Question

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

• A. $s p^{2} > s p^{3} > s p$
• B. $s p^{3}< s p^{2} > s p$
• C. $ sp^{3}>sp^ {2}>sp $
• D. $ sp>sp^{2}>sp^{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
$\underset{50}{s p} > \underset{33.3}{s p^{2}} > \underset{25}{s p^{3}}$
s-character Thus, order of bond angle is
$\underset{\left(180^{\circ}\right)}{s p}>\underset{\left(120^{\circ}\right)}{s p^{2}}>\underset{\left(109.5^{\circ}\right)}{s p^{3}}$