2024 H2 Chem P1 Q4 — Chemical bonding Solution

The answer is D (x: \(109^\circ\), y: \(109^\circ\), z: \(120^\circ\)).
• Angle x (H-N-H) is centred on the \(\text{N}^+\) atom. N has 4 bond pairs and no lone pair, so the arrangement is tetrahedral and the bond angle is \(109^\circ\).
• Angle y (H-C-C) is centred on the \(\text{CH}_2\) carbon. This C has 4 bond pairs and no lone pair, so the bond angle is \(109^\circ\).
• Angle z (C-C-O) is centred on the carboxylate carbon. This C has 3 regions of electron density (C-C, C=O and C-\(\text{O}^-\)) and no lone pair, so it is trigonal planar and the bond angle is \(120^\circ\).
• A is wrong: neither x nor y is \(90^\circ\); both centres are tetrahedral.
• B is wrong: \(104^\circ\) belongs to a centre with 2 bond pairs and 2 lone pairs (as in water), y is not \(90^\circ\), and z is \(120^\circ\), not \(117^\circ\).
• C is wrong: x is at \(\text{N}^+\), which has no lone pair, so it is \(109^\circ\), not \(104^\circ\).