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

Hint: We know that B.D.E. depends on a lot of factors . N-H molecules have hydrogen bonding so they are more stable than P-H. And as we know P-F shows back-bonding and hence is more stable than N-F.

Complete Step by step answer: B.D.E. here stands for bond dissociation enthalpy and it actually depends on a few factor like – the structure of the molecule, number of lone pair of electrons and so on. Bond Dissociation Enthalpy is the amount of energy needed to break down a particular bond .

We have been asked for the order of Bond Energy . Bond Energy refers to the average amount of energy needed to break down all the bonds that exist .
The order of bond energy will depend on the strength of individual bonds.

At first we have $N{H_3},P{H_3}$ . We will compare these two molecules. Nitrogen is one of the most electronegative elements. So when it reacts with hydrogen , it will form hydrogen bonds. Because of hydrogen bonding in ammonia molecules . Its bonds will have greater energy.
Hence , bond dissociation energy of N-H> bond dissociation energy of P-H.
Now , we will have to compare $N{F_3},P{F_3}$ .
Phosphorus has empty 3d orbitals which can accept electron pairs from fluorine forming back-bonds (though not very strong as phosphorus is large , backbonding becomes difficult).
Nitrogen and fluorine are small atoms so their nuclei repel more effectively than in case of phosphorus .
Hence, the bond dissociation enthalpy of P-F> the bond dissociation of N-H .

Hence, the correct options are A and D.

Note: By going through all the explanations it may be noted and concluded that the explanation of both is different from each other because various factors are dominant in various molecules and we need to find the most relevant ones.