Question

Order of the bond strength of $C-H$ bonds involving sp, $s{p}^2$ and $s{p}^3$ hybridized carbon atoms is:

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

Answer 1

The correct option is A $sp>s{p}^2>s{p}^3$

$\equiv \stackrel{sp}{C}-H$ $=\stackrel{s{p}^2}{CH}-H$ $-\stackrel{s{p}^3}{C{H}_2}-H$

In $sp,s{p}^2$ and $s{p}^3$ hybridized central carbon atoms, the percentage of $s$ character in bonding orbitals (i.e. $C-H$ bonds) are $50$%, $33.3$% and $25$% respectively. We know that percentage of $s$ character increases the bond strength because the shape of $s$ orbitals is spherical and it is more closer to nucleus of an atom. Therefore order of bond strength of $C-H$ bonds will be:

$sp>s{p}^2>s{p}^3$