Question

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

• A. $\left[Co\right(H_{2}O{)}_6{]}^{2+}$ is transformed into $[CoC{l}_6{]}^{4–}$
• B. $\left[Co\right(H_{2}O{)}_6{]}^{2+}$ is transformed into $[CoC{l}_4{]}^{2–}$
• C. tetrahedral complexes have smaller crystal field splitting than octahedral complexes.
• D. tetrahedral complexes have larger crystal field splitting than octahedral complex.

Answer 1

Answer: (B), (C)

The correct options are
B $\left[Co\right(H_{2}O{)}_6{]}^{2+}$ is transformed into $[CoC{l}_4{]}^{2–}$
C tetrahedral complexes have smaller crystal field splitting than octahedral complexes.

Aqueous pink solution of cobalt (II) chloride is due to electronic transition of electron from $t_{2}g$ to $e_g$ energy level of $\left[Co\right(H_{2}O{)}_6{]}^{2+}$ complex. When excess of HCl is added to this solution (i) $\left[Co\right(H_{2}O{)}_6{]}^{2+}$ is transformed into $[CoC{l}_4{]}^{2-}$.
(ii) Tetrahedral complexes have smaller crystal field splitting than octahedral complexes because ${△}_t=\left(4/9\right){△}_O$
Hence, options (b) and (c) are correct choices.