Q.
For the following pattern of hybridization shown by the central atom,
$sp$
$sp^2$
$sp^3$
$sp^3d$
(1)
(2)
(3)
(4)
which of the following options represent the correct sequence of hybridisation, i.e. according to the sequence mentioned above?
| $sp$ | $sp^2$ | $sp^3$ | $sp^3d$ |
|---|---|---|---|
| (1) | (2) | (3) | (4) |
NTA AbhyasNTA Abhyas 2020Chemical Bonding and Molecular Structure
Solution:
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 $+ lp$
$CO _{2}$
$sp$
(number of $\sigma$ bonds $=0$ number of $lp$ of electrons $=0$ )
$BF _{3}$
$sp^2$
(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 $=2)$
$PCl _{5}$
$sp^3 d$
(number of $\sigma$ bonds $=5$ number of $lp$ of electrons $=0$ )
| Molecule | Hybridisation | Number of $\{\sigma\}$ bond $+ lp$ |
|---|---|---|
| $CO _{2}$ | $sp$ | (number of $\sigma$ bonds $=0$ number of $lp$ of electrons $=0$ ) |
| $BF _{3}$ | $sp^2$ | (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 $=2)$ |
| $PCl _{5}$ | $sp^3 d$ | (number of $\sigma$ bonds $=5$ number of $lp$ of electrons $=0$ ) |