Would you expect the nonbonding electron-pair domain in \(\mathrm{NH}_{3}\) to be greater or less in size than the corresponding one in \(\mathrm{PH}_{3}\) ?

First, we need to draw the Lewis structure for both NH₃ and PH₃ molecules. In NH₃, nitrogen (N) has 5 valence electrons and hydrogen (H) has 1 valence electron, making a total of 8 valence electrons. In the Lewis structure, nitrogen is the central atom and needs to form three single bonds with three hydrogen atoms, leaving one lone pair of electrons (nonbonding electron-pair domain) on nitrogen. Similarly, in PH₃, phosphorus (P) has 5 valence electrons and hydrogen (H) has 1 valence electron, making a total of 8 valence electrons. In the Lewis structure, phosphorus is the central atom and needs to form three single bonds with three hydrogen atoms, leaving one lone pair of electrons (nonbonding electron-pair domain) on phosphorus.

Nonbonding electron-pair domains are regions where the electron pair is not involved in any bond formation. The size of the nonbonding electron-pair domain depends on the electronegativity of the central atom and the size of its atomic orbitals. In our case, the electronegativity of nitrogen (3.04) is greater than that of phosphorus (2.19). So, the nitrogen in NH₃ attracts the electron pair more effectively, and consequently, the nonbonding electron-pair domain is more contracted and smaller in size as compared to the one in PH₃. On the other hand, the atomic orbitals of phosphorus are larger than that of nitrogen due to its increased size and the presence of more shells. Thus, the electron cloud is more dispersed for PH₃ and the size of the nonbonding electron-pair domain is greater for PH₃ than NH₃.

After comparing the electronegativity and size of atomic orbitals, we can conclude that the nonbonding electron-pair domain in NH₃ will be smaller in size than the corresponding one in PH₃ due to the greater electronegativity and smaller atomic orbitals of nitrogen.