RBSE Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

Rajasthan Board RBSE Class 12 Chemistry Chapter 9 Coordination Compounds

RBSE Class 12 Chemistry Chapter 9 Text Book Type Questions

RBSE Class 12 Chemistry Chapter 9 Multiple Choice Questions

Question 1.
The oxidation state of Fe in K₄ [Fe(CN)₆] is
(a) +2
(b) 3
(c) O
(d) None of these

Question 2.
Which of the following compound has tetrahedral shape?
(a) [Ni(CN₂)^2-
(b) [NiCl₄]^2-
(c) [PdCl₄]^2-
(d) [Ni(CN)₄]^2-

Question 3.
The coordination number of (EDTA)^4- is
(a) 3
(b) 6
(c) 4
(d) 5

Question 4.
The number of geometrical isomers of the [Pt(NH₃)₂Cl₂] is
(a) 3
(b) 2
(c) 4
(d) 1

Question 5.
A complex which is formed by chloride and nitrate ligand gives two moles of precipitate of AgCl with AgNO₃. Its molecular formula will be
(a) [CO(NH₃)₅NO₃]Cl₂
(b) [CO(NH₃)₅CI]CINO₃ CI
(c) (CO(NH₃)₅Cl]NO₂
(d) None of the above

Question 6.
Which of the complex shows optical isomerism ?
(a) [CO(CN)₆]³⁺
(b) [ZnCl₄]^2-
(c) [CO(CN)₂Cl₂]
(d) [Cu(NH₃)₄]²⁺

Question 7.
The hybridisation of [Ni(CO)₄] is
(a) sp
(b) sp²
(c) dsp²
(d) sp³

Question 8.
Chlorophyll contains
(a) cobalt
(b) magnesium
(c) iron
(d) nickel

Answers:
1. (a) 2. (b) 3. (b) 4. (a) 5. (c) 6. (c) 7. (d) 8. (b)

RBSE Class 12 Chemistry Chapter 9 Very Short Answer Type Questions

Question 1.
Calculate the oxidation state and coordination number of central metal atom in the K₃ [Fe(C₂O₄)₃]
Answer:
As K⁺ ion carries charge = +1, charge on 3K ions = +3.
Hence charge on complex ion = +3. As each oxalate ion C₂O^2-₄ has charge = +2, charge on three C₂O^2-₄ = -6. Therefore charge on Fe should be = +3 i.e., oxidation number of Fe in the given complex = +3.
Coordination number of ligand = No. of ligand x Denticity
= 3 x 2 = 6 .

Question 2.
Name of the suitable ligand used to determine hardness of water.
Answer:
[EDTA]^4- Ethylene diaminetetraacetate ion

Question 3.
Write the IUPAC name of Li[AIH₄].
Answer:
Lithium Tetrahyridoaluminate (III)

Question 4.
Draw the mirror image of [CO(en)₂Cl₂]⁺
Answer:
[COCl₂(en)₂]⁺ exhibits geometrical isomerism as under

Question 5.
Calculate the magnetic moment of Ni²⁺ ion.
Answer:
Configuration of d orbitals for Ni²⁺ is 3d⁸ so the number of unpaired electrons are two therefore the magnetic moment of Ni²⁺ given by
µ = \(\sqrt { n(n+2) }\)
= \(\sqrt { 2(2+2) }\)
= √8 = 2√2
= 2 x 1.414 = 2.828 B.M.

Question 6.
Write the IUPAC name of [Mn(CO)₁₂]
Answer:
Dodecyl caorbonyl manganese (0).

Question 7.
With the help of an example explain why ambidentate ligands are named so ?
Answer:
Ligands which can ligate (link) through two or more different donor atoms present in it, are called ambidentate ligands. Example: NO^–₂ and SCN^–, NO^–₂ can link through N as well as O while SCN^– can link through S as well as N atoms.

Question 8.
Classify the following ligands as mono, bidentate etc:
(i) en
(ii) CN
(iii) acac
(iv) dmg
Answer:
Ligand                                  Denticity
en                                   Bidentate ligand.
CN^–                              Monodentate ligand
асас                                Bidentate ligand
dmg                               Bidentate ligand

RBSE Class 12 Chemistry Chapter 9 Short Type Answer Questions

Question 1.
What is the chelate effect ? Give an example.
Answer:
When a polydentate ligand simultaneously coordinates to a metal ion through more than one donor sites, a ring like structure is obtained which is known as chelate ring and the ligand is known as chelating ligand. Chelate complexes are more stable than ordinary complexes. The stability of complexes is greatly increased by the chelation. This is known as chelate effect.
eg. [Fe(EDTA)]^– is more stable as it has five ring structure than [Cu(en)₂]²⁺, it has two ring structure. Greater the number of rings present in a complex, greater is the stability of complex.

