Answers and Explanations

DISCRETE PRACTICE QUESTIONS

1 . A

The energies    of  the subshells   within  a   principle   quantum number  are as  follows:    s   <   p   <   d   <   f

2 . D
All single bonds are σ bonds; double and triple bonds each contain one σ bond and one or two π
bonds, respectively. The compounds CH 4 , C 2 H 2 , and C 2 H 4 all contain at least one single bond and
therefore contain at least one σ bond.

3 . B
In a carbon with one double bond, sp^2 hybridization occurs—that is, one s-orbital hybridizes with
two p-orbitals to form three sp^2 -hybridized orbitals. The third p-orbital of the carbon atom
remains unhybridized and takes part in the formation of the π bond of the double bond. Although
there is an unhybridized p-orbital, there are no unhybridized s-orbitals, eliminating choice (D).

4 . D
The carbon and nitrogen atoms are connected by a triple bond in CN– (:C≡N:–). A triple-bonded
atom is sp hybridized; one s-orbital hybridizes with one p-orbital to form two sp-hybridized
orbitals. The two remaining unhybridized p-orbitals take part in the formation of two π bonds.

5 . A
Beryllium has only two electrons in its valence shell. When it bonds to two hydrogens, it requires
two hybridized orbitals, meaning that its hybridization must be sp. Note that the presence of only
single bonds does not mean that the hybridization must be sp^3 ; this is a useful assumption for
carbon, but does not apply to beryllium because of its smaller number of valence electrons. The
two unhybridized p-orbitals around beryllium are empty in BeH 2 , which takes on the linear
geometry characteristic of sp-hybridized orbitals.