Question

${{[Ni{{(CN)}_{4}}]}^{2-}}$and ${{[Ni{{(Cl)}_{4}}]}^{2-}}$have similarity but not in:
a.) Magnetic moment
b.) C.N. and O.N.
c.) Structure
d.) Both (a) and (c)

Answer 1

Hint: We know that both compounds are complex compounds and both have the same central atom. So, the basis of the nature of ligands we will see all properties of complexes.
Step by step solution:
${{[Ni{{(CN)}_{4}}]}^{2-}}$ is a complex compound. In this complex compound the central atom is attached with four ligands. And these ligands are $-CN$ . Overall compound has 2- charge on it and the central atom has +2 charges (this we will calculate by charge balancing on the compound).
Similarly,${{[Ni{{(Cl)}_{4}}]}^{2-}}$ is a complex compound. In this complex compound the central atom is attached with four ligands. And these ligands are $Cl$. Overall compound has 2- charge on it and the central atom has +2 charges (this we will calculate by charge balancing on the compound).
$N{{i}^{2+}}-[Ar]3{{d}^{8}}$
So, when the ligand is a strong field ligand it will pair all the unpaired electrons and hence be diamagnetic. When the ligand is a weak field ligand it won't pair the unpaired electrons and hence paramagnetic. And we know that $-CN$is a strong field ligand and $Cl$ is a weak field ligand. So, ${{[Ni{{(CN)}_{4}}]}^{2-}}$ is a diamagnetic and ${{[Ni{{(Cl)}_{4}}]}^{2-}}$ is a paramagnetic. So, both have different magnetic fields. Both have coordination numbers of 4. ${{[Ni{{(CN)}_{4}}]}^{2-}}$is $ds{{p}^{2}}$ hybridised and ${{[Ni{{(Cl)}_{4}}]}^{2-}}$is $s{{p}^{3}}$ hybridised. So, both have different structures one is tetrahedral and the other is square planar.
So, option “A” and “B” both are correct. Then the answer is “D”.

Note: Structure, hybridisation and magnetic behaviour all these depend on the nature of the ligands that they are strong or weak fields.