Topic 14: Chemical Bonding and Structure (HL)

Question 1

Formula for Formal Charge

Answer 1

FC = (number of valence electrons) - 1/2(number of bonding electrons) - (number of non-bonding electrons)

Question 2

How to calculate partial charge based on electronegative values?

Answer 2

Based on electronegative values.
For each element in a compound do:
PC = (electronegative value for M)/(electronegative sum of values for the whole compound)
Then for the higher PC multiply it by the number of electrons in the bond(s) ie. 2e.
Then this value - 1 gives the partial charge

Question 3

What is the molecular geometry based on five electron domains with no lone pairs?

Answer 3

trigonal bipyramidal

Between equatorial and axial position the angle is 90° between two equatorial it’s 180°

Question 4

What is the molecular geometry based on five electron domains with one lone pair?

Answer 4

trigonal bipyramidal

Between equatorial and axial position the angle is <90° between two equatorial it’s <180°

AB4E

Question 5

What is the molecular geometry based on five electron domains with two lone pairs?

Answer 5

trigonal bipyramidal

Between equatorial and axial position the angle is <90°

AB3E2

Question 6

What is the molecular geometry based on five electron domains with three lone pairs?

Answer 6

trigonal bipyramidal

Between two axial positions the angle is 180°

AB2E3

Question 7

What is the molecular geometry based on six electron domains no lone pairs?

Answer 7

octahedral

Bond angle 90°

Question 8

What is the molecular geometry based on six electron domains with one lone pair?

Answer 8

octahedral

Question 9

What is the molecular geometry based on five electron domains with two lone pairs?

Answer 9

octahedral

Bond angle 90°

Question 10

Sigma bond

Answer 10

In the formation of a sigma bond, there is a direct head-on overlap of the atomic orbitals along the internuclear axis and the electron density is located along this axis.

Question 11

Pi bond

Answer 11

In the formation of a pi bond, there is a sideways overlap of the atomic orbitals and the electron density is located above and below the internuclear axis.

Question 12

Delocalisation

Answer 12

Delocalisation involves
electrons that are shared by more than two atoms in a molecule or ion as opposed to being localized between a pair of atoms.

Question 13

Ozone - bonds and shape

Answer 13

V-shaped (bent) molecule with a bond angle of 116.8° and its two O-O bond lengths are equal.

Question 14

bond order

Answer 14

e.g. for O-O bond in ozone

BO = (total number of O-O bonding pairs)/(total number of O-O positions)

Question 15

How are the bonds in ozone broken?

Answer 15

The bonds in ozone can be broken by UV radiation (hv) . The bond order in ozone (1 .5) is lower than the bond order in oxygen (2) , so the O=O double bond in oxygen is stronger. Radiation of shorter wavelength is required to break the stronger bond in oxygen. The reason is that the energy, E, of a photon of light is inversely proportional to the wavelength λ, so the greater the energy, the shorter the wavelength and vice versa.

Question 16

Valence Bond Theory (VBT)

Answer 16

• considers that atoms approach each other to form a molecule
• bond formed results from the overlap of atomic orbitals
• the greater the degree of orbital overlap, the stronger the bond

Question 17

Molecular Orbital Theory (MOT)

Answer 17

• atomic orbitals overlap, but the overlap results in the formation of new orbitals, called the molecular orbitals.
• the two atomic orbitals combine to form two new molecular orbitals:
  1) bonding molecular orbital sigma, which is of lower energy
  2) anti-bonding molecular orbital sigma star, which has higher energy.

Question 18

Hybridization

Answer 18

Mixing of atomic orbitals to generate a set of new hybrid orbitals that are equivalent.

Hybridization type can be determined from the number of electron domains around the central atom rather than the molecular geometry

Question 19

Hybrid orbital

Answer 19

Results from the mixing of different types of atomic orbital on the same atom.

Question 20

Formation of sp3 hybrid orbitals in methane

Answer 20

Carbon: [He]2s2.2p2
Hydrogen: 1s1

The tetrahedral nature of methane involves the hybridization of its one 2s and three 2p orbitals on the C atom in an excited state

Step 1: One of the electrons in the 2s orbital of the ground-state configuration of carbon is promoted to the vacant 2pz
orbital, to form an excited-state.
But methane has a tetrahedral molecular geometry with H-C-H bond angle of 109.5°, but the three 2p orbitals are at 90° to each other.

Step 2: The four atomic orbitals 2s, 2px, 2py, and 2pz
combine to form a set of four new sp3 hybrid orbitals.
The shape of each sp3 hybrid orbital will have 75% of the characteristics of a p orbital (dumbbell shaped) combined with 25% of the spherical s orbital.

Step 3: Overlap of each carbon sp3 orbital with a hydrogen 1s atomic orbital.

Any molecule with a tetrahedral electron domain geometry on its interior central atom (based on four electron domains) would be predicted to have sp3
hybridization.

Question 21

Formation of sp2 hybrid orbitals in ethene

Answer 21

Step 1: One of the electrons in the 2s orbital of the ground-state configuration of carbon is promoted to the vacant 2pz
orbital to form an excited-state

Step 2: To account for the approximate 120° bond angles (based on three electron domains) , the next step involves hybridization of three of the atomic orbitals 2s, 2px and 2py. These combine to form a set of three
new sp2 hybrid orbitals. The 2pz
orbital remains
unhybridized

Step 3: formation of a sigma bond along the internuclear axis by the overlap of two sp2 hybrid orbitals, one on each carbon atom. A pi bond is formed from the sideways overlap of the two pz unhybridized atomic orbitals, with the
overlap regions above and below the internuclear axis

Step 4: overlap of each remaining sp2
orbital on carbon with a hydrogen 1 s atomic orbital.

Any molecule with a trigonal-planar electron domain geometry will be predicted to have sp2
hybridization

Question 22

Formation of sp hybrid orbitals in ethyne

Answer 22

Step 1: One of the electrons in the 2s orbital of the ground-state con?guration of carbon is promoted to the vacant 2pz
orbital to form an excited-state.

Step 2: To account for the 180° bond angle (based on two electron domains) , the next step involves hybridization of two of the atomic orbitals 2s and 2px. The remaining 2py
and 2pz
orbitals remain unhybridized.

Step 3: formation of a sigma bond along the internuclear axis by the overlap of two sp hybrid orbitals, one on each carbon atom. Two pi bonds are formed from the sideways overlap of the two py and two pz unhybridized atomic orbitals, with the overlap regions above and below the internuclear axis

Step 4: overlapping of each remaining sp orbital on carbon with a hydrogen 1 s atomic orbital

Any molecule with a linear electron domain geometry on its interior central atom would be predicted to have sp hybridization.