Coordination Compounds Important Questions
Que 1. (i) Write down the IUPAC name of the following complex:
[Pt(NH3)(H2O)Cl2] and K3[Fe(CN)6]
(ii) Write the formula for the following complex:
tris(ethane-l,2-diamine)chromium(III) chloride and Pentaaminenitrito-N-cobalt(III)
Ans 1. (i) Ammineaquadichloridoplatinum (II) and Potassiumhexacyanoferrate(III) ion
(ii) [Cr(en)3]Cl3 and [CO(NH3)5NO2]2+
Que 2. Ni(CO)4 possesses tetrahedral geometry, while [Pt(NH3)4]2+ is square planar. Why?
Ans 2. Ni(CO)4 possesses sp3 hybridization and then tetrahedral, whereas [Pt(NH3)4]2+ possesses dsp2 hybridization, thus square planar.
Que 3. Explain the following:
(i) NH3 act as a ligand but NH4+ does not.
(ii) CN− is a ambidentate ligand.
Ans. (i) NH3 has one lone pair while NH4 + does not.
(ii) Because it has two donor atoms in a monodentate ligand.
Que 4. A solution of [Ni(H2O)6]2+ is green, but a solution of [Ni(CN)4]2- is colourless. Explain.
Ans 4. In [Ni(H2O)6]2+, H2O is a weak field ligand. Therefore, there are unpaired electrons in Ni2+. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e. the possibility of d-d transition is present. Hence, [Ni(H2O)6]2+ is coloured. In[Ni(CN)]2–, the electrons are all paired as CN– is a strong field ligand. Therefore, the d-d transition is not possible in [Ni(CN)4]2–. Hence, it is colorless. As there are no unpaired electrons, it is diamagnetic.
Coordination Compounds Important Questions
Que 5. Mn2+ (aq.) ion is light pink colored while [Mn(CN)6]4– is blue in color. Explain.
Ans 5. In complexes, where Mn (II) is present, the configuration of a metal ion is d5. There may be two types of spin arrangements in the presence of different kinds of ligands.
(A) High spin complex (with weak field ligands) and (B) Low spin complex (with strong field ligands).
The arrangement of electrons in these complexes can be depicted as:
In high spin complex compounds, it is observed that d–d transition requires reversion of spin which is against the spin selection rules and this makes them spin forbidden and the intensity of the color is only about 1/100 when the transition is allowed. In [Mn(CN)6] 4– on the other hand, d–d transitions do not have any such restrictions and are spin allowed. Intense color also is observed when the transition takes place.
Que 6. What is meant by the chelate effect?
Ans 6. Chelate effect: When a bidentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or six-membered ring is formed. This effect is called the Chelate effect. As a result, the stability of the complex increases. For example, the complex of Ni2+ with ‘+ion’ is more stable than NH3.
Que 7. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Ans 7. Spectrochemical series gives the arrangement of ligands in the increasing order of crystal field splitting.
Weak field ligands cause less crystal field splitting. They form high spin complexes. For Example, chloride ions (Cl–), fluoride ion (F–) etc.
Strong field ligands cause greater crystal field splitting. They form low spin complexes. Examples include cyanide ion (CN–) and CO.
I- < Br – < S2- < SCN– < Cl– < N3 < F– < OH– < C2O42- ~ H2O < NCS– ~ H– < CN– < NH3 < en ~ SO32- < NO2– < phen < CO.
Coordination Compounds Important Questions
Que 8. [CuCl4]2– exists but [CuI4]2– does not. Why?
Ans 8. [CuI4]2– decomposed as [CuI4]2– ⎯→ 2CuI + 3I2. The instability may be explained as it is complex of oxidizing agent (Cu2+) and reducing agent (I–). Also, I– is a poor electron donor than Cl– and steric effect may also play some role.
Que 9. State a reason for each of the following situations:
(i) Co2+ is easily oxidized to Co3+ in presence of a strong ligand.
(ii) CO is a stronger complexing reagent than NH3.
(iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)4]2-.
Ans 9. (i) Strong ligands provide energy that overcomes 3rd ionization enthalpy and Co2+ gets oxidized to Co3+.
(ii) CO can form σ as well as π bond, therefore, it is a stronger ligand than NH3which can form only a bond.
(iii) Ni(CO)4 has sp3 hybridization, tetrahedral shape, whereas [Ni(CN)4]2– has dsp2 hybridization, therefore, it has a square planar shape.
Que 10. [Fe(H2O)6]3+ is strongly paramagnetic while [Fe(CN)6]3- is weakly paramagnetic.
Ans 10. In both cases, Fe is oxidation state +3. Outer electronic configuration of Fe3+ is
In presence of CN-1, the 3d-electrons pair up leaving only one unpaired electron. The hybridization involved is d2sp3 forming an inner orbital complex that is weakly paramagnetic. In presence of H2O (a weak ligand), 3d-electrons do not pair up. The hybridization involved is sp3d2 forming an outer orbital complex. As it contains five unpaired electrons, it is strongly paramagnetic.
Que 11. Write all the geometrical isomers of [Pt(NH3)(Cl)(py)(Br)]. How many of these will exhibit optical isomerism?
Ans 11. The oxidation state of Pt = +2 Complex is square planar. Square planar molecules of M(ABCD) type will not show optical isomerism. Geometrical isomers are
Coordination Compounds Important Questions
Que 12. What are t2g and eg orbitals?
Ans 12. In a free transition metal ion, the d-orbitals are degenerate. When it form complex, the degeneracy is split and d-orbitals split into t2g and eg orbitals.
Que 13. The magnetic moment of [MnCl4]2– is 5.92 B.M. On the basis of its magnetic moment, write the configuration of Mn2+ in this complex.
Ans 13. For an atom/ion
Thus in this complex, Mn contains 5 unpaired electrons and so its possible configuration may be Mn2+ in [MnCl4]2– = [Ar] 3d5 4s0 So hybridization of Mn2+ in the given complex must be sp3.
Que 14. Discuss the nature of bonding in metal carbonyls.
Ans 14. (i) Formation of M ← C sigma bond through the donation of lone pair of electrons of carbon (of CO) into empty orbital of metal atom. This is dative overlap.
(ii) Formation of M → C π bond through the donation of electrons from filled metal d-orbitals into vacant antibonding π∗ molecular orbitals of CO.
It is back donation or back bonding. It creates a synergic effect and strengthens the bond between CO and the metal.
Que 15. The magnetic moment of [MnBr4]2– is 5.9 B.M. What is the geometry of this complexion?
Ans 15. Since the coordination number of Mn2+ ion in this complexion is 4, it may be either tetrahedral (sp3 hybridization) or square planar (dsp2 hybridization) as shown below at (a) and (b). But the fact that the magnetic moment of the complexion is 5.9 B.M. shows that it should be tetrahedral in shape rather than square-planar.
Coordination Compounds Important Questions
