Question

Barium ions, $C{N}^{-}$ and $C{o}^{+2}$ form an ionic complex. If this complex is 75% ionized in aqueous solution with Vant Hoff factor(i) equal to four and paramagnetic moment is found to be 1.73 BM(due to spin only) then hybridization state of 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

$ds{p}^3$

as $\mu =1.73,$ no of unpaired $e^{-}=1,\alpha$ of complex =0.75, i=4
Therefore no.of particles in the solution $=5(i=1+(n-1\left)\alpha \right)$
$B{a}^{+2}\left[Co\right(CN{)}_y{]}^{-2}\Rightarrow B{a}_x\left[Co\right(CN{)}_y{]}_2\therefore x=3andy=5C{o}^{+2}\left[Ar\right]4S^{\circ }3d^7;$ get paired ; $\therefore dS{p}^3$