Ammonia, \(\mathrm{NH}_{3}\), reacts with incredibly strong bases to produce the amide ion, \(\mathrm{NH}_{2}\) -. Ammonia can also react with acids to produce the ammonium ion, \(\mathrm{NH}_{4}^{+}\). (a) Which species (amide ion, ammonia, or ammonium ion) has the largest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle? (b) Which species has the smallest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle?

Let's start determining the number of electron pairs around the nitrogen atom for each species: 1. Ammonia (NH₃): Nitrogen has 5 valence electrons, and it is bonded to 3 hydrogen atoms. So, there is one lone pair and three bonding pairs, which gives a total of four electron pairs. 2. Amide ion (NH₂⁻): Nitrogen has 5 valence electrons like in ammonia, but it gains one extra electron due to the negative charge. In this species, nitrogen has two lone pairs and two bonding pairs, totaling four electron pairs. 3. Ammonium ion (NH₄⁺): Nitrogen has 5 valence electrons but loses one electron as it has a positive charge. It is bonded to 4 hydrogen atoms, and thus, it has 0 lone pairs and four bonding pairs, totaling four electron pairs. For each of the species mentioned above, there are four electron pairs around the nitrogen atom. Therefore, the electron pair geometry for each one of them is tetrahedral.

Using the VSEPR theory, we can determine the molecular shape of the species: 1. Ammonia (NH₃): It has one lone pair and three bonding pairs. The molecular shape is trigonal pyramidal. 2. Amide ion (NH₂⁻): It has two lone pairs and two bonding pairs. The molecular shape is bent or V-shaped. 3. Ammonium ion (NH₄⁺): It has no lone pair and four bonding pairs, and that gives it a tetrahedral molecular shape with four equivalent H-N-H bond angles.

Now, based on the molecular shapes, we can predict the H-N-H bond angles: 1. Ammonia (NH₃): In a trigonal pyramidal molecular shape, the H-N-H bond angles are typically 107°. 2. Amide ion (NH₂⁻): In a bent or V-shaped molecular shape with two lone pairs, the H-N-H bond angles are typically smaller than 107°, ranging from 104.5° to close to 109.5° for different molecules. 3. Ammonium ion (NH₄⁺): In a tetrahedral molecular shape, the H-N-H bond angles are precisely 109.5°. Now we can answer the original questions: (a) Ammonium ion, NH₄⁺, has the largest H-N-H bond angle (109.5°). (b) Amide ion, NH₂⁻, has the smallest H-N-H bond angle (typically smaller than 107°).