Question

The orbitals involved in case of $s{p}^3{d}^2$ hybridisation is

• A. $s+p_x+p_y+d_{xy}+p_z+d_{z^2}$
• B. $s+p_x+p_y+d_{z^2}+p_z+d_{yz}$
• C. $s+p_x+p_y+p_z+d_{z^2}+d_{x^2-y^2}$
• D. $s+p_x+p_y+p_z+d_{yz}+d_{xz}$

Answer 1

The correct option is C $s+p_x+p_y+p_z+d_{z^2}+d_{x^2-y^2}$

$s{p}^3{d}^2$ is hybridisation for octahedral geometry.
In octahedron, bonds are formed along three mutually perpendicular directions (like coordinate axes $x,y$ and $z$). Hence, $d_{z^2}$ and $d_{x^2-y^2}$ are involved to form hybrid orbitals due to their alignment along axes similar to $p$-subshell orbitals.