An Introduction to Hybrid Orbitals (10.2.2)

Question 1

• Hybridization involves mathematically combining orbitals that are suitable for bonding.

Answer 1

• Hybridization involves mathematically combining orbitals that are suitable for bonding.

Question 2

• Hybridization of orbitals accounts for the expected geometries of trigonal planar and linear molecules.

Answer 2

• Hybridization of orbitals accounts for the expected geometries of trigonal planar and linear molecules.

Question 3

• Molecules with expanded octets require hybridization of orbitals s, p, and d.

Answer 3

• Molecules with expanded octets require hybridization of orbitals s, p, and d.

Question 4

• Lone pair electrons can be accounted for with hybrid orbitals.

Answer 4

• Lone pair electrons can be accounted for with hybrid orbitals.

Question 5

Hybridization involves mathematically combining
atomic orbitals that are suitable for bonding.
In the example of methane, in the carbon atom one
electron in the 2s orbital is promoted to the 2p
orbital, resulting in four electrons available for
sigma bonding.

Determining the outcome of hybridization entails
assigning all orbitals the same energy level, after
which they are all equally likely to form a bond. The
hybrid energy level is a weighted average of the
energy levels of the component orbitals. These four
orbitals are called sp3 hybrids.

The resulting CH4 molecule, after hybridization
shows four hydrogen atoms at angles of 109.5°.

Answer 5

Hybridization involves mathematically combining
atomic orbitals that are suitable for bonding.
In the example of methane, in the carbon atom one
electron in the 2s orbital is promoted to the 2p
orbital, resulting in four electrons available for
sigma bonding.

Determining the outcome of hybridization entails
assigning all orbitals the same energy level, after
which they are all equally likely to form a bond. The
hybrid energy level is a weighted average of the
energy levels of the component orbitals. These four
orbitals are called sp3 hybrids.

The resulting CH4 molecule, after hybridization
shows four hydrogen atoms at angles of 109.5°.

Question 6

Trigonal planar molecules are formed as a result of
the formation of hybrid orbitals in steric number 3
(SN 3) atoms.
The example shows the formation of three sp2
hybrid orbitals in an atom of boron. One of the 2s
electrons is promoted to an empty 2p orbital and the
2s and two 2p orbitals containing an electron
available for sigma bonding are hybridized. They
take on a hybrid energy level and the hybrid shape
of a lop-sided dumbbell. These hybrid orbitals will
form sigma bonds with fluorine atoms in a trigonal
planar molecule with bond angles at 120°.
The unpictured empty p orbital is coming out of the
page, at 90° to the plane of the page.

Answer 6

Trigonal planar molecules are formed as a result of
the formation of hybrid orbitals in steric number 3
(SN 3) atoms.
The example shows the formation of three sp2
hybrid orbitals in an atom of boron. One of the 2s
electrons is promoted to an empty 2p orbital and the
2s and two 2p orbitals containing an electron
available for sigma bonding are hybridized. They
take on a hybrid energy level and the hybrid shape
of a lop-sided dumbbell. These hybrid orbitals will
form sigma bonds with fluorine atoms in a trigonal
planar molecule with bond angles at 120°.
The unpictured empty p orbital is coming out of the
page, at 90° to the plane of the page.

Question 7

Linear molecules are formed as a result of the
formation of hybrid orbitals in steric number 2 (SN 2)
atoms.
The example shows the formation of two sp hybrid
orbitals through the promotion of a 2s electron to a p
orbital and subsequent hybridization of the orbitals
with one electron. The hybrid orbitals acquire a
hybrid energy level and a hybrid shape. These
hybrid orbitals will sigma bond with fluorine atoms in
a linear molecule.
Atoms with SN 5 form trigonal bipyramidal
molecules with 5 hybrid orbitals (sp3
d), and SN 6
atoms form octahedral molecules with six hybrid
orbitals (sp3
d2
)

Answer 7

Linear molecules are formed as a result of the
formation of hybrid orbitals in steric number 2 (SN 2)
atoms.
The example shows the formation of two sp hybrid
orbitals through the promotion of a 2s electron to a p
orbital and subsequent hybridization of the orbitals
with one electron. The hybrid orbitals acquire a
hybrid energy level and a hybrid shape. These
hybrid orbitals will sigma bond with fluorine atoms in
a linear molecule.
Atoms with SN 5 form trigonal bipyramidal
molecules with 5 hybrid orbitals (sp3
d), and SN 6
atoms form octahedral molecules with six hybrid
orbitals (sp3
d2
)

Question 8

Hybridization also accounts for atoms such as
oxygen with lone pair electrons. In this case, the
lone pair of 2s electrons is promoted and the
orbitals are hybridized to form four sp3
hybrid
orbitals but with only two electrons available for
bonding.

