Chapter 5 Coordination Compounds Intext Questions NCERT Solutions Class 12 Chemistry- PDF Download
Page No. 124
5.1. Write the formulas for the given coordination compounds:
(i) Tetraamminediaquacobalt(III) chloride
(ii) Potassium tetracyanonickelate(II)
(iii) Tris(ethane−1,2−diamine) chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane−1,2−diamine)platinum(IV) nitrate
(vi) Iron(III) hexacyanoferrate(II)
Solution
(i) [Co(H2O)2(NH3)4]Cl3
(ii) K2[Ni(CN)4]
(iii) [Cr(en)3]Cl3
(iv) [Pt (NH3) Br Cl (N02)]–
(v) [PtCl2(en)2](NO3)2
(vi) Fe4[Fe(CN)6]3
5.2. Write the IUPAC names of the following coordination compounds
(i) [Co(NH3)6]Cl3
(ii) [Co(NH3)5Cl]Cl2
(iii) K3[Fe(CN)6]
(iv) K3[Fe(C2O4)3]
(v) K2[PdCl4]
(vi) [Pt(NH3)2Cl(NH2CH3)]Cl
Solution
(i) Hexaamminecobalt(III) chloride
(ii) Pentaamminechloridocobalt(III) chloride
(iii) Potassium hexacyanoferrate(III)
(iv) Potassium trioxalatoferrate(III)
(v) Potassium tetrachloridopalladate(II)
(vi) Diamminechlorido(methylamine)platinum(II) chloride
Page No. 128
5.3. Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:
(i)K[Cr(H2O)2(C2O4)2]
(ii)[CO(en)3]Cl3
(iii)[CO(NH3)5(NO2)(NO3)2]
(iv)[Pt(NH3)(H2O)Cl2]
Solution
(i) Both geometrical (cis-, trans-) isomers for K[Cr(H2O)2(C2O4)2 can exist. Also, optical isomers for cis-isomer exist.
Trans-isomer is optically inactive. On the other hand, cis-isomer is optically active.
(ii) Two optical isomers for [Co(en)3]Cl3 exists:
Two optical isomers are possible for this structure.
(iii) A pair of optical isomers exist for [Co(NH3)5(NO2)](NO3)2:
It can also show linkage isomerism.
[Co(NH3)5(NO2)](NO3)2 and [Co(NH3)5(ONO)](NO3)2
It can also show ionization isomerism.
[Co(NH3)5(NO2)](NO3)2 = [CO(NH3)5(NO3)](NO3)(NO2)
(iv) Geometrical (cis-, trans-) isomers of [Pt(NH3)(H2O)Cl2 can exist.
5.4. Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionization isomers.
Solution
When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react differently with different reagents to give different products.
[CO(NH3)5Cl]SO4 + Ba2+ → BaSO4↓
(BaSO4↓: White precipitate)
[CO(NH3)5Cl]SO4 + Ag+ → No reaction
[CO(NH3)5SO4]Cl + Ba2+ → No reaction
[CO(NH3)5SO4]Cl + Ag+ → AgCl↓
(AgCl↓: White precipitate)
Page No. 135
5.5. Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic.
Solution
Outer electronic configuration of nickel (Z = 28) in ground state is 3d84s2. Nickel in this complex is in + 2 oxidation state. It achieves + 2 oxidation state by the loss of the two 4s-electrons. The resulting Ni2+ ion has outer electronic configuration of 3d8. Since CN– ion is a strong field, under its attacking influence, two unpaired electrons in the 3d orbitals pair up.
Outer electronic configuration of nickel (Z = 28) in ground state is 3d84s2 Nickel in this complex is in +2 oxidation state. Nickel achieves +2 oxidation state by the loss of two 4s-electrons. The resulting Ni2+ ion has outer electronic configuration of 3d8. Since Cl- ion is a weak field ligand, it is not in a position to cause electron pairing.
Since, there are 2 unpaired electrons in [NiCl4]2−, it is paramagnetic in nature.
5.6. [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
Solution
Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands.
In [NiCl4]2− , Ni is in +2 oxidation state.
Ni(28): 3d84s2
Ni2+: 3d84s0
Cl- is weak field ligand. It does not pair up e-s. Hence, it is paramagnetic.
In [Ni(CO)4], Ni is in 0 oxidation state.
Ni(28): 3d84s2
CO is strong field ligand, as it pairs the 4s e-s with 3d e-s to give 3d104s0. So, no unpaired e-s. Hence, the complex is diamagnetic.
5.7. [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain.
Solution
In both [Fe(H2O)6]3+ and [Fe(CN)6]3−, Fe exists in the +3 oxidation state i.e., in d5 configuration.
Since CN− is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital.
Therefore,
= √3
= 1.732 BM
On the other hand, H2O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5.
Therefore,
= √35
≈ 6 BM
Thus, it is evident that [Fe(H2O)6]3+ is strongly paramagnetic, while [Fe(CN)6]3- is weakly paramagnetic.
5.8. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Solution
In [Co(NH3)6]3+,
Co(27) 3d74s2
Here, Co is in +3 oxidation state, so
Co3+: 3d64s2
In presence of NH3, two d e-s pair up leaving two d-orbitals empty. Hence, hybridisation is d2sp3 i.e., inner orbital complex.
In [Ni(NH3)6]2+, Ni is +2 oxidation state so,
Ni2+: 3d84s0
In presence of NH3, d e-s do not pair up. The hybridisation is sp3d2 i.e., outer orbital complex.
5.9. Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion.
Solution
In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp2hybridization. Now, the electronic configuration of Pt(+2) is 5d8.
CN− being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in [Pt(CN)4]2−
5.10. The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
Solution
|
[Mn(H2O)6)]2+ |
[Mn(CN)6]4- |
|
Mn is in the +2 oxidation state The electronic configuration is d5 |
Mn is in the +2 oxidation state. The electronic configuration is d5 |
|
The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in [Mn(H2O)6]2+ is t2g3eg2. |
The crystal field is octahedral. Cyanide is a strong field ligand. Therefore, the arrangement of the electrons in [Mn(CN)6]4- is t2g5eg0. |
Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron.
5.11. Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ ion, given that β4 for this complex is 2.1 × 1013.
Solution
β4 = 2.1 × 1013
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β4.
= 4.7×10-14