Answer the question on the basis of the passage given below:

There are some deposits of nitrates and phosphates in the earth's crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under laboratory conditions, but microbes do it easily. Ammonia forms a large number of complexes with transition metals. Hybridization easily explains the ease of sigma donation of \(NH_3\) and \(PH_3\).

Phosphine is an inflammable gas, it is prepared from white phosphorus.

\(PH_3\) is a better reducing agent than \(NH_3\) because:

Bond dissociation enthalpy of \(P-H\) is low

The correct answer is option 2. Bond dissociation enthalpy of \(P-H\) is low.

The bond dissociation enthalpy is the energy required to break a bond in a molecule, and it is an indicator of bond strength. In the case of \(PH_3\) (phosphine) and \(NH_3\) (ammonia), the relevant bonds are \(P-H\) and \(N-H\), respectively.

A lower bond dissociation enthalpy indicates that the bond is weaker and can be more easily broken. In this context:

The bond dissociation enthalpy of \(P-H\) in \(PH_3\) is lower than the bond dissociation enthalpy of \(N-H\) in \(NH_3\).

This lower bond dissociation enthalpy means that \(PH_3\) can more easily donate its hydrogen atom (and the associated electron) compared to \(NH_3\), making \(PH_3\) a better reducing agent. The ease with which \(PH_3\) can release its lone pair of electrons is related to the lower energy required to break the \(P-H\) bond.