Question

For the following pattern of hybridisation shown by the central atom,
$\begin{array}{c}sp,\\ \left(1\right)\end{array}\begin{array}{c}{sp}^2,\\ \left(2\right)\end{array}\begin{array}{c}{sp}^3,\\ \left(3\right)\end{array}\begin{array}{c}{sp}^{3}d\\ \left(4\right)\end{array}$
which of the following will represent the correct sequence of hybridisation, i.e., according to the sequence mentioned above?

• A. $H_{2}O,C{O}_2,B{F}_3,P{Cl}_5$
• B. $C{O}_2,H_{2}O,B{F}_3,P{Cl}_5$
• C. $C{O}_2,B{F}_3,H_{2}O,P{Cl}_5$
• D. $H_{2}O,C{O}_2,P{Cl}_6,B{F}_3$

Answer 1

The correct option is C $C{O}_2,B{F}_3,H_{2}O,P{Cl}_5$

The orbitals of the central atom that goes for hybridisation are:
(a) orbitals used for a $\sigma$ bond formation.
(b) orbitals contain $lp$ (lone pair) of electrons.
(c) if required empty orbitals.
Note:- Orbitals involved in $\pi$-bond formation and having odd electron will not get hybrid.
Thus,

Molecule	Hybridisation	Number of $\sigma$ bonds $+lp$
$C{O}_2$	$sp$	(number of $\sigma$ bonds $=2$, number of $lp$ of electrons $=0$)
$B{F}_3$	${sp}^2$	(number of $\sigma$ bonds $=3$, number of $lp$ of electrons $=0$)
$H_{2}O$	${sp}^3$	(number of $\sigma$ bonds $=2$, number of $lp$ of electrons $=2$)
$P{Cl}_5$	${sp}^{3}d$	(number of $\sigma$ bonds $=5$, number of $lp$ of electrons $=0$)