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Crystal Field Split in Octahedral Complexes

Which of the ...

Question

Which of the following options are correct for $K_{3} [F e (C N)_{6}]$ complex?

A

d^{2} s p^{3}

hybridisation state

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B

s p^{3} d^{2}

hybridisation state

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C

Paramagnetic

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D

Diamagnetic

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Solution

The correct option is C Paramagnetic
$C N^{-}$ is a strong field ligand that will cause the pairing of the electrons in the 3d orbital. Therefore, it undergoes $d^{2} s p^{3}$ hybridization and 1 electron in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.

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Similar questions

Q. Which of the following options are correct regarding the complex

[F e (C N)_{6}]^{3 -}

[F e (C N)_{6}]^{4 -}

Q. Which of the following options are correct for

K_{3} [F e (C N)_{6}]

Q. Identify the correct order of flocculation value for

F e (O H)_{3}

Q. The equilibrium constants for the two complex formation are :

4 K^{+} + F e {(C N)}_{6}^{4 -} ⟶ K_{4} [F e {(C N)}_{6}]; K = 3.8 \times 10^{- 38}

3 K^{+} + F e {(C N)}_{6}^{3 -} ⟶ K_{3} [F e {(C N)}_{6}]; K = 5 \times 10^{- 18}

Q. Which one is more stable in the following pairs of complexes? Give reasons.
i)

[C o (N O_{2})_{6}]^{4 -}

[C o (N O_{2})_{6}]^{3 -}

K_{4} [F e (C N)_{6}]

K_{3} [F e (C N)_{6}]

[C O (H_{2} O)_{6}]^{2 +}

[C O (N H_{3})_{6}]^{2 +}

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Crystal Field Split in Octahedral Complexes

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