Question

Consider the following compounds with their regular geometries.

Compounds	Geometry
$A$	Square planar
$B$	Trigonal bipyramidal
$C$	Square pyramidal
$D$	Octahedral
$E$	Pentagonal bipyramidal

Calculate value of $(x+y-z{)}^2$, where $x$ is total number of axial d-orbitals having zero nodal plane, $y$ and $z$ are total no. of non-axial and axial d-orbitals respectively each one having two nodal planes used in hybridization of central atoms of compounds $A$ to $E$.

Answer 1

Compounds	Geometry	Hybridization	Used d-orbital
$A$	Square planar	$ds{p}^2$	$d_{x^2-y^2}$
$B$	Trigonal bipyramidal	$s{p}^{3}d$	$d_{z^2}$
$C$	Square pyramidal	$s{p}^{3}d$	$d_{x^2-y^2}$
$D$	Octahedral	$s{p}^3{d}^2$	$d_{x^2-y^2,d_{z^2}}$
$E$	Pentagaonal bipyramidal	$s{p}^3{d}^3$	$d_{x^2-y^2},d_{z^2},d_{xy}$

$x=d_{z^2}=3$
$y=d_{xy}=1$z = d_{x^{2} - y^{2}} = 4$.

So, (x + y - z)^{2} = (3 + 1 - 4)^{2} = 0$