absorbed when an electron is promoted to a higher energy orbital.
• Incorrect use of terms, for example stating that the orbital absorbs energy.

Answer.

Compound A is [Co(NH₃)₅SO₄]Br

Compound B is [Co(NH₃)₅Br]SO₄

The fact that the complexes are neutral means that the NH₃ is bonded to cobalt in both compounds; otherwise a solution of ammonia in water would be basic. In any case, ammonia bonds strongly to cobalt(III) and a compound containing both is expected to have all the ammonia molecules coordinated to cobalt.

The white precipitate produced when AgNO₃ is added to Compound A is due to the formation of insoluble silver bromide. Thus bromide is ionic. The failure to get a precipitate upon addition of BaCl₂ means that the sulfate is coordinated to cobalt.

The white precipitate produced when BaCl₂ is added to Compound B is due to the formation of insoluble barium sulfate. Thus sulfate is ionic in Compound B. On the other hand, bromide must be coordinated to cobalt since there is no precipitate when silver nitrate is added.

Answer

 

Compound I: cis-[Cr(en)2(NCS)2]SCN	Compound II: trans-[Cr(en)2(NCS)2]SCN

The red color developed upon addition of iron(III) shows that at least one thiocyanate is uncoordinated. Cr(III) typically has a coordination number of 6. Thus each complex ion is bonded to two ethylenediamine molecules and two two thiocyanates. The two compounds are geometric isomers.

Compound I is the cis-isomer. It lacks a plane of symmetry, and is thus expected to have optical isomers (enantiomers). The second compound has a plane of symmetry so would not have optical isomers.
 
 

Answer. There are 3 isomers, 2 mer and one fac. None are chiral.

Answer

The picolinate anion is a good chelating agent and forms a five-membered ring with metal ions when it uses its nitrogen and oxygen atoms for coordination. Thus chromium(III) achieves its usual coordination number of 6 with three bidentate ligands. There are two isomers, "mer" and "fac"; both are chiral as there are no planes of symmetry.

(a) Draw appropriate d-orbital splitting diagrams for each.

(b) Calculate the CFSE for the first four. Include the pairing energy.

(c) Explain why complexes of cobalt have so many colours, whereas the complexes of titanium(IV) and zinc(II) are colourless.

Complex	Magnetic moment, B.M.	Colour
[Co(NH3)6]3+	0.0	yellow
[CoF6]3-	4.9	blue
[Co(H2O)6]2+	3.8	pink
[CoCl4]2-	3.8	blue
[Co(oaph)2] (ligand structure shown below)	1.7	red

The magnetic moment data show that there are no unpaired electrons in the first complex, four in the second, three in the third and fourth, and one in the fifth complex.

The first three complexes are octahedral (6 ligands).

[CoCl₄]^2- and Co(oaph)₂ must be four-coordinate since there are not six donor atoms (oaph is a bidentate chelating ligand). They could be either tetrahedral or square. Tetrahedral is most common and would lead to 3 unpaired electrons; one unpaired electron requires square-planar geometry.

Complex	d-orbital diagram	CFSE
[Co(NH3)6]3+		2.4 DeltaO - 2 P
[CoF6]3-		0.4 Deltaoct
[Co(H2O)6]2+		0.8 Deltaoct
[CoCl4]2-		0.6 Deltatet
[Co(oaph)2]

The reason that the complexes of cobalt have so many colors is that Co(II) and Co(III) both have unfilled d-shells. Electronic transitions are possible when electrons in lower-energy orbitals are promoted to vacancies in higher-energy orbitals; energy required for these transitions falls in the visible range.

On the other hand, Zn(II) complexes are colorless because the d-orbitals are completely filled and any transitions to higher energy orbitals correspond to energies outside the visible range; hence no color. Ti(IV) has no d-electrons so there are no transitions possible involving them.
 
 

 

Complex	B2
[Mg(ox)2(H2O)2]2-	2.4 x 104
[Co(ox)2(H2O)2]2-	5.0 x 106
[Ni(ox)2(H2O)2]2-	4.0 x 107

Mg²⁺ is d⁰, Co²⁺ is d⁷high-spin, and Ni²⁺ is d⁸. The corresponding CFSE's are:

Complex	Electronic configuration of metal ion	CFSE
[Mg(ox)2(H2O)2]2-	no d electrons	zero
[Co(ox)2(H2O)2]2-	t2g5eg2	0.8 Deltaoct
[Ni(ox)2(H2O)2]2-	t2g6eg2	1.2 Deltaoct

The stability is related to the CFSE. The higher the CFSE, the greater the stability.
 
 

Assume that V is 6-coordinate and that both ethylenediamines are chelated to the vanadium((III) ion.

(a) geometric isomers

(b) optical isomers

(c) linkage isomers

(d) ionization isomers

(a) 
 
 

(b) 
 
 

(c) 
 
 

(d) [V(en)₂Cl₂]NCS and [V(en)₂Cl(NCS)]Cl
 
 

This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem232/2P32_Typical_Questions_MT1.html
Created January 14, 2001 by M. F. Richardson
© Brock University, 2001