Question

Bond dissociation enthalpy of E-H (E = element) bonds is given below. Which of the compounds will act as the strongest reducing agent?
Compound and its ${{\Delta }_{diss\text{ (E-H}{{\text{)}}^{H}}\text{ (KJ/mol)}}}$ are given below:
$N{{H}_{3}}-398$,
$P{{H}_{3}}-322$,
$As{{H}_{3}}-297$,
$Sb{{H}_{3}}-255$

(a)- $N{{H}_{3}}$
(b)- $P{{H}_{3}}$
(c)- $As{{H}_{3}}$
(d)- $Sb{{H}_{3}}$

Answer 1

Hint: Bond dissociation enthalpy is the energy to break the bond in the molecule. The lesser the bond dissociation enthalpy of the molecule, the easier is to break the bond of the molecule. The reducing character of the molecule is measured by the ease of release of the hydrogen atom.

Complete step by step answer:
Bond dissociation enthalpy is the amount of energy required to dissociate one mole of bonds presents between the atoms in the gaseous molecules.
The reducing character of the molecule is decided by the ease of removal of hydrogen ions. This means that it could easily donate the proton and make the other molecule to be reduced.
So, the bond dissociation energy is inversely proportional to the reducing character of the molecule. As the bond dissociation decreases the reducing character of the molecule increases.
So, in the question the reducing character of the molecules are given as:
The bond dissociation of $N{{H}_{3}}$ (ammonia) = 389
The bond dissociation of $P{{H}_{3}}$ (phosphine) = 322
The bond dissociation of $As{{H}_{3}}$ (arsine) = 297
The bond dissociation of $Sb{{H}_{3}}$ (stibine) = 255

So, it can be seen that the bond dissociation of stibine is the least and the bond dissociation of ammonia is the most.
Hence the stibine will easily donate the proton and will act as the strongest reducing agent.
So, the correct answer is “Option D”.

Note: Ammonia, phosphine, arsine, and stibine belong to the same group. We can also say that while moving down the group the stability of the hydrides decreases hence the reducing character increases.