Question

Identify the correct descending order of strengths of bonds formed by the overlapping of the respective orbitals.

• A. $s-s>s-p>p-p\sigma >p-p\pi$
• B. $p-p\sigma >s-p>s-s>p-p\pi$
• C. $s-p>s-s>p-p\sigma >p-p\pi$
• D. $s-s>p-p\sigma >s-p>p-p\pi$

Answer 1

The correct option is A $s-s>s-p>p-p\sigma >p-p\pi$

Smaller bonds are stronger bonds. Then sigma bonds are stronger than pi bonds as sigma bonds are formed by head on overlapping which is more effective than sideways overlapping of pi bonds. Sigma bonds are further classified as $s-s,s-p,p-p,p-pr,p\pi ,$ and so on.

Bond strength $s-s>s-p>p-p$