JEE Main 2023 — Chemistry Inorganic Chemistry

Cobalt chloride when dissolved in water forms pink coloured is$[\mathrm{Co}{({H}_{2}O)}_{6}]{\mathrm{Cl}}_{2}$[Co(H2​O)6​]Cl2​ . This is due to the electronic transition of an electron from ${t}_{2g}$t2g​ to ${e}_{g}$eg​ energy level of ${[\mathrm{Co}{({H}_{2}O)}_{6}]}^{2+}$[Co(H2​O)6​]2+ complex. When excess of $\mathrm{HCl}$HCl is added to the solution ${[\mathrm{Co}{({H}_{2}O)}_{6}]}^{2+}$[Co(H2​O)6​]2+ is transformed into deep blue colour complex $[{\mathrm{CoCl}}_{4}{]}^{2-}$[CoCl4​]2−.

$${\triangle }_{t}=\frac{4}{9}{\triangle }_{o}$$△t​=94​△o​

$${\triangle }_{t}<{\triangle }_{o}$$△t​<△o​

Tetrahedral complexes have smaller crystal field splitting than octahedral complexes.

$${\mathrm{CoCl}}_{2}+6{H}_{2}O\longrightarrow \underset{ }{[\mathrm{Co}{({H}_{2}O)}_{6}]{\mathrm{Cl}}_{2}}$$CoCl2​+6H2​O⟶[Co(H2​O)6​]Cl2​​

$${[\mathrm{Co}{({H}_{2}O)}_{6}]}^{2+}\overset{\mathrm{Conc}.\mathrm{HCl}}{\rightarrow }[{\mathrm{CoCl}}_{4}{]}^{2-}$$[Co(H2​O)6​]2+→Conc.HCl[CoCl4​]2−

Hence, $X={[\mathrm{Co}{({H}_{2}O)}_{6}]}^{2+},Y={[{\mathrm{CoCl}}_{4}]}^{2-},Z=\mathrm{Tetrahedral}$X=[Co(H2​O)6​]2+,Y=[CoCl4​]2−,Z=Tetrahedral