Question 2.
Two complexes having molecular formula CO(NH₃)₅SO₄Br are taken in the two bottles A and B. One complex will give white precipitate with BaCl₂ and other light yellow precipitate with AgNO₃. Calculate the molecular formula of complexes A and B.
Answer:
[CO(NH₃)_5sBr]SO₄ gives sulphate ions in aqueous solution which reacts with Ba⁺² on addition of BaCl₂ to give precipitate of BaSO₄.
[CO(NH₃)₅Br]SO₄ ⇄ (CO(NH₃)₅Br]²⁺ + SO^2-₄
SO^2-₄ + BaCl₂ → BaSO₄ ↓ + 2Cl^–
However, it does not react with AgNO₃ solution.

The complex [CO(NH₃)₅SO₄]Br gives bromide ions in aqueous solution which gives yellow ppt with silver nitrate.

However, it does not react with BaCl₂ solution.
[Co(NH₃)₅SO₄] Br \(\underrightarrow { Bacl_{ 2 } }\) No reaction So, molecular formula of A complex will be [CO(NH₃)₅Br] SO₄ and molecular formula of B complex will be [Co(NH₃)₅SO₄] Br.

Question 3.
Calculate the oxidation state of central metal • atom in the following complexes :
(i) K₂[Fe(C₂O₄)₃]
(ii) [Fe(CN)₆]^3-
Answer:
(i) 1 x 2 + x – 2 x 3 = 0
2 + x – 6 = 0
x = 6 – 2 = + 4
Oxidation state of central metal atom will be + 4.

(ii) x + ( – 1 x 6) = -3
x= -3+6= 3
Oxidation state of central metal atom will be + 3.

Question 4.
What will be geometry of complexes having hybridisation sp³ and dsp². Give an example of each.
Answer:
The geometry of complex having hybridisation sp³ will be tetrahedral and geometry for dsp².
hybridisation will be square planar.
e.g., [NiCl₄]^2- having tetrahedral geometry and hybridization will be sp³ and [Pt(CN)₄]^2- having square planar geometry and hybridisation will be dsp on central metal atom.

Question 5.
Explain the importance of coordination compounds in extraction of metals.
Answer:
KCN is used to extract silver from Ag₂S in the form of cynide complex and then reducing it with zinc as show by the following equations :
4KCN + Ag₂S → 2K[Ag(CN)₂] + K₂S
2K[Ag(CN)₂] + Zn → K₂[Zn(CN)₄] + 2Ag

RBSE Class 12 Chemistry Chapter 9 Long Answer Type Questions

Question 1.
Explain the hybridisation of the central metal atom in the complex [Ni(CN)₄]^2- with the help of diagram.
Answer:
[Ni(CN)₄]^2- – Electronic configuration of
Ni : [Ar]4s²3d⁸ Electronic configuration of Ni²⁺ : [Ar] 4s⁰ 3d⁸
Outer electronic configuration of Ni²⁺ :

Pairing of electrons takes place in the presence of strong field ligand CN^–. Thus, one 3d orbital becomes vacant.

As the hybridisation involved is dsp’, we get square · planar complex. As there are no unpaired electrons the complex is diamagnetic.

Question 2.
Compare the complex [Fe(H₂O)₆]²⁺ and [Fe(CN)₆]^4- with the help of crystal field theory.
Answer:
H₂Ois a weak field ligand and it will have low value of ∆₀. As a result, the electrons can fill lower set of dorbitals as well as upper set as : t⁴_2ge²_g

On the other hand, CN^– is strong field ligand, so that ∆₀. will be high. As a result, electron will try to remain only in the lower set and will therefore pair up i.e., t⁶_2g.

Question 3.
Explain ionisation isomerism. Write IUPAC name of [CO(NH₃)₅Cl]SO₄ and [CO(NH₃)₅SO₄]Cl. Give reasons to show that these are ionisation isomers.
Answer:
Ionisation isomerism-Ionisation isomerism occurs when same molecular formula gives different ions in the solution. For example, [Pt(NH₃)₄Cl₂]Br₂ and [Pt(NH₃)₄Br₂]Cl₂ are ionisation isomers.
[CO(NH₃)₄Cl]SO₄ – IUPAC
Name – Pentaammine chloridocobalt (III) sulphate
[CO(NH₃)₅SO₄]Cl – IUPAC
Name – Pentaammine sulphato cobalt (III) chloride
On dissolving in water, the two compounds will give different ions in the solution which can be tested by adding AgNO₃ solution and BaCl₂ solution. When Cl^– ions are the counter ions, a white ppt. will be obtained with the AgNO₃ and BaCl₂ solution. If SO^2-₄ ions are the counter ions, a white ppt is obtained. It is shown by the following equations :

Question 4.
Write the IUPAC name of the following complexes :
(a) [Pt(NH₃ )₂Cl(NO₂)]
(b) Na[BH₄]
(c) [CO(NH₃)₅CO₃]Cl
(d) Zn₂[Fe(CN)₆]
Answer:
(a) Diammine chloridonitrito – N – platinum (II)
(b) Sodium tetrahydrido borate (III)
(c) Pentaammine carbonato cobalt (III) chloride
(d) Zinc hexacynoferrate (II)

RBSE Solutions for Class 12 Chemistry