NCERT Questions for Class 12 Chemistry Chapter 5 – Coordination Compounds

Important Questions with Solutions of Class 12 Chemistry Chapter 5 – Coordination Compounds

Short Answer Questions

1) Describe Werner’s Theory’s key principles.

Ans – Werner’s Theory’s primary principles are:

(i) The core metal ion exhibits two distinct types of valencies, namely the primary valency and the secondary valency.

(ii) A compound’s primary valency is its oxidation state, and its secondary valency is a complex’s coordination number.

(iii) Since each species’ secondary valency is fixed, the coordination number is also fixed.

(iv) A metal atom is able to satisfy both its primary and secondary valencies. A negative ion satisfies the primary valency. Conversely, neutral molecules and negative ions satisfy secondary valencies.

2) Using an example of each, list two distinctions between a coordination compound and a double salt.

Ans – A coordination compound and double salt differ in the following ways:

Double Salt: A double salt is a compound composed of two different salt components. In the water, they totally separate into their ions.
eg – K₂SO₄.Al₂(SO₄)₃.24H₂O – Potash Alum

Coordination Compound: A complex salt is composed of ligands that establish coordination bonds with a core metal atom. Coordination compounds do not entirely separate into their ions in water.
eg – K₄Fe(CN)₆ – Potassium ferrocyanide

3) Which kind of structural isomerism does each of the following complexes represent?

(i) [Co(NH₃)₅ (NO₃)]SO₄ and [Co(NH₃)₅ (SO₄)]NO₃

Ans – Ionisation isomers are isomers that, even though they share the same chemical makeup, generate different ions in solution. In the former case, the nitrate ion is coordinated to the cobalt ion, whilst in the latter case, the sulphate ion is out of the coordination sphere; in the latter case, it is inside the coordination sphere. Its ionisation isomers are the outcome.

(ii) [Mn(CO)₅(SCN)] and [Mn(CO)₅(NCS)]

Ans – The provided complexes demonstrate structural isomerism of the linkage type. Linkage isomerism is the existence of coordination compounds with the same composition but varying metal-to-ligand connection. SCN− and NCS− are the ligands exhibiting linkage isomerism.

3 Mark Questions

1) Describe how the structures of the two nickel complexes [Ni(CN)₄ ]^2- and Ni(CO)₄ differ although their magnetic behaviours are the same. (Atomic no. of Ni = 28).

Ans – The ground state valence shell electronic configuration of the nickel atom in [Ni(CN)₄ ]^2- is Ar[18]3d⁸ 4s². Since nickel has an oxidation state of +2, Ni²⁺ has the electronic configuration Ar[18]3d⁸ 4s².

When a strong field ligand CN is present, electron pairing begins, and [Ni(CN)₄ ]^2- undergoes dsp² hybridisation with a square planar shape. The diamagnetic character of [Ni(CN)₄ ]^2- is caused by the existence of two unpaired electrons in the 3d subshell.

Since there is no oxidation state of nickel in Ni(CO)₄, its electronic configuration will be Ar[18]3d⁸ 4s². The electron pairing begins when CO, a strong field ligand, pushes all ten valence electrons to the 3d subshell and pairs them. Consequently, Ni(CO)₄ has a tetrahedral shape and is sp³ hybridised. Since Ni(CO)₄ has no unpaired electrons, it is diamagnetic.

2) Determine the following compounds’ magnetic moments:

(i) [CoF₆]^3-

(ii)[Fe(CN)₆ ]^4-

Ans – (i) Central cobalt has an oxidation state of +3. The Co⁺³ is hence a 3d⁵ complex.

There won’t be any electron pairing because fluoride is a weak field ligand. Thus, n = 5. So,

(ii) Central iron has an oxidation state of +2. Fe²⁺ is therefore a 3d⁶ complex. Since CN⁻ is a strong field ligand, there are no unpaired electrons since the electron pairs up. The magnetic moment can be calculated as:

3) Two forms, “A” and “B,” of a metal complex with the composition Cr(NH₃ )₄Cl₂Br have been isolated. A white precipitate that is easily soluble in diluted aqueous ammonia is produced when form “A” combines with AgNO₃ solution, while form “B” produces a pale yellow precipitate that dissolves in concentrated ammonia solution. Write the “A” and “B” formulas. Mention the isomerism that develops between “A” and “B” as well.

Ans – [Cr(NH₃)₄ClBr]Cl is compound A. AgNO₃‘s reaction with it:

[Cr(NH₃)₄ClBr]Cl + AgNO₃ → AgCl + [Cr(NH₃)₄ClBr]⁺NO₃⁻

Silver chloride (AgCl) makes up the whitish precipitate. In an ammonia solution, this white silver chloride dissolves as follows:

AgCl + 2NH₄OH → [Ag(NH₃)₄Cl] + 2H₂O

The compound [Cr(NH3)4Cl2]Br is compound B. AgNO3’s reaction with it:

[Cr(NH₃)₄Cl₂]Br + AgNO₃ → AgBr + [Cr(NH₃)₄Cl₂]⁺NO₃⁻

Silver bromide (AgBr), a pale yellow precipitate that dissolves in a concentrated ammonia solution, is produced:

AgBr + 2NH₄OH → [Ag(NH₃)₄Br] + 2H₂O

Compounds A and B are ionization isomers of each other.