Question

For the following pattern of hybridization shown by the central atom,

$sp$	$s{p}^2$	$s{p}^3$	$s{p}^{3}d$
(1)	(2)	(3)	(4)

which of the following options represent the correct sequence of hybridisation, i.e. according to the sequence mentioned above?

• A. $H_{2}O,C{O}_2,B{F}_3,PC{l}_5$
• B. $C{O}_2,H_{2}O,B{F}_3,PC{l}_5$
• C. $C{O}_2,B{F}_3,H_{2}O,PC{l}_5$
• D. $H_{2}O,C{O}_2,PC{l}_5,B{F}_3$

Answer 1

Answer: (C)

The orbitals of the central atom that goes for hybridization are:
(a) Orbitals used for a $\sigma$ bond formation
(b) Orbitals that 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 $\{\sigma \}$ bond $+lp$
$C{O}_2$	$sp$	(number of $\sigma$ bonds $=0$ number of $lp$ of electrons $=0$ )
$B{F}_3$	$s{p}^2$	(number of $\sigma$ bonds $=3$ number of $lp$ of electrons $=0$ )
$H_{2}O$	$s{p}^3$	(number of $\sigma$ bonds $=2$ number of $lp$ of electrons $=2)$
$PC{l}_5$	$s{p}^{3}d$	(number of $\sigma$ bonds $=5$ number of $lp$ of electrons $=0$ )