Use VSEPR to predict the geometry of these ions. (a) \(\mathrm{NH}_{2}{ }^{-}\) (b) \(\mathrm{NO}_{2}{ }^{-}\) (c) \(\mathrm{NO}_{2}^{+}\) (d) \(\mathrm{NO}_{3}^{-}\)

For each ion, we'll draw the Lewis structure, which will show us the arrangement of the atoms and the valence electrons involved in bonding. (a) \(\mathrm{NH}_{2}{ }^{-}\) 1. Nitrogen has 5 valence electrons, and each hydrogen atom contributes 1 valence electron. The negative charge means there is an extra electron, summing up to 8 valence electrons. 2. Arrange the atoms with nitrogen as the central atom, and connect each hydrogen atom to nitrogen with a single bond (each bond accounting for 2 electrons). 3. Distribute the remaining electrons as lone pairs around the nitrogen atom. In this case, nitrogen will have one lone pair. (b) \(\mathrm{NO}_{2}{ }^{-}\) 1. Nitrogen has 5 valence electrons, each oxygen atom contributes 6 valence electrons, and the negative charge adds 1 more electron, totaling 18 valence electrons. 2. Arrange the atoms with nitrogen as the central atom, and connect each oxygen atom to nitrogen with a single bond (each bond accounting for 2 electrons). 3. Distribute the remaining electrons as lone pairs around the oxygen atoms and then nitrogen to complete their octets. In this case, one oxygen atom will form a double bond with nitrogen. (c) \(\mathrm{NO}_{2}^{+}\) 1. Nitrogen has 5 valence electrons, each oxygen atom contributes 6 valence electrons, and the positive charge means 1 electron is missing, totaling 16 valence electrons. 2. Arrange the atoms with nitrogen as the central atom, and connect each oxygen atom to nitrogen (each bond accounting for 2 electrons). One of the oxygen atoms will form a double bond with nitrogen in this case. 3. Distribute the remaining electrons as lone pairs around the oxygen atoms to complete their octets. (d) \(\mathrm{NO}_{3}^{-}\) 1. Nitrogen has 5 valence electrons, each oxygen atom contributes 6 valence electrons, and the negative charge adds 1 more electron, totaling 24 valence electrons. 2. Arrange the atoms with nitrogen as the central atom, and connect each oxygen atom to nitrogen with a single bond (each bond accounting for 2 electrons). 3. Distribute the remaining electrons as lone pairs around the oxygen atoms and then nitrogen to complete their octets. In this case, one of the oxygen atoms will form a double bond with nitrogen (due to resonance structures).

Using the Lewis structures from Step 1 and VSEPR theory, we will find the electron pair geometry, which considers both bonding and lone pairs, and molecular geometry, which considers only the bonding pairs. (a) \(\mathrm{NH}_{2}{ }^{-}\) Electron pair geometry: trigonal planar (3 electron pairs around the central atom) Molecular geometry: bent (2 bonding pairs) (b) \(\mathrm{NO}_{2}{ }^{-}\) Electron pair geometry: trigonal planar (3 electron pairs around the central atom) Molecular geometry: bent (2 bonding pairs) (c) \(\mathrm{NO}_{2}^{+}\) Electron pair geometry: linear (2 electron pairs around the central atom) Molecular geometry: linear (2 bonding pairs) (d) \(\mathrm{NO}_{3}^{-}\) Electron pair geometry: trigonal planar (3 electron pairs around the central atom) Molecular geometry: trigonal planar (3 bonding pairs) So, the geometries of the ions are as follows: (a) \(\mathrm{NH}_{2}{ }^{-}\) has a bent geometry. (b) \(\mathrm{NO}_{2}{ }^{-}\) has a bent geometry. (c) \(\mathrm{NO}_{2}^{+}\) has a linear geometry. (d) \(\mathrm{NO}_{3}^{-}\) has a trigonal planar geometry.