Question

Which of the following shows ${\text{dsp}}^2$ hybridisation and a square planar geometry?

• A. $S{F}_6$
• B. $Br{F}_5$
• C. $PC{l}_5$
• D. $\left[Ni\right(CN{)}_4{]}^{2-}$

Answer 1

The correct option is D $\left[Ni\right(CN{)}_4{]}^{2-}$

Rules to determine hybridisation of a coordination compound:-

$1.$ Central metal ion release the electron equal to its oxidation state,

$2.$ Metal ion arrange empty orbitals equal to its coordination number,

$3.$ These empty orbital arrange in a specfic geometry,

$4.$Hybridised orbitals of metal ion forms bonds with the lone pair orbitals of ligand.

Apply these rules for $\left[Ni\right(CN{)}_4{]}^{2-}$,

So the correct option is [D]