Question

Select the correct order of bond energy for $P{H}_3,N{H}_3$ and $N{F}_3,P{F}_3$.

• A. $B.D.E_{N-H}>B.D.E_{P-H}$
• B. $B.D.E_{N-H}<B.D.E_{P-H}$
• C. $B.D.E_{N-F}>B.D.E_{P-F}$
• D. $B.D.E_{N-F}<B.D.E_{P-F}$

Answer 1

Answer: (A), (D)

The correct options are
A $B.D.E_{N-H}>B.D.E_{P-H}$
D $B.D.E_{N-F}<B.D.E_{P-F}$

In ${NH}_3$, there are 4 electron pairs(3 bonding pairs and 1 lone pair) in the outer most shell of N. The repulsion between lone pair and a bond pair of electrons always exceeds to that of two bond pairs. Thus the bond angles are reduced from 109degree. 27' to 107degree.48'.

Considering the PH, would be expected to be similar. But, the bond pairs of electrons are much further away from the central atom due to its larger size than they are in ${NH}_3$. Thus the lone pair causes even greater distortion in ${PH}_3$. So, the bond angle decreases to 91degree.18'

Thus, the bond angle of ${PH}_3$ molecule is lesser than that in ${NH}_3$ molecule.

The NF3 bond angle is 102degree. There is more distortion than for NH3 because the single bonds are taking up less room, close to the nitrogen. Fluorine is more electronegative than hydrogen and the electron density in the N-F bond is skewed towards the fluorine.
The PF3 bond angle will be about 109degree since it has a trigonal pyramidal molecular geometry.

Thus the bond angle of ${Pf}_3$ molecule is greater than that in ${NF}_3$ molecule