Question

Barium ions, $C{N}^{-}$ and $C{o}^{2+}$ form an ionic complex. If this complex is 75% ionised in aqueous solution with Van't Hoff factor (i) equal to four and paramagnetic moment is found to be 1.73 BM (due to spin only) then the hybridisation state to Co (II) in the complex will be :

• A. $s{p}^{3}d$
• B. $d^{2}s{p}^3$
• C. $s{p}^3{d}^2$
• D. $ds{p}^3$

Answer 1

The correct option is D $ds{p}^3$

The dissociation reaction is $B{a}_{(x-2)}\left[Co\right(CN{)}_x{]}_2\to (x-2)B{a}^{2+}+2\left[Co\right(CN{)}_x{]}^{-(x-2)}$
At equilibrium the number of moles of $B{a}_{(x-2)}\left[Co\right(CN{)}_x],B{a}^{2+}and[Co(CN{)}_x{]}^{-(x-2)}$ are $1-\alpha ,(x-2)\alpha ,and2\alpha$ respectively.
The van't Hoff factor is $i=1-\alpha +x\alpha -2\alpha +2\alpha$
Thus, $(x-1)\alpha =3$ or $x-1=\frac{3}{\alpha }$
Substitute $\alpha =0.75;i=4$ in the above expression.
Hence, $x=5$
The formula of the complex is $B{a}_3\left[Co\right(CN{)}_5{]}_2$
It is paramagnetic with spin only magnetic moment of 1.73 BM . Thus it contains one unpaired electron.
It inner orbital complex with strong field ligand and undergoes $ds{p}^3$ hybridization.