Pick out the correct statement with respect to [ Mn ( CN ) 6 ] 3 - .

1. It is ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridised and octahedral.
2. It is ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridised and tetrahedral.
3. It is ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridised and octahedral.
4. 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
${\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]}^{3-}=$

Asked in: NEET 2017

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