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

• A

  It is ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridised and octahedral.
• B

  It is ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridised and tetrahedral.
• Correct answer
  C

  It is ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridised and octahedral.
• D

  It is ${\mathrm{dsp}}^2$ hybridised and square planar.

Solution

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

$\mathrm{Mn}\left(\mathrm{III}\right)=\left[\mathrm{Ar}\right]3{\mathrm{d}}^4$

${\mathrm{CN}}^{-}$ being strong field ligand forces pairing of electrons.

This gives ${\mathrm{t}}_{2\mathrm{g}}^4\mathrm{ }{\mathrm{e}}_{\mathrm{g}}^0$



$\because$The coordination number of $\mathrm{Mn}=6$

$\therefore$ Structure$=$octahedral

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

Hence, the correct answer is option: 'C'.