Phosphorous hydride's bond angle is closer to 90° while the bond angle of ammonium hydride is closer to 104.5°. Among the following, which one explains this structural feature correctly?
Correct answer: C. Due to quite high energy difference between 3s and 3p orbitals, the lone pair on phosphorous prefers to occupy unhybridized s-orbital rather than hybridized sp3 hybridized orbital which causes its s-orbital energy to increase.
- A. Nitrogen bond pair electron cloud concentrates near the central atom because of its higher electronegativity, thus the bond pair - bond pair repulsion increases which in turn decreases the bond angle in NH3.
- B. Due to larger size of the lone pair electron cloud, there is larger lone pair - bond pair repulsion in PH3 compared to NH3.
- C. Due to quite high energy difference between 3s and 3p orbitals, the lone pair on phosphorous prefers to occupy unhybridized s-orbital rather than hybridized sp3 hybridized orbital which causes its s-orbital energy to increase.
- D. Phosphorous forms pπ - dπ bonds while nitrogen doesn't.
Explanation
The energy difference between 3s and 3p orbitals is quite high in group 15 and group 16 hydrides (except for NH3 and H2O). The orbital energy of 3s increases so much by hybridization so lone pair tends to occupy an unhybridized s orbital. . For example in PH3, 600 kJ mol-1 of energy is required to hybridize the central atom. In order to avoid such energy demanding hybridization, by leaving the lone pair in the spherical s orbital, P forms bonds with unhybridized p orbitals. Thus, the bond angle in PH3 is closer to 90° while that in NH3 is 104.5°.
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