Consider the molecule \(\mathrm{HNF}_{2}(\mathrm{~N}\) is the central atom in the molecule). (a) Draw the dot diagram. (b) Draw the molecule's three-dimensional shape, and label the numeric value of all bond angles. (c) What is the shape of this molecule? (d) Draw in the individual bond dipole moments. (e) Is the molecule polar? If yes, draw the molecular dipole moment vector.

First, we need to determine the number of valence electrons present in the molecule. Here we have one hydrogen atom (H), one nitrogen atom (N), and two fluorine atoms (F). We use the periodic table to determine the number of valence electrons for each atom: • Hydrogen (H) has \(1\) valence electron. • Nitrogen (N) has \(5\) valence electrons. • Fluorine (F) has \(7\) valence electrons each. Therefore, there is a total of \(1 + 5 + 7 \times 2 = 20\) valence electrons in the HNF2 molecule.

According to electron sharing rules, we'll draw a Lewis structure with Nitrogen as the central atom, as mentioned in the exercise. First, connect atoms with single bonds to fulfill the octet rule (except for hydrogen), then distribute the remaining electrons as lone pairs. 1. Place Nitrogen (central atom) in the middle and connect it to Hydrogen and two Fluorine atoms with single bonds. 2. Distribute the remaining valence electrons as lone pairs to fulfill the octet rule (except for hydrogen). Nitrogen has \(2\) remaining lone pairs, while each Fluorine atom keeps the \(6\) lone pairs. The Lewis structure of HNF2 is represented by: ``` F H \ / N = = = F ```

Using VSEPR (Valence Shell Electron Pair Repulsion) theory, we can determine the three-dimensional shape of the HNF2 molecule. Here, the central Nitrogen atom has \(3\) atoms bonded to it and \(1\) lone pair of electrons. This corresponds to an electronic geometry of "tetrahedral" with \(4\) electron domains.

The three-dimensional shape of a molecule with an electronic geometry of "tetrahedral" and \(1\) lone pair of electrons forming \(3\) single bonds is "trigonal pyramidal." The bond angles between the three atoms bonded to the central atom in a tetrahedral geometry are approximately \(109.5^{\circ}\). However, lone pair-bonding pair repulsion causes a slight decrease, creating bond angles slightly less than \(109.5^{\circ}\). Thus, the shape of the HNF2 molecule is trigonal pyramidal with bond angles slightly less than \(109.5^{\circ}\).

Bond dipole moments occur due to the difference in electronegativity between two atoms in a bond. Using the electronegativity values, we can estimate the bond dipole moments of H-N and N-F bonds. Electronegativity values: • Hydrogen (H): \(2.1\) • Nitrogen (N): \(3.0\) • Fluorine (F): \(3.98\) The individual bond dipole moments: • H-N bond: \(\Delta EN = 3.0 - 2.1 = 0.9\) • N-F bond: \(\Delta EN = 3.98 - 3.0 = 0.98\) The H-N bond dipole moment points from hydrogen to nitrogen, and the N-F bonds dipole moment points from nitrogen to both fluorine atoms.

Since the molecule's shape is "trigonal pyramidal," the bond dipole moments do not cancel each other out. Therefore, the molecule is polar, with a dipole moment originating from the nitrogen atom and directed to the three outer atoms. In conclusion: a) The Lewis structure (dot diagram) of HNF2 is: ``` F H \ / N = = = F ``` b) The three-dimensional shape of the molecule is trigonal pyramidal with bond angles slightly less than \(109.5^{\circ}\). c) The shape of this molecule is trigonal pyramidal. d) The individual bond dipole moments are \(0.9\) for H-N and \(0.98\) for N-F bonds. e) Yes, the molecule HNF2 is polar, with the molecular dipole moment vector pointing from the nitrogen atom to the outer atoms (hydrogen and fluorine).