Question

Which is correct statement?
As the $s$-character of a hybrid orbital decreases
(I) The bond angle decreases
(II) The bond strength increases
(III) The bond length increases
(IV) Size of orbitals increases

• A. (I), (III) and (IV)
• B. (II), (III) and (IV)
• C. (I) and (II)
• D. All are correct

Answer 1

The correct option is A (I), (III) and (IV)

s-character is the contribution of sigma type bond in a hybridization: $s{p}^3$=25% s-character, 75% p-character $s{p}^2$=33% s-character, 66% p-character $sp$=50% p-character. The more s-character a bond has, the stronger and shorter the bond is. Hence the bond length decrease with increase in s character. An sp-sp bond is strongest and $s{p}^3-s{p}^3$ bond is weakest.

The bond angle of $s{p}^3$ is 109.5, $s{p}^2$ is 120 and $sp$ is 180. An sp orbital is half s character, $s{p}^2$ is 1/3 s character and $s{p}^3$ is 1/4 s character, so increasing the s character corresponds to increasing the bond angle.

The size of the orbital depends upon the value of principal quantum number(n). Greater the value of n, larger is the size of the orbital and lesser the s-character