Determine the shape or geometry and bond angle of PH3.

Question

BlackTom AI · Accepted Answer

Phosphorus has five valence electrons, so in PH3 (three H atoms) the valence electron count is 5 + 3×1 = 8 electrons, which form four electron pairs around P. Placing those pairs: three are used in P–H bonding and one remains as a lone pair. That gives an AX3E1 VSEPR situation (three bonding pairs, one lone pair).

Step-by-step reasoning:
1. Count valence electrons: P = 5, 3 H = 3, total = 8 e− → 4 electron pairs.
2. Assign pairs: three bonding pairs (P–H) + one lone pair on P.
3. Electron-domain geometry for four electron pairs is tetrahedral (electron-domain geometry depends on number of domains, not just bonds).
4. Molecular geometry (shape) considers only atom positions: with three bonds and one lone pair, the shape is trigonal pyramidal (lone pair occupies one vertex of the tetrahedron, pushing the three H atoms into a pyramid).
5. Bond angle: the ideal tetrahedral angle is 109.5°. Lone pairs occupy more space than bonding pairs, increasing lone pair–bond pair repulsion and compressing bond angles. Thus H–P–H bond angles are less than 109.5°. (In fact PH3’s experimental H–P–H angle is significantly less than 109.5°, ~93–94°, but the important point here is that it is less than 109.5°.)

Why other options are incorrect:
- “Tetrahedral, 109.5°” is the electron-domain geometry, not the molecular geometry for AX3E1; PH3 is not a full tetrahedral arrangement of atoms.
- “Bent, <120°” or “trigonal planar, 120°” correspond to different numbers of electron domains (AX2E1 or AX3 respectively) and do not match PH3’s three bonded atoms plus one lone pair.
- The duplicate “bent, <120°” is likewise incorrect for the same reason.

Therefore the correct description is trigonal pyramidal, with bond angles < 109.5°, matching the provided answer.

Determine the shape or geometry and bond angle of PH3. 单项选择题

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