Answer 8

Hybridization also accounts for atoms such as
oxygen with lone pair electrons. In this case, the
lone pair of 2s electrons is promoted and the
orbitals are hybridized to form four sp3
hybrid
orbitals but with only two electrons available for
bonding.

Question 9

Valence bond theory predicts that the following molecule should result when a carbon atom joins two hydrogen atoms. Which statement about this predicted molecule is not true?

Answer 9

We use valence bond theory to explain that the bonding occurs when an orbital that contains a pair of electrons overlaps with another orbital that has one pair of electrons. (D)

We do use valence bond theory, but this theory states that bonding occurs when a half-filled orbital (i.e., one containing an unpaired electron) overlaps with another half-filled orbital.

Question 10

What is hybridization of electron orbitals?

Answer 10

Hybridization of electron orbitals occurs when several atomic orbitals (such as s and p orbitals) of an atom combine to form (the same number of) new, hybrid orbitals. A promotion energy has to be involved in order for this event to occur. (C)

Question 11

The CH2 molecule, called carbene, is very unstable. This is because carbon “wants” to have four C–H bonds instead of just two C–H bonds. When carbon has four C–H bonds, the molecule methane, CH4, is formed. It is much more stable than carbene. In order to rectify the instability that exists in the CH2 molecule, the carbon atom realigns its electrons so that four hydrogen atoms can be bonded to it. Which statement about this rearrangement of electrons in the carbon atom is not true?

Answer 11

VSEPR theory predicts the correct geometry of methane. (B)

VSEPR theory predicts that the three hydrogen atoms bonded to the 2p orbitals of carbon would all be arranged along x -, y -, and z-axes, at right angles to each other. The fourth hydrogen atom would try to arrange itself as far away as possible from those atoms. What is observed, however, is a perfect tetrahedral geometry. Therefore, there is another concept, hybridization, that is required to explain the geometry of methane that contradicts VSEPR theory.

Question 12

Which statement about the formation of a boron trifluoride molecule, BF3, from one boron atom and three fluorine atoms is NOT correct?

Answer 12

The flourine atoms are hybridized and this hybridization of the fluorine atoms occurs as a result of the use of promotion energy. (C)

Fluorine does not undergo hybridization. The boron atom is hybridized in order to accommodate the three fluorine atoms that are ready to bond.

Question 13

Which of the following statements correctly describes the geometry of the sp3 hybrids for molecules that have SN = 4?

Answer 13

tetrahedral (C)

This is really just a question about the predicted geometry of a molecule. Because we learned that SN and hybridization are intimately related, we can predict the geometry of a molecule that has a hybridized atom by simply looking at the SN. If the SN = 4, the predicted molecular geometry is a structure pointing to the four corners of a tetrahedron.

Question 14

Which statement best describes the relationship between hybridization and steric number (SN)?

Answer 14

Hybridization and steric number are intimately related. For each steric number for a molecule, there is an observed molecular arrangement that can be directly explained by the hybridization event. (C)

This is a very thorough and accurate answer. Using SN, you can predict molecular geometry. Using hybridization (when appropriate), you can explain the geometry.

Question 15

Carbon forms numerous compounds with hydrogen, called hydrocarbons. Which of the following statements is not correct.

Answer 15

Through hybridization of the carbon atom, a total of four lone electron orbital sites are established. These sites are called p4 orbitals. (B)

This statement is false.

Question 16

Which statement about the hybridization of atoms with SN values of 5 or 6 is not true?

Answer 16

The nomenclature for the hybrid orbitals for SN = 6 molecules is sp4d. (C)

Because SN = 6, the sum of the orbitals has to equal 6. In this case, sp4d can be written as s1p4d1. The sum of the orbitals = 1 + 4 + 1 = 6 = SN. However, the p orbital can not be assigned an integer value greater than three, since there are only x, y, and z orbitals. The extra orbital here should actually be assigned to a new orbital site in the d orbital. This is correctly expressed as s1p3d2.

Question 17

Which of the following correctly summarizes the exact relationships between the SN values of 2 to 6, the hybrid orbital names, and the predicted geometries for molecules that have hybridized central atoms?

Answer 17

SN = 2; sp; linear
SN = 3; sp2; trigonal planar
SN = 4; sp3; tetrahedral
SN = 5; sp3d; trigonal bipyramidal
SN = 6; sp3d2; octahedral (C)

Notice that each SN value equals the sum of the exponents (in the hybrid orbital names terms). Also, notice that the geometries get more complicated with increasing SN values.

Question 18

Which of the following statements is not a main step in the hybridization event?

Answer 18

Take all the original electron pairs and break them into single electron sites. (D)

Sometimes (e.g., H2O) the electron pairs are preserved. What does happen is that the orbitals are altered (to hybrid orbitals), resulting in new orbitals, each at a new energy level. Remember, though, that the overall energy level remains the same.