Comparison of Electron Donating Ability of Ammonia and Phosphine

To determine the relative electron-donating ability of ammonia ($\text{NH}_3$) and phosphine ($\text{PH}_3$), we consider their electronic structure and hybridization:

1. Hybridization and Orbital Character in $\text{NH}_3$:
   In ammonia ($\text{NH}_3$), the nitrogen atom is $sp^3$ hybridized. The bond angle of $\text{H}-\text{N}-\text{H}$ is approximately $107.8^\circ$, which is close to the ideal tetrahedral angle ($109.5^\circ$). Consequently, the lone pair of electrons on the nitrogen atom resides in an $sp^3$ hybrid orbital. An $sp^3$ hybrid orbital is strongly directional and oriented outward in space, making the lone pair readily accessible for donation to an electron acceptor (such as a proton or a transition metal ion).

2. Orbital Character in $\text{PH}_3$ (Drago's Rule):
   In phosphine ($\text{PH}_3$), the bond angle of $\text{H}-\text{P}-\text{H}$ is around $93.5^\circ$. According to Drago's rule, elements of the third period and below with central atoms having lower electronegativity do not undergo significant hybridization when bonded to less electronegative terminal atoms like hydrogen. Thus, the $\text{P}-\text{H}$ bonds are formed predominantly by almost pure $3p$ orbitals, and the non-bonding lone pair of electrons resides primarily in an almost pure, spherical $3s$ orbital.

3. Comparison of Donor Ability:

   • Electrons in a spherical $s$-orbital have greater penetrating power, are held more tightly by the phosphorus nucleus, and lack directional character, rendering them poor donors for $\sigma$-bond formation.
   • In contrast, the lone pair in $\text{NH}_3$ occupies a directional $sp^3$ hybrid orbital, allowing for effective orbital overlap with an acceptor's empty orbital.

Therefore, between $\text{NH}_3$ and $\text{PH}_3$, $\text{NH}_3$ is a better electron donor because its lone pair of electrons occupies an $sp^3$ orbital and is more directional.

Thus, the correct option is C.