Question

Pick out the correct statement with respect $\left[Mn\right(CN{)}_6{]}^{3-}:$

• A. It is $s{p}^3{d}^2$ hybridised and octahedral
• B. It is $s{p}^3{d}^2$ hybridised and tetrahedral
• C. It is $d^{2}s{p}^3$ hybridised and octahedral
  
• D. It is $ds{p}^2$ hybridised and square planar

Answer 1

The correct option is C It is $d^{2}s{p}^3$ hybridised and octahedral


$\left[Mn\right(CN{)}_6{]}^{3-}Mn\left(III\right)=\left[Ar\right]3d^4$
$C{N}^{–}$ being a strong field ligand, forces pairing of electrons
This gives $t_{2g}^4{e}_g^0$
$\therefore$ $Mn\left(III\right)=\left[Ar\right]$

$\because$ Coordination number of $Mn$ = 6
$\because$ Structure = octahedral
$\left[Mn\right(CN{)}_6{]}^{3-}=$