10.2.2 An Introduction to Hybrid Orbitals

Question 1

An Introduction to Hybrid Orbitals

Answer 1

• Hybridization involves mathematically combining orbitals that are suitable for bonding.
• Hybridization of orbitals accounts for the expected geometries of trigonal planar and linear molecules.
• Molecules with expanded octets require hybridization of orbitals s, p, and d.
• Lone pair electrons can be accounted for with hybrid orbitals.

Question 2

note

Answer 2

• 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 sp 3 hybrids.
• The resulting CH 4 molecule, after hybridization shows four hydrogen atoms at angles of 109.5°.
• 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 sp 2 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 3

note 2

Answer 3

• 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 (sp 3 d), and SN 6 atoms form octahedral molecules with six hybrid orbitals (sp 3 d 2 )
• 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 sp 3 hybrid orbitals but with only two electrons available for bonding.

Question 4

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

Answer 4

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

Question 5

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 5

VSEPR theory predicts the correct geometry of methane

Question 6

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

Answer 6

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

Question 7

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

Answer 7

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

Question 8

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 8

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

Question 9

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

Answer 9

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

Question 10

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

Answer 10

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.

Question 11

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 11

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.

Question 12

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

Answer 12

tetrahedral

Question 13

What is hybridization of electron orbitals?

Answer 13

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.