Question
The hybridisation involved in complex $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is ( Atomic number of $$Ni=28$$ )
A.
$$ds{p^2}$$
B.
$$s{p^3}$$
C.
$${d^2}s{p^2}$$
D.
$${d^2}s{p^3}$$
Answer :
$$ds{p^2}$$
Solution :
$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
Let oxidation state of $$Ni$$ in $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is $$x.$$
$$\eqalign{ & \therefore \,\,x - 4 = - 2 \cr & {\text{or}}\,\,\,\,x = 2 \cr} $$
Now, $$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}$$
$$\because \,\,C{N^ - }$$ is a strong field ligand. Hence, all unpaired electrons are paired up.
∴ Hybridisation of $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is $$ds{p^2}$$
$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
Let oxidation state of $$Ni$$ in $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is $$x.$$
$$\eqalign{ & \therefore \,\,x - 4 = - 2 \cr & {\text{or}}\,\,\,\,x = 2 \cr} $$
Now, $$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}$$
$$\because \,\,C{N^ - }$$ is a strong field ligand. Hence, all unpaired electrons are paired up.
∴ Hybridisation of $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is $$ds{p^2}$$