Coordination Compounds intext questions solutions: Class 12 Chemistry Chapter 5 intext questions solutions
| Textbook | Ncert |
| Class | Class 12 |
| Subject | Chemistry |
| Chapter | Chapter 5 |
| Chapter Name | Coordination Compounds Class 12 intext questions solutions |
| Category | Intext questions |
| Medium | English |
Are you looking for Ncert Chemistry Class 12 intext questions solutions chapter 5 Coordination Compounds? Now you can download Coordination Compounds intext questions solutions from here.
Question 5.1: Write the formulas for the following coordination compounds:
- (i) Tetraamminediaquacobalt(III) chloride
- (ii) Potassium tetracyanidonickelate(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) hexacyanidoferrate(II)
Solution 5.1:
- (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
Question 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 5.2:
- (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
Question 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 5.3: 1. 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 exist.
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.
Question 5.4: Give evidence that [Co(NH3)5Cl]S04 and [Co(NH3)5S04]Cl are ionisation isomers.
Solution 5.4: 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)
Question 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 5.5: 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.
Question 5.6: [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
Solution 5.6: 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. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic.
In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2.
But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic.
Question 5.7: [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain.
Solution 5.7: 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,
\(\mu = \sqrt{n(n+2)} \)
\(\mu = \sqrt{1(1+2)}\)
= √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,
\(\mu = \sqrt{n(n+2)} \)
\(\mu = \sqrt{5(5+2)}\)
= √35
≈ 6 BM
Thus, it is evident that [Fe(H2O)6]3+ is strongly paramagnetic, while [Fe(CN)6]3– is weakly paramagnetic.
Question 5.8: Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Solution 5.8:
| [Co(NH3)6]3+ | |
| Oxidation state of cobalt = +3 | Oxidation state of Ni = +2 |
| Electronic configuration of cobalt = d6 | Electronic configuration of nickel = d8 |
 NH3 being a strong field ligand causes the pairing. Therefore, Cobalt can undergo d2sp3 hybridization. Hence, it is an inner orbital complex. |  If NH3 causes the pairing, then only one 3d orbital is empty. Thus, it cannot undergo d2sp3 hybridization. Therefore, it undergoes sp3d2 hybridization.  Hence, it forms an outer orbital complex. |
Question 5.9: Predict the number of unpaired electrons in the square planar [Pt(CN)4]2- ion.
Solution 5.9: 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−
Question 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 5.10:
| [Mn(H2O)6)]2+ | [Mn(CN)6]4- |
| Mn is in the +2 oxidation state. | Mn is in the +2 oxidation state. |
| The electronic configuration is d5 | 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.