Question

According to the valence bond theory, the hybridization of central metal atom is dsp² for which one of the following compounds?

विकल्प

• \[\ce{Na2[NiCl4]}\]

• \[\ce{NiCl2 * 6H2O}\]

• \[\ce{K2[Ni(CN)4]}\]

• \[\ce{[Ni(CO)4]}\]

Answer 1

\[\ce{K2[Ni(CN)4]}\]

Explanation:

We must first determine the coordination number of the core metal ion in order to find the complex that contains the central metal atom with dsp-2 hybridization. From there, we can determine that the main metal ion in the coordination complex has hybridised.

In \[\ce{Na2[NiCl_4]}\], Ni exists in +2 oxidation state. The `"Cl"^-` is a weak field ligand. The coordination number of Ni is 4 hence, it undergoes `sp^3` hybridization.

For \[\ce{NiCl2 * 6H2O}\]

Ni	[Ar]3d8 4s2
Ni2+	[Ar] 3d8 4s0

In \[\ce{NiCl2 * 6H2O}\], Ni occurs in +2 oxidation state. The coordination number of Ni is 6.

As H₂O is a weak ligand, Ni⁺² undergoes sp³d² hybridisation as follows:

Ni	[Ar]3d8 4s2
Ni2+	[Ar] 3d8 4s0

In K₂[Ni(CN)₄], Ni occurs in +2 oxidation state. The CN^- is a strong field ligand. The Ni undergoes dsp² hybridization as shown below.

In Ni(CO)₄, the Ni atom undergoes sp³ hybridization as shown below.

Therefore, according to valence bond theory, the hybridization of central metal atom is dsp² for K₂[Ni(CN)₄].