The CFSE value of tetrahedral $[C o C l_{4}]^{2 -}$ complex is:

+ 2.4 Δ_{t}

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- 1.8 Δ_{t}

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+ 1.2 Δ_{t}

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- 1.2 Δ_{t}

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Solution

The correct option is D $- 1.2 Δ_{t}$
In $[C o C l_{4}]^{2 -}$ , the oxidation state of metal complex is +2.
Thus, its configuration is $d^{7}$ .
$C l^{-}$ is a weak field ligand so it will give high spin complex.

$CFSE = 4 \times (- 0.6) Δ_{t} + 3 \times (+ 0.4) Δ_{t}$

$CFSE = - 2.4 Δ_{t} + 1.2 Δ_{t}$
$CFSE = - 1.2 Δ_{t}$

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The CFSE for octahedral $[C o C l_{6}]^{4 -}$ is 18000 $c m^{- 1}$ . The CFSE for tetrahedral $[C o C l_{4}]^{2 -}$ will be

An aqueous pink solution of cobalt(II) chloride to deep blue on addition of excess of HCL. This is because ..........
(a) $[C o (H_{2} O)_{6}]^{2 +}$ is transformed into $[C o C l_{6}]^{4 -}$
(b) $[C o (H_{2} O)_{6}]^{2 +}$ is transformed into $[C o C l_{4}]^{2 -}$
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