Predict the geometry of the following molecules and ion using the VSEPR model: (a) \(\mathrm{CBr}_{4},\) (b) \(\mathrm{BCl}_{3}\) (c) \(\mathrm{NF}_{3}\), (d) \(\mathrm{H}_{2} \mathrm{Se},\) (e) \(\mathrm{NO}_{2}^{-}\).

For molecule (a) \(\mathrm{CBr}_{4}\), Carbon (C) is the central atom. Carbon has four valence electrons and each Br atom donates an electron to form a bond with Carbon. This makes four electron groups, all bonding, giving a shape of Tetrahedral. For (b) \(\mathrm{BCl}_{3}\), Boron (B) has three valence electrons which form bonds with each Cl atom. This makes for three electron groups, all bonding, resulting in a trigonal planar shape. In molecule (c) \(\mathrm{NF}_{3}\), Nitrogen (N) has five valence electrons, three of which form bonds with F atoms and one pair is left unbonded. Therefore, it has four electron groups, three bonding, one lone pair, resulting in a Trigonal Pyrimidal shape. Similarly for (d) \(\mathrm{H}_{2} \mathrm{Se},\) Selenium (Se) is the central atom with six valence electrons. Two of these form bonds with each H atom, leaving two pairs unbonded. Therefore, it has four electron groups, two bonding, two lone pair resulting in a Bent shape. Lastly for the ion (e) \(\mathrm{NO}_{2}^{-}\), Nitrogen has five valence electrons, two of which form bonds with each O atom, while it also carries a negative charge contributing an additional electron, thus one pair is left unbonded. It has three electron groups, two bonding, one lone pair resulting in a Bent shape.