Question

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

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

Answer 1

Answer: (B)

${\left[Mn(CN{)}_6\right]}^{3-}$

$Mn(III)=[Ar]3d^4$

$C{N}^{-}$ being strong field ligand forces pairing of electrons

This gives $t_{2g}^4{e}_g^0$

$\therefore Mn(III)=[Ar]$



$\because$ Coordination number of $Mn=6$

$\therefore$ Structure $=$ octahedral

${\left[Mn(CN{)}_6\right]}^{3-}=$