The anion $[N i C l_{4}]^{2 -}$ is paramagnetic but when $C N^{-}$ ions are added, the product $[N i (C N)_{4}]^{2 -}$ is diamagnetic because in $[N i C l_{4}]^{2 -}$ , $C l^{-}$ is a weak ligand and thus, electrons are unpaired in $N i^{2 +}$ making it paramagnetic, $s p^{3}$ hybridised and tetrahedral.
If true enter 1, else enter 0.

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Solution

$[N i C l_{4}]^{2 -} (a q) + 4 C N^{-} (a q) \to [N i (C N)_{4}]^{2 -} (a q) + 4 C l^{-} (a q)$
In $[N i C l_{4}]^{2 -} C l^{-}$ is a weak ligand. Thus, electrons are unpaired in $N i^{2 +}$ making it paramagnetic. It is a high spin complex. It is $s p^{3}$ hybridised and has tetrahedral structure.
When strong ligand like $C N^{-}$ is added, substitution takes place and electrons are paired in $[N i (C N)_{4}]^{2 -}$ . It is a low spin complex. It is $d s p^{2}$ hybridised and has square planar structure.
Hence, the given statement is true.

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