1. Draw the Lewis structure of nitrogen triiodide, NI3. 2. The two N-O bonds in the nitrite ion, NO2, are the same length. Draw two resonance forms of the nitrite ion that supports this hypothesis. 3. Draw a Lewis structure of sulfur trioxide, SO3, in which the formal charges of all atoms are zero.
The correct answer and explanation is:
Explanation:
1. Nitrogen triiodide (NI₃):
The Lewis structure of nitrogen triiodide (NI₃) shows a central nitrogen atom bonded to three iodine atoms. Nitrogen has five valence electrons, and each iodine atom has seven valence electrons. Nitrogen forms three single covalent bonds with iodine atoms, using three of its five electrons, while the remaining two electrons form a lone pair on nitrogen. Each iodine atom achieves stability by sharing one electron with nitrogen, completing its octet with three lone pairs. This arrangement leads to a trigonal pyramidal geometry due to the lone pair on nitrogen, slightly distorting bond angles to less than 109.5°.
2. Resonance structures of nitrite ion (NO₂⁻):
The nitrite ion (NO₂⁻) has resonance structures, indicating that its electrons are delocalized across multiple atoms. Nitrogen (five valence electrons) forms bonds with two oxygen atoms, each initially having six valence electrons. In one resonance structure, nitrogen forms a double bond with one oxygen atom and a single bond with the other oxygen atom carrying a negative charge. In the other resonance structure, the positions of single and double bonds interchange. Due to resonance, neither structure individually accurately describes the actual ion. Instead, electrons become delocalized across both N-O bonds, resulting in equal bond lengths and strengths, intermediate between single and double bonds.
3. Sulfur trioxide (SO₃):
In SO₃, the sulfur atom (six valence electrons) is bonded to three oxygen atoms (each with six valence electrons) through double bonds. Each oxygen atom forms one double bond with sulfur and possesses two lone pairs. The sulfur atom uses all six valence electrons to form three double bonds, leading to no lone pairs remaining on sulfur. This structure results in zero formal charges for all atoms, as sulfur shares electrons equally with the oxygen atoms, each bond representing equal electron sharing. Consequently, SO₃ adopts a trigonal planar structure with equal bond angles of 120°, reflecting the symmetrical electron distribution and the absence of formal charges.
The provided diagrams visually reinforce these explanations, clearly depicting the electron-sharing arrangements and the resonance structures involved.