Chemical bonding

Question 1

Define valence electrons.

Answer 1

Valence electrons are those electrons furthest from the nucleus (i.e. in the outer shell) - which are typically involved in covalent bonding.

Question 2

Describe a reliable method to determine the number of valence electrons for an element.

Answer 2

The number of valence electrons can be determined by referring to which main-group column the element is found in. For example nitrogen is found in main-group 15 and therefore nitrogen has 5 valence electrons. These are the electrons described by the [He]2s²2p³ electron configuration.

Question 3

List the steps in drawing a Lewis structure.

Answer 3

The steps are (according to the method used in class): Draw a molecular skeleton generally with the least electronegative element in the centre. Draw Lewis Dot diagrams for each atom. Count the total number of valence electrons (add 1 for a negative charge, subtract 1 for a positive charge). Draw bonds between the central and outer atoms by pairing up lone electrons. Move around electrons, if required, to give each atom a stable octet. Assign formal charges to all atoms. Draw all necessary resonance structures. Make sure the final structure(s) looks neat and count the total VE’s as a check.

Question 4

Define formal charge.

Answer 4

Formal charge is an apparent charge which results if an atom “owns” more or less electrons than its normal count of valence electrons. All lone pairs plus 1/2 of the shared pairs are “owned”. Formal charge is therefore calculated by subtracting the number of electrons “owned” from the normal count of valence electrons.

Question 5

Define resonance structure.

Answer 5

Resonance structures are two or more Lewis structures drawn (separated by double headed arrows) which have the same arrangement of atoms and number of electrons, but a different location of the electrons. The electrons that “move” are said to be delocalized.

Question 6

Explain VSEPR theory.

Answer 6

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts that electron pairs about a central atom will arrange themselves so as to minimize repulsion. Electron pairs include both bonding and lone pairs.

Question 7

Compare and contrast electronic versus molecular geometry.

Answer 7

Electronic geometry describes the arrangement of the electron pairs about the central atom. There are five basic geometries: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral. The electronic geometry is essentially determined by the central atom’s bonds and lone pairs. Molecular geometry describes the arrangement of atoms around a central atom. Molecular geometry is identical to electronic geometry when there are no lone pairs of electrons. For example, the electronic and molecular geometry of CH₄ is tetrahedral (EG = 4, AX₄-type molecule). By contrast the electronic geometry of NH₃ is tetrahedral (EG = 4), but the electronic geometry is trigonal pyramidal (AX₃E-type molecule). This is because NH₃ has one lone pair of electrons. [Note EG = electron groups]

Question 8

Give the name of the electronic and molecular geometries for an AX₃E₂-type molecule.

Answer 8

AX₃E₂-type molecule has an electronic geometry essentially the same as an AX₅-type molecule - that is trigonal bipyramidal. Its molecular geometry is T-shaped.

Question 9

Give the name of the electronic and molecular geometries for an AX₄E₂-type molecule.

Answer 9

AX₄E₂-type molecule has an electronic geometry essentially the same as an AX₆-type molecule - that is octahedral. Its molecular geometry is square planar.

Question 10

Give the name of the electronic and molecular geometries for an AX₄E-type molecule.

Answer 10

AX₄E-type molecule has an electronic geometry essentially the same as an AX₅-type molecule - that is trigonal bipyramidal. Its molecular geometry is see saw.

Question 11

Define electronegativity.

Answer 11

Electronegativity is the attraction the nucleus has for a shared pair of electrons in a bond.

Question 12

Define bond polarity.

Answer 12

Bond polarity refers to the degree to which there is an unequal attraction for a shared pair of electrons in a covalent bond. An equal attraction occurs when both atoms involved in a bond are identical or if both atoms have similar electronegativities. This would result in a non-polar covalent bond. An unequal attraction occurs when the two atoms involved in a bond have different electronegativities. The greater the difference in electronegativity, the more polar is the bond.

Question 13

Compare and contrast bond polarity versus molecular polarity.

Answer 13

Bond polarity occurs when electrons are “shared” between atoms of different electronegativities. Molecular polarity occurs as a result of bond polarity and when there is an unequal distribution of electron density throughout the molecule (or at least in a portion of the molecule). In very simple cases (e.g. H-F), bond polarity is the same as molecular polarity. In other cases, molecular polarity can only be determined by considering bond polarity and the molecular geometry (shape). It is possible for bonds to be polar, but because of the molecular geometry this effect is cancelled (e.g. CO₂). In other cases the bonds are polar and their effect is reinforced and as a result the molecule is polar (e.g. SO₂, bent shape).

Question 14

Describe atomic orbital hybridization.

Answer 14

Atomic orbital hybridization is the result of the original atomic orbitals “