Questions & Answers on Coordination Chemistry (D Ray)
1. What are the geometries of the following two complexes?
(i) [AlCl4]‐
(ii) [Ag(NH3)2]+
Ans. tetrahedral, linear
2. What is the respective central‐metal oxidation state, coordination number, and the
overall charge on the complex ion in NH4[Cr(NH3)2(NCS)4]?
Ans. +3; 6; 1‐
3. The formula for lithium iodotris(trifluorophosphine)nickelate(0) is: (Note:
trifluorophosphine is a neutral ligand with the formula PF3.)
Ans. Li[Ni(PF3)3I3]
4. Which is more likely to form a high‐spin complex—en, F‐, or CN‐?
Ans. F‐
5. Name the following compound: K2[CrCO(CN)5].
Ans. potassium carbonylpentacyanochromium(III)
6. What is macrocyclic effect?
Ans. Macrocyclic ligands of appropriate size form more stable complexes than chelate
ligands. A competition between a noncyclic chelating ligand and a macrocyclic
(chelating) ligand having the same number and type of donor atoms will generally lead
to complex formation predominantly by the macrocyclic ligand. This is known as the
macrocyclic effect.
The entropy term drives this type of reaction to the right. The long noncyclic ligand is
more flexible than the corresponding macrocyclic ligand and can adopt many more
conformations than the macrocyclic ligand when it is not coordinated.
7. Predict the no. of unpaired electrons, the spin‐only magnetic moments at 25 C for
, each of the following.
a) [Fe(CN)6]4‐
b) [Ru(NH3)6]3+
c) [Cr(NH3)6]2+
d) [EuCl6]4‐
Ans. a) Fe2+ is 3d6. Has 0 unpaired electron in l.s. complexes and thus the magnetic
moment would be close to 0 µB.
b) Ru3+ is 4d5. Has 1 unpaired electron in l.s. complexes and thus the magnetic moment
would be close to 1.73 µB.
c) Cr2+ is 4d4. Have 4 and 2 unpaired electrons in h.s. and l.s. complexes and thus the
magnetic moment would be close to 4.90 and 2.83 µB, respectively.
d) Eu2+ is 4f7. Has 7 unpaired electrons in h.s. complexes and thus the magnetic moment
would be close to 7.94 µB. The magnetic moments deviate considerably from the spin‐
only values because of strong spin‐orbit coupling. The f‐orbitals have so little overlap
and interactions with ligand orbitals.
8. With reference to the 3d elements in the periodic table identify the elements and
associated oxidation numbers that form square planar complexes. Give formulas for
three examples of such complexes.
Ans. The square‐planar geometry is primarily associated with the d8 electron
configuration, and the following elements and oxidation states commonly form the d8
configuration:
3 4 5 6 7 8 9 10 11 12
2+
Ni
+
Rh Pd2+
Ir+ Pt2+ Au3+
Examples of such complexes: RhCl(PPh3)3, IrCl(CO)(PPh3)2, Ni(CN)42–, PdCl42‐, cis‐ and
trans‐PtCl2(NH3)2 and AuCl4–. Note however that NiCl42– is not square planar, but is
instead tetrahedral. In general, the preference for square planar vs. tetrahedral increases
with increasing ligand field strength; the ligand fields for the 2nd and 3rd transition series
are intrinsically larger than for the first series metals. Note however, that Ni(CO)4 is
also tetrahedral, not square planar, but that is because this complex is d10, rather than
d8!
9. Sketch the two structures that describe most six‐coordinate complexes. Which is
more common?
Ans. Octahedral Trigonal prismatic
1. What are the geometries of the following two complexes?
(i) [AlCl4]‐
(ii) [Ag(NH3)2]+
Ans. tetrahedral, linear
2. What is the respective central‐metal oxidation state, coordination number, and the
overall charge on the complex ion in NH4[Cr(NH3)2(NCS)4]?
Ans. +3; 6; 1‐
3. The formula for lithium iodotris(trifluorophosphine)nickelate(0) is: (Note:
trifluorophosphine is a neutral ligand with the formula PF3.)
Ans. Li[Ni(PF3)3I3]
4. Which is more likely to form a high‐spin complex—en, F‐, or CN‐?
Ans. F‐
5. Name the following compound: K2[CrCO(CN)5].
Ans. potassium carbonylpentacyanochromium(III)
6. What is macrocyclic effect?
Ans. Macrocyclic ligands of appropriate size form more stable complexes than chelate
ligands. A competition between a noncyclic chelating ligand and a macrocyclic
(chelating) ligand having the same number and type of donor atoms will generally lead
to complex formation predominantly by the macrocyclic ligand. This is known as the
macrocyclic effect.
The entropy term drives this type of reaction to the right. The long noncyclic ligand is
more flexible than the corresponding macrocyclic ligand and can adopt many more
conformations than the macrocyclic ligand when it is not coordinated.
7. Predict the no. of unpaired electrons, the spin‐only magnetic moments at 25 C for
, each of the following.
a) [Fe(CN)6]4‐
b) [Ru(NH3)6]3+
c) [Cr(NH3)6]2+
d) [EuCl6]4‐
Ans. a) Fe2+ is 3d6. Has 0 unpaired electron in l.s. complexes and thus the magnetic
moment would be close to 0 µB.
b) Ru3+ is 4d5. Has 1 unpaired electron in l.s. complexes and thus the magnetic moment
would be close to 1.73 µB.
c) Cr2+ is 4d4. Have 4 and 2 unpaired electrons in h.s. and l.s. complexes and thus the
magnetic moment would be close to 4.90 and 2.83 µB, respectively.
d) Eu2+ is 4f7. Has 7 unpaired electrons in h.s. complexes and thus the magnetic moment
would be close to 7.94 µB. The magnetic moments deviate considerably from the spin‐
only values because of strong spin‐orbit coupling. The f‐orbitals have so little overlap
and interactions with ligand orbitals.
8. With reference to the 3d elements in the periodic table identify the elements and
associated oxidation numbers that form square planar complexes. Give formulas for
three examples of such complexes.
Ans. The square‐planar geometry is primarily associated with the d8 electron
configuration, and the following elements and oxidation states commonly form the d8
configuration:
3 4 5 6 7 8 9 10 11 12
2+
Ni
+
Rh Pd2+
Ir+ Pt2+ Au3+
Examples of such complexes: RhCl(PPh3)3, IrCl(CO)(PPh3)2, Ni(CN)42–, PdCl42‐, cis‐ and
trans‐PtCl2(NH3)2 and AuCl4–. Note however that NiCl42– is not square planar, but is
instead tetrahedral. In general, the preference for square planar vs. tetrahedral increases
with increasing ligand field strength; the ligand fields for the 2nd and 3rd transition series
are intrinsically larger than for the first series metals. Note however, that Ni(CO)4 is
also tetrahedral, not square planar, but that is because this complex is d10, rather than
d8!
9. Sketch the two structures that describe most six‐coordinate complexes. Which is
more common?
Ans. Octahedral Trigonal prismatic
