• # 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.

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