Consider the following reaction: MX_4 + X'_2 | vedclass

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(C) The central atom $M$ is Tellurium ($Te$,atomic number $52$).In $MX_4$ (e.g.,$TeCl_4$),the hybridization of $Te$ is $sp^3d$,which involves the $d_{

Vedclass · Accepted Answer

Central atom $M$ does not use any valence non-axial set of $d-$ orbital in hybridization of final product.

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(C) The central atom $M$ is Tellurium ($Te$,atomic number $52$).In $MX_4$ (e.g.,$TeCl_4$),the hybridization of $Te$ is $sp^3d$,which involves the $d_{z^2}$ orbital. The structure is a see-saw shape with one lone pair.In $MX_4X'_2$ (e.g.,$TeCl_4F_2$),the hybridization of $Te$ is $sp^3d^2$,which involves $d_{x^2-y^2}$ and $d_{z^2}$ orbitals.The axial positions in the octahedral geometry of $MX_4X'_2$ are formed by hybrid orbitals $(sp^3d^2)$.Since $X'$ is more electronegative,it occupies the axial positions to minimize repulsion.In $sp^3d^2$ hybridization,the $d-$ orbitals used are $d_{x^2-y^2}$ and $d_{z^2}$. Both are axial $d-$ orbitals. Therefore,the central atom does not use any non-axial $d-$ orbitals (like $d_{xy}, d_{yz}, d_{zx}$) in the hybridization of the final product.Thus,statement $C$ is correct.

Consider the following reaction: $MX_4 + X'_2 \to MX_4X'_2$. If the atomic number of $M$ is $52$ and $X$ and $X'$ are halogens and $X'$ is more electronegative than $X$. Then choose the correct statement regarding the given information.

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