Question

What hybridization is required for central atom that have a tetrahedral arrangement of electron pairs? A trigonal planar arrangement of electron pairs? A linear arrangement of electron pairs? How many unhybridized p atomic orbitals are present when a central atom exhibits tetrahedral geometry? Trigonal planar geometry? Linear geometry? What are the unhybridized p atomic orbitals used for.

Short answer

Hybridization is the mixing of pure atomic orbitals of nearly same energy to form hybrid orbitals. The table below shows the hybridization corresponding to a particular geometry.

Geometry	Hybridization	Number of unhybridized p orbitals
Tetrahedral	$s{p}^3$	0
Trigonal Planar	$s{p}^2$	1
Linear	$sp$	2

Step-by-step solution

Step 1

• Tetrahedral geometry is obtained when hybridization is $s{p}^3$.
• Trigonal planar arrangement is obtained when hybridization is $s{p}^2$.
• A linear arrangement is obtained when hybridization is sp.

Step 2

• No p atomic orbital is unhybridized when central atom has tetrahedral geometry.
• One p orbital is unhybridized when there is trigonal planar geometry.
• Two p orbitals are unhybridized when there is linear geometry.

Step 3

The unhybridized p orbital helps in forming the pi bond.

For example;

• Methane$\left({\text{CH}}_{\text{4}}\right)$ is a tetrahedral molecule having ${\text{sp}}^{\text{3}}$hybridization. It has no pi bond because all the p orbitals are hybridized.
• Ethene$\left({\text{CH}}_{\text{2}}\text{=} {\text{CH}}_{\text{2}}\right)$ is a trigonal planar molecule having ${\text{sp}}^{\text{2}}$ hybridization. It has one pi bond because one p orbital is unhybridized.
• Ethyne $\left(\text{CH}\equiv \text{CH}\right)$is a linear molecule having sp hybridization. It has two pi bond because two p orbitals are unhybridized.