Ncert Exercises & Solutions

An aqueous pink solution of cobalt(II) chloride changes to deep blue on the addition of excess of HCl. This is because:

a.	$[Co(H_2O)_6]^{2-}$ is transformed into $[CoCl_6]^{4-}$
b.	$[Co(H_2O)_6]^{2-}$is transformed into $[CoCl_4]^{2-}$
c.	Tetrahedral complexes have smaller crystal field splitting than octahedral complexes
d.	Tetrahedral complexes have larger crystal field splitting than octahedral complexes

Choose the correct option:
1. a and b
2. b and c
3. c and d
4. a and c

Hint: $∆_t = \dfrac {4}{9} ∆_0$

Explanation:
Aqueous pink solution of cobalt (II) chloride is due to the electronic transition of electron fom

t_{2 g} to e_{g}

, the energy level of

{[Co (H_{2} O_{6})]}^{2 +}

complex. When an excess of HCI is added to this solution

(i) ${[Co (H_{2} O_{6})]}^{2 +}$ is transformed into ${[C o C l_{4}]}^{2 -}$

(ii) Tetrahedral complexes have smaller crystal field splitting than octahedral complexes because $∆_{t} = \frac{4}{9} ∆_{0}$

Hence, options (b) and (c) are correct choices.

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a.	\([Co(H_2O)_6]^{2-}\) is transformed into \([CoCl_6]^{4-}\)
b.	\([Co(H_2O)_6]^{2-}\)is transformed into \([CoCl_4]^{2-}\)
c.	Tetrahedral complexes have smaller crystal field splitting than octahedral complexes
d.	Tetrahedral complexes have larger crystal field splitting than octahedral complexes