Question

In the valence bond theory, hybridization of orbitals is an integral part of the bond formation. Hybridisation consists of mixing or a linear combination of the "pure" atomic orbitals in such a way as to form new hybrid orbitals such as $sp,s{p}^2,s{p}^3,s{p}^{3}d,s{p}^3{d}^3$, etc.

Which of the following atomic orbitals does not participate in trigonal bipyramidal hybridization i.e. $s{p}^{3}d$ hybridization?

• A. $s$
• B. $d_{x^2-y^2}$
• C. $p_x$
• D. $p_y$

Answer 1

The correct option is B $d_{x^2-y^2}$

Hybridisation	Atomic orbitals	Bond angle(s)	Geometry
$sp$	$s+$ arbitrary $p^n$	${180}^o$	Linear
$s{p}^2$	$s+$ arbitrary $p^n$	${120}^o$	Trigonal
$s{p}^3$	$s+$ arbitrary $p^n$	${109.5}^o$	Tetrahedral
$s{p}^{3}d$	$d_{z^2}+s+p_x+p_y+p_z$	${90}^o,{120}^o$	Trigonal bipyramidal
$s{p}^3{d}^2$	$d_{z^2}+d_{x^2-y^2}+s+p_x+p_y+p_z$	${90}^o$	Octahedral

Hence, we can see that the orbital, $d_{x^2-y^2}$ does not participate in trigonal bipyramidal hybridisation i.e. $s{p}^{3}d$ hybridisation.

Here, orbital involved is $d_{z^2}$.