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  • Try this! - Coordination compounds: - NEET

Pick out the correct statement with respect \left [ Mn(CN)_{6} \right ]^{3-}

  • Option 1)

    It is sp^{3}d^{2} hybridised and tetrahedral

  • Option 2)

    It is d^{2}sp^{3} hybridised and octahedral

  • Option 3)

    It is dsp^{2} hybridised and square planar

  • Option 4)

    It is sp^{3}d^{2} hybridised and octahedral

 

Answers (1)

As we learnt in 

Hybridisation -

sp3d2 - square bipyramidal or octahedral 

d2sp3 - octahedral 

sp3 - tetradedral 

dsp2 - square planar

- wherein

sp3d2 - outer complex

d2sp3 - inner complex

sp3 - [Ni(Cl)_{4}]^{2-}

dsp2 - [Pt(CN)_{4}]^{2-}

 

 e- configuration of Mn3+  in [Mn(CN)6]3- 

Mn3+ = 3d4 4s0

As CN- is a strong field ligand, so,  e- in 3d orbital will pair up and make way for d2 sp3 hybridization

The geometry will be octahedral.

 


Option 1)

It is sp^{3}d^{2} hybridised and tetrahedral

This option is incorrect

Option 2)

It is d^{2}sp^{3} hybridised and octahedral

This option is correct

Option 3)

It is dsp^{2} hybridised and square planar

This option is incorrect

Option 4)

It is sp^{3}d^{2} hybridised and octahedral

This option is incorrect

Posted by

subam

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