SATHEE: Coordination Compounds - Result Question 38

Coordination Compounds - Result Question 38

####40. The Crystal Field Stabilisation Energy (CFSE) for $[C o C l_{6}]^{4 -}$ is $18000 c m^{- 1}$ . The CFSE for $[C o C l_{4}]^{2 -}$ will be

[NEET Odisha 2019]

(a) $8000 c m^{- 1}$

(c) $6000 c m^{- 1}$

(c) $16000 c m^{- 1}$

(d) $18000 c m^{- 1}$

Show Answer

Solution:

(a) $∵ Δ_{t} = \frac{4}{9} Δ_{0}$

$Δ_{t} = \frac{4}{9} \times 18000 c m^{- 1} = 8000 c m^{- 1}$

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