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Chapter 9: Problem 56

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plust (rotate) around a single \(\sigma\) bond same?

Short Answer

Expert verified
(a) In sp hybridization, there is one unhybridized p orbital remaining, and the atom can form one π bond. (b) It would be easier to twist (rotate) around a single σ bond than around a double bond (σ plus π) due to the π bond overlap restriction in the double bond.

Step by step solution

01

Understanding sp hybridization and counting unhybridized p orbitals

An atom with sp hybridized orbitals has one s orbital and one p orbital combining to form two equivalent sp hybrid orbitals. Therefore, the atom has one remaining unhybridized p orbital, as there are initially three p orbitals in the valence shell of most elements (p_x, p_y, p_z).
02

Calculating the number of π bonds possible

The unhybridized p orbital can overlap with another unhybridized p orbital from another atom to form a π bond. Since there is only one unhybridized p orbital, the atom can form a maximum of one π bond. Answer for part (a): There is one unhybridized p orbital remaining and the atom can form one π bond.
03

Comparing the ease of rotation around single σ and double bonds

A single bond, which is a σ bond, allows for free rotation around the bond axis because its electron density is distributed symmetrically between the two bonded atoms. On the other hand, a double bond consists of one σ bond and one π bond. The π bond electron density is distributed above and below the plane formed by the two atoms and the σ bond. This distribution restricts the rotation around the bond axis because rotating the bond breaks the π bond overlap. Answer for part (b): It would be easier to twist (rotate) around a single σ bond than around a double bond (σ plus π) because of the π bond overlap restriction in the double bond.

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Most popular questions from this chapter

(a) Does SCl\(_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does BeCl\(_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

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