(a) Which geometry and central atom hybridization would you expect in the series $\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}{ }^{+} ?(\mathbf{b})$ What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

We start by determining the number of electron groups around the central atom for each molecule: \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}{ }^{+}\). Electron groups include lone pairs of electrons and bonds (single, double, or triple). - \(\mathrm{BH}_{4}^{-}\): Boron has 3 valence electrons, and it has 4 single bonds with 4 Hydrogen atoms. Therefore, it has 4 electron groups in total. - \(\mathrm{CH}_{4}\): Carbon has 4 valence electrons, and it has 4 single bonds with 4 Hydrogen atoms. Therefore, it has 4 electron groups in total. - \(\mathrm{NH}_{4}{ }^{+}\): Nitrogen has 5 valence electrons, but due to the positive charge, it loses 1 electron. It has 4 single bonds with 4 Hydrogen atoms. Therefore, it has 4 electron groups in total.

Since all three molecules have 4 electron groups around the central atom with no lone pairs, they have a tetrahedral molecular geometry. For a tetrahedral geometry, the central atom's hybridization is sp3. Therefore, the hybridization in \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4},\) and \(\mathrm{NH}_{4}{ }^{+}\) is sp3.

To determine the bond dipoles, we need to examine the electronegativity difference between the central atom and the surrounding atoms in each molecule. In general, as we go from B to N in the periodic table, the electronegativity increases. In all three molecules, Hydrogen is bonded to the central atom. Thus, the bond dipoles have the following trends: - For \(\mathrm{BH}_{4}^{-}\), since B has a lower electronegativity than H, the bond dipoles will be directed from B to H. - For \(\mathrm{CH}_{4}\), the electronegativity of C and H is almost equal; thus, the bond dipoles are small and there is no net dipole moment. - For \(\mathrm{NH}_{4}{ }^{+}\), N has a higher electronegativity than H, so the bond dipoles will be directed from H to N.

The analogous species of the elements in Period 3 for \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4},\) and \(\mathrm{NH}_{4}{ }^{+}\) are: - \(\mathrm{AlH}_{4}^{-}\) - \(\mathrm{SiH}_{4}\) - \(\mathrm{PH}_{4}{ }^{+}\)

For each Period 3 species, the central atom has similar electron groups as the corresponding species in Period 2. Thus, we can expect the hybridization of each central atom to be the same: - The hybridization at the central atom Aluminum in \(\mathrm{AlH}_{4}^{-}\) is sp3. - The hybridization at the central atom Silicon in \(\mathrm{SiH}_{4}\) is sp3. - The hybridization at the central atom Phosphorus in \(\mathrm{PH}_{4}{ }^{+}\) is sp3.