AP Chemistry: Problem Set 8

Phosphorus forms two chlorides, PCl₃ and PCl₅; nitrogen forms only NCl₃. Explain.

Phosphorus is located in the third period, meaning it has d orbitals. Thus, it can share more than 8 electrons. Conversely, N, located in the second period, can only have 8 valence electrons. Thus, it is only able to make compounds with up to 4 bonds.

Which of the three species (CO₃²⁻, CO₂, or CO) has the shortest C-O bond length? Explain your answer.

Since there is a triple bond solely in CO, it has the shortest bond length. In other words, the bond order for CO is the greatest (3 > 2 > 4/3).

Draw A Lewis dot structure for SF₂ (not all of them). Identify the shape and give the hybridization of the central atom.

bent; sp³

Draw A Lewis dot structure for NNO (not all of them). Identify the shape and give the hybridization of the central atom.

linear; sp

Draw A Lewis dot structure for I₃⁻ (not all of them). Identify the shape and give the hybridization of the central atom.

linear; sp³d

Draw A Lewis dot structure for PF₄⁻ (not all of them). Identify the shape and give the hybridization of the central atom.

see-saw; sp³d

Draw A Lewis dot structure for XeF₄ (not all of them). Identify the shape and give the hybridization of the central atom.

square planar; sp³d²

Draw A Lewis dot structure for NO₃⁻ (not all of them). Identify the shape and give the hybridization of the central atom.

trigonal planar; sp²

Draw A Lewis dot structure for SO₃²⁻ (not all of them). Identify the shape and give the hybridization of the central atom.

trigonal pyramidal; sp³

The H-C-H bond angle in CH₄ is 109.5. Explain.

C, the central atom in CH₄, has four bonds attached to it and no lone pairs. The electrons repel each other and separate as much as possible. Thus, it bears a tetrahedral geometry, translating to bond angles of 109.5°.

Predict the molecular shapes for CO₃²⁻, CO₂, and CO. Give the hybridization for each carbon.

CO₃²⁻: trigonal planar, sp² CO₂: linear, sp CO: linear, sp

The H‑N‑H bond angle in NH₃ is 107. Explain.

N, the central atom in NH₃, has three bonds and one lone pair. The electrons repel each other and separate as much as possible. Thus, it bears a trigonal pyramidal geometry with bond angles of 107° (less than those of a tetrahedral molecule because the lone pair pushes the other bonds downward).

Use formal charges to predict and explain whether NCO⁻ or CNO⁻ is more stable.

NCO⁻ is more stable because the number of formal charges in the resonance structures is far less than that of CNO⁻ (4 vs. 8). Also, in one of the resonance structures for NCO⁻, O is the only atom that bears a negative formal charge, which fits because O is electronegative.

Carbon-carbon single bonds (as in ethane) rotate freely while carbon-carbon double bonds (as in ethene) do not rotate. Explain.

Single bonds can rotate because their rotation doesn't change the identity of the orbital used to make the bond, so the bond itself doesn't change or break. A double bond consists of a sigma bond and a pi bond. p orbitals have to be parallel to participate in a pi bond with one another (i.e. overlap). If a double bond is rotated, the pi bond will break because the p orbitals will no longer be parallel.

Draw A Lewis dot structure for ClF₃ (not all of them). Identify the shape and give the hybridization of the central atom.

T-shaped; sp³d

Draw A Lewis dot structure for XeF₃⁺ (not all of them). Identify the shape and give the hybridization of the central atom.

T-shaped; sp³d

The bond lengths in SO₃ are all identical and are shorter than a sulfur‑oxygen single bond. Explain.

The bond lengths in SO₃ are identical because the double bond spends an equal amount of time in each of the three positions. Some resonance forms of SO₃ have all double bonds and some have a combination of double and single bonds. Thus, the average strength of the bond between S and O is greater than that of a single bond and less than that of a double bond. Since the length of a bond is inversely proportional to its strength, it will be shorter than an S-O bond.

Xenon is capable of forming compounds (e.g. XeF₄) while argon is not. Explain.

The xenon atom is larger than the other noble gases so its outer electrons are farther away from the nucleus and held more loosely, making them subject to interaction with strong electron acceptors (ex. F). Argon's electrons are closer to the nucleus, so it is not as reactive.