question_answer

Consider the following reaction: \[M{{X}_{4}}+X{{'}_{2}}\xrightarrow{{}}M{{X}_{4}}X{{'}_{2}}\]

If atomic number of M is 52 and X and X' are halogens and X' is more electronegative than X. Then choose correct statement regarding given information:

A) Both X' atoms occupy axial positions which are formed by overlapping of p and \[d\]-orbitals only

B) All M - X bond lengths are identical in both \[{{\operatorname{MX}}_{4}}\] and \[{{\operatorname{MX}}_{4}}X'{{  }_{2}}\] compounds

C) Central atom 'M' does not use non-axial set of \[d\]-orbital in hybridization of final product.

D) Hybridization of central atom 'M' remains same in both reactant and final product.

Correct Answer: C

Solution :

\[{{\operatorname{MX}}_{4}}+{{X}_{2}}\xrightarrow{{}}M{{X}_{4}}X_{2}^{'}(M=Te)\]

Hybridization of \[\operatorname{Te}:s{{p}^{3}}d\]

In the above compound orbitals used hybridization of Te are oxial t.e\[s,{{p}_{x}}^{,}{{p}_{y}},{{p}_{z}}^{,}\]

\[{{d}_{{{x}^{2}}-{{y}^{2}}}},{{d}_{{{z}^{2}}}}\]

All M?X bond lengths are identical.

Now,

Hybridization of Te: \[s{{p}^{3}}{{d}^{2}}\]

In the above comopound orbitals used in the hybridization of Te are \[s,{{p}_{x}},{{p}_{y}},{{p}_{z}},{{d}_{{{z}^{2}}}}.\]

All M?X bond lengths are not identical,