Question

Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code.

$\begin{array}{cc}\text{Column~I}& \text{Column~II}\\ \left(Complexion\right)& (Hybridisation,numberofunpairedelectrons)\\ A.\left[Cr\right(H_{2}O{)}_6{]}^{3+}& 1.ds{p}^2,1\\ B.\left[Co\right(CN{)}_4{]}^{2-}& 2.s{p}^3{d}^2,5\\ C.\left[Ni\right(N{H}_3{)}_6{]}^{2+}& 1.d^{2}s{p}^3,3\\ D.[Mn{F}_6{]}^{4-}& 2.s{p}^3{d}^2,2\end{array}$

Codes

$\begin{array}{ccccc}& A& B& C& D\\ \left(a\right)& 3& 1& 4& 2\\ \left(b\right)& 4& 3& 2& 1\\ \left(c\right)& 3& 2& 4& 1\\ \left(d\right)& 4& 1& 2& 3\end{array}$

Answer 1

$A.\to \left(3\right)B.\to \left(1\right)C.\to \left(4\right)D.\to \left(2\right)$
Formation of inner orbital complex and outer orbital complex determines hybridisation of molecule which inturn depends upon field strength of ligand and number of vacant d orbitals.
(i) Strong field ligand forms inner orbital complex with hybridisation $d^{2}s{p}^3.$
(ii) Weak field ligand forms outer orbital complex with hybridisation $s{p}^3{d}^2.$
According to VBT, hybridisation and number of unpaired electrons of coordination compounds can be calculated as
A. $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$
MOEC (Molecular orbital electronic configuration) of $C{r}^{3+}$ in $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$ is



Hybridisation = $d^{2}s{p}^3$
n (number of unpaired electrons) = 3

B. $\left[Co\right(CN{)}_4{]}^{2-}$
MOEC of $C{o}^{2+}$ is



Hybridisation = $ds{p}^2$
n=1

C. $\left[Ni\right(N{H}_3{)}_6{]}^{2+}$
MOEC of $N{i}^{2+}$ in $\left[Ni\right(N{H}_3{)}_6{]}^{2+}$ is



Hybridisation = $s{p}^3{d}^2$
n=2

D. $[Mn{F}_6{]}^{4-}$
MOEC of $M{n}^{2+}$ in $[Mn{F}_6{]}^{4-}$



Hybridisation = $s{p}^3{d}^2$
n=5
Hence, correct choice can be represented by (a).