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(A) The atomic number of $Ti$ is $22$,and its electronic configuration is $[Ar] 3d^2 4s^2$. In both $TiCl_3$ and $[Ti(H_2O)_6]Cl_3$,the oxidation stat

Vedclass · Accepted Answer

$X =$ Colourless,$Y =$ Coloured

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(A) The atomic number of $Ti$ is $22$,and its electronic configuration is $[Ar] 3d^2 4s^2$. In both $TiCl_3$ and $[Ti(H_2O)_6]Cl_3$,the oxidation state of $Ti$ is $+3$,which corresponds to a $3d^1$ configuration.In the complex $[Ti(H_2O)_6]^{3+}$,the presence of ligands $(H_2O)$ causes crystal field splitting of the $d$-orbitals into $t_{2g}$ and $e_g$ levels. The single electron can undergo a $d-d$ transition by absorbing light,making the complex coloured (violet).In the anhydrous compound $TiCl_3$,there are no ligands to cause crystal field splitting. Without splitting,$d-d$ transitions cannot occur,rendering the substance colourless.Therefore,$X$ is colourless and $Y$ is coloured.

Which of the following is correct related to the colours of $TiCl_3$ $(X)$ and $[Ti(H_2O)_6]Cl_3$ $(Y)$?

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