Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the (a) \(\mathrm{NH}_{4}^{+},(\mathbf{b}) \mathrm{NO}_{2}^{-}\), nitrogen: (c) \(\mathrm{N}_{2} \mathrm{O},(\mathrm{d}) \mathrm{NO}_{2}\)

For each species, we need to calculate the total number of valence electrons. Valence electrons are found in the outermost shell of each element, and their number can be determined from the periodic table. (a) NH4+ Valence electrons for N: 5 Valence electrons for 4 H atoms: 4(1) = 4 Since it has a positive charge, subtract 1 electron: Total valence electrons: 5+4 - 1 = 8 (b) NO2- Valence electrons for N: 5 Valence electrons for 2 O atoms: 2(6) = 12 Since it has a negative charge, add 1 electron: Total valence electrons: 5+12 + 1 = 18 (c) N2O Valence electrons for 2 N atoms: 2(5) = 10 Valence electrons for O: 6 Total valence electrons: 10 + 6 = 16 (d) NO2 Valence electrons for N: 5 Valence electrons for 2 O atoms: 2(6) = 12 Total valence electrons: 5 + 12 = 17

Now we will distribute the valence electrons to each atom in the species, connecting them with a single bond. (a) NH4+: H - N - H has a trigonal pyramidal shape with 3 single bonds and a lone pair on nitrogen. \[ \mathrm{H} - \mathrm{N}^{-} - \mathrm{H} \] \[ \mid \mathrm{H} \] (b) NO2-: O = N - O has a bent shape with a single bond between N and one O atom and a double bond between N and the other O atom. There is also a lone pair on nitrogen. \[ \mathrm{O} = \mathrm{N}^+ - \mathrm{O}^- \] (c) N2O: N = N = O has a linear shape with a triple bond between the two N atoms and a double bond between the N and the O atom. \[ \mathrm{N} = \mathrm{N} = \mathrm{O} \] (d) NO2: O = N - O has a bent shape with a single bond between N and one O atom, a double bond between N and the other O atom, and an unpaired electron on nitrogen. \[ \mathrm{N}^+ - \mathrm{O}^{-} \] \[ \mathrm{O} = \mathrm{N} \cdot \]

Now that we have the Lewis structures, we can determine the shape of each molecule and the oxidation state of nitrogen. (a) NH4+ Shape: Tetrahedral Oxidation state of N: +5 (each H contributes -1, so +5 is needed to offset) (b) NO2- Shape: Bent Oxidation state of N: +3 (each O contributes -2, so +3 is needed to offset) (c) N2O Shape: Linear Oxidation state of the left N: +1 (N triple-bonded to N contributes -3, so +1 is needed to offset) Oxidation state of the right N: +4 (triple-bonded N has an oxidation state of +4) (d) NO2 Shape: Bent Oxidation state of N: +4 (each O contributes -2, so +4 is needed to offset)