Question

Match Column-I with Column-II:
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{a.}{\left[Ni\right(CN{)}_4]}^{2-}& \text{p.}{d}^{2}s{p}^3\\ \text{b.}\left[Ni\right(CO{)}_4]& \text{q.}s{p}^3{d}^2\\ \text{c.}{\left[Co\right(en{)}_3]}^{3+}& \text{r.}ds{p}^2\\ \text{d.}{\left[Co\right(Br{)}_{3}C{l}_3]}^{3-}& \text{s.}s{p}^3\\ & \text{t. Inner orbital complex}\\ & \text{u. Outer orbital complex}\end{array}$

• A. $(a-r,t),(b-s,u)$
  $(c-p,t),(d-p,t)$
• B. $(a-s,t),(b-r,u)$
  $(c-p,t),(d-p,t)$
• C. $(a-r,t),(b-s,u)$
  $(c-q,t),(d-p,t)$
• D. $(a-s,t),(b-s,u)$
  $(c-q,u),(d-q,u)$

Answer 1

The correct option is A $(a-r,t),(b-s,u)$

$(c-p,t),(d-p,t)$
When the complex formed involves the inner `'(n-1)d'` orbital for hybridisation `(d^2sp^3)`, the complex is called inner orbital complex. They are also called as low spin complex.

When the complex formed involves the outer `'nd'` orbital for hybridisation `(sp^3d^2)`, the complex is called outer orbital complex. They are also called as high spin complex.

$\left[Ni\right(CN{)}_4{]}^{2-}$
The ligand is $C{N}^{-}$
It is a strong field ligand for $N{i}^{2+}$
$N{i}^{2+}\Rightarrow 3d^{8}4s^{0}4p^{0}4d^0$
Pairing of electron happens due to the presence of a strong field ligand. Thus, the hybridization is $ds{p}^2$ and the geometry is square plannar and it forms an inner orbital complex.

$\left[Ni\right(CO{)}_4]$
The ligand is $CO$
It is a strong field ligand for $Ni$
$Ni\Rightarrow 3d^{8}4s^{2}4p^{0}4d^0$
In the presence of a strong field ligand, the electrons in `s`-orbitals gets shifted to the `d`-orbital and thus the hybridisation is $s{p}^3$ and it forms an outer orbital complex.

$\left[Co\right(en{)}_3{]}^{3+}$
The ligand is $Ethylenediammine$
It is a strong field ligand for $C{o}^{3+}$
$C{o}^{3+}\Rightarrow 3d^{6}4s^{0}4p^{0}4d^0$
Pairing of electrons takes place because of the presence of a strong field ligand. Thus the hybridisation will be it $d^{2}s{p}^3$ and it forms an inner orbital complex.

$\left[Co\right(Br{)}_3(Cl{)}_3{]}^{3-}$
The ligand is $C{l}^{-}$ and $B{r}^{-}$
They are strong field ligands for $C{o}^{3+}$
$C{o}^{3+}\Rightarrow 3d^{6}4s^{0}4p^{0}4d^0$
Pairing of electrons takes place due to the presence of strong field ligands. Thus the hybridisation will be $d^{2}s{p}^3$ and it forms an inner orbital complex .