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Question
An aqueous pink solution of cobalt (II) chloride changes to deep blue on addition of excess of HCl. This is because:
(i) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl6]}^{4-}\]
(ii) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl4]}^{2-}\]
(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.
(iv) tetrahedral complexes have larger crystal field splitting than octahedral complex.
Solution
(ii) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl4]}^{2-}\]
(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.
Explanation:
Aqueous pink solution of cobalt (II) chloride is due to electronic transition of electron from t2g to eg energy level of \[\ce{[Co(H2O)6]^{2+}}\] complex. When excess of HCl is added to this solution
(i) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl4]^{2-}}\].
(ii) Tetrahedral complexes have smaller crystal field splitting than octahedral complexes because Δr = `4/9` Δ0
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