Question

The bond angle formed by different hybrid orbitals are in the order

• A. $s{p}^2>s{p}^3>sp$
• B. $s{p}^3<s{p}^2>sp$
• C. $s{p}^3>s{p}^2>sp$
• D. $sp>s{p}^2>s{p}^3$

Answer 1

Answer: (D)

As the s-character increases, electronegativity increases and thus, the bond angle increases.
The order of s-character in different hybrid orbitals is
$\underset{50}{\mathrm{sp}}>\underset{33.3}{s{p}^2}>\underset{25}{s{p}^3}$
s-character Thus, order of bond angle is
$\underset{\left(180^{\circ }\right)}{\mathrm{sp}}>\underset{\left(120^{\circ }\right)}{s{p}^2}>\underset{\left(109.5^{\circ }\right)}{s{p}^3}$