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KCET · Chemistry · Coordination Compounds

Match List-I with List-II
List-I (Complex)List-II (Geometry)
(a)\([\text{Co(NH}_3\text{)}_6]^{3+}\)(i)Trigonal bipyramidal
(b)\([\text{NiCl}_4]^{2-}\)(ii)Octahedral
(c)\([\text{Ni(CN)}_4]^{2-}\)(iii)Tetrahedral
(d)\([\text{Fe(CO)}_5]\)(iv)Square planar
Choose the correct answer from the options given below.

  1. A a-ii, b-iii, c-iv, d-i
  2. B a-ii, b-i, c-iii, d-iv
  3. C a-iii, b-ii, c-iv, d-i
  4. D a-i, b-iii, c-iv, d-ii
Verified Solution

Answer & Solution

Correct Answer

(A) a-ii, b-iii, c-iv, d-i

Step-by-step Solution

Detailed explanation

For \([\text{Co(NH}_3\text{)}_6]^{3+}\), the coordination number is \(6\), which corresponds to an octahedral geometry. Thus, (a) matches with (ii).

For \([\text{NiCl}_4]^{2-}\), \(\text{Ni}\) is in the \(+2\) oxidation state with a \(3d^8\) configuration. Since \(\text{Cl}^-\) is a weak field ligand, no pairing occurs, leading to \(sp^3\) hybridization and a tetrahedral geometry. Thus, (b) matches with (iii).

For \([\text{Ni(CN)}_4]^{2-}\), \(\text{Ni}\) is in the \(+2\) oxidation state with a \(3d^8\) configuration. Since \(\text{CN}^-\) is a strong field ligand, pairing of electrons occurs, leading to \(dsp^2\) hybridization and a square planar geometry. Thus, (c) matches with (iv).

For \([\text{Fe(CO)}_5]\), the coordination number is \(5\). \(\text{Fe}\) is in the \(0\) oxidation state with a \(3d^6 4s^2\) configuration. The strong field ligand \(\text{CO}\) causes pairing of electrons, resulting in \(dsp^3\) hybridization and a trigonal bipyramidal geometry. Thus, (d) matches with (i).

The correct matching is a-ii, b-iii, c-iv, d-i.

Answer: a-ii, b-iii, c-iv, d-i