Draw the Lewis structure. (1 mark) For the following questions, use NF4+ b.) Assign a VSEPR shape. (1 mark) c.) Use electronegativity values to classify each bond. Draw bond dipoles where appropriate. (1 mark) d.) State whether the molecule is polar or nonpolar overall. Draw dipoles where appropriate. (1 mark)
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The Correct Answer and Explanation is:
Sure! Let’s go step by step with NF₄⁺ (tetrafluoroammonium ion):
a) Lewis Structure (1 mark):
- Nitrogen (N) has 5 valence electrons.
- Each fluorine (F) has 7 valence electrons × 4 = 28
- Total valence electrons before considering the +1 charge = 5 + 28 = 33
- Since the ion has a +1 charge, subtract 1 electron → Total = 32 electrons
Now draw:
- Nitrogen is the central atom.
- Form single bonds from nitrogen to each of the four fluorine atoms (4 bonds × 2 electrons = 8 electrons).
- Distribute remaining 24 electrons as lone pairs on the 4 fluorine atoms (6 electrons per fluorine).
✔️ Lewis structure:
F
|
F - N⁺ - F
|
F
Each F has 3 lone pairs. Nitrogen has 4 bonding pairs and no lone pairs.
b) VSEPR Shape (1 mark):
- Central atom (N) has 4 bonding groups, 0 lone pairs → AX₄ type
✔️ VSEPR shape = Tetrahedral
c) Bond Classification & Dipoles (1 mark):
- Electronegativity:
- Fluorine (F) = 3.98
- Nitrogen (N) = 3.04
- ΔEN = 3.98 – 3.04 = 0.94 → Polar covalent bonds
✔️ Draw bond dipoles pointing from nitrogen → fluorine (towards more electronegative F):
δ- δ-
↑ ↑
F ← N⁺ → F
↓
F δ-
Each N–F bond is polar.
d) Overall Polarity (1 mark):
- The molecule is symmetrical (tetrahedral), and dipoles cancel out due to equal magnitude and opposite directions.
✔️ Overall molecule is nonpolar
Explanation:
NF₄⁺, or tetrafluoroammonium ion, contains a central nitrogen atom bonded to four fluorine atoms. To construct its Lewis structure, we start by determining the total valence electrons: nitrogen contributes 5 electrons, and each of the four fluorine atoms contributes 7, totaling 33. Since the molecule has a +1 charge, we remove one electron, leaving 32. These are distributed by forming single bonds between nitrogen and each fluorine, using 8 electrons. The remaining 24 electrons are placed as lone pairs on the fluorine atoms. Nitrogen ends up with four bonding pairs and no lone pairs.
Based on VSEPR theory, a central atom with four bonding groups and no lone pairs has a tetrahedral electron and molecular geometry. This means the atoms are arranged symmetrically around the nitrogen in three dimensions, with bond angles close to 109.5°.
To classify the bonds, we look at the difference in electronegativity. Fluorine (3.98) is more electronegative than nitrogen (3.04), resulting in polar covalent bonds (ΔEN = 0.94). Bond dipoles point from nitrogen toward fluorine. While each N–F bond is polar, the overall molecular shape causes these dipoles to cancel out due to the tetrahedral symmetry.
Thus, despite containing polar bonds, the NF₄⁺ ion is nonpolar overall because of its symmetrical geometry. There’s no net dipole moment across the molecule. This distinction between polar bonds and overall molecular polarity is a key concept in molecular chemistry, showing how geometry influences physical and chemical behavior.
