Question

The compounds having tetrahedral geometry is:
A. ${[Ni{(CN)_4}]^{ - 2}}$
B. ${[Pd{(CN)_4}]^{ - 2}}$
C. ${[PdC{l_4}]^{ - 2}}$
D. ${[NiC{l_4}]^{ - 2}}$

Answer 1

Hint: In the presence of strong field ligands, large splitting takes place and the geometry obtained is square planar. In the presence of weak field ligands, small splitting takes place and the geometry of the molecule is tetrahedral geometry.

Complete step by step answer:
Crystal field theory (CFT) explains the breaking of orbital degeneracy in transition metal complexes due to the presence of ligands. Initially, the five d-orbitals have the same energy.
In the tetrahedral geometry, ${d_{xy}}$, ${d_{xz}}$ and ${d_{yz}}$ orbitals are raised in energy while the ${d_{{x^2} - {y^2}}}$ and ${d_{{z^2}}}$ orbitals are lowered in energy. For the square planar complexes, the greatest interaction is with the ${d_{{x^2} - {y^2}}}$ orbital and thus it has the highest energy. The second greatest interaction is with the ${d_{xy}}$ orbital, followed by ${d_{{z^2}}}$. The orbitals with the lowest equivalent energy are ${d_{xz}}$ and ${d_{yz}}$.
In the presence of strong field ligands, always high splitting takes place and the geometry obtained is square planar. The hybridization of the central atom is $ds{p^2}$. In the presence of weak field ligands, always low splitting takes place and the geometry of the complex is tetrahedral. The hybridization of the central atom in this case is $s{p^3}$. Transition elements of fifth period and larger always have square planar geometry irrespective of the ligand.

Now, in the question
${[Ni{(CN)_4}]^{ - 2}}$
Ni is a fourth period element and has four strong field ligands CN. Hence, it has high splitting and its geometry is square planar.
${[Pd{(CN)_4}]^{ - 2}}$
Pd is a fifth period element. Hence, it has high splitting irrespective of the ligand and the geometry is square planar.
${[PdC{l_4}]^{ - 2}}$
Pd is a fifth period element. Hence, it has high splitting irrespective of the ligand and the geometry is square planar.
${[NiC{l_4}]^{ - 2}}$
Ni is a fourth period element and has four weak field ligands Cl. Hence, it has low splitting and its geometry is tetrahedral.

Hence, the correct answer is (D) ${[NiC{l_4}]^{ - 2}}$

Note: If the compound has an equal number of weak field and strong field ligands, then it is a data-based question or the compound is of period greater than or equal to 5. Such questions are rarely asked in JEE. If the period is greater than or equal to 5, high splitting takes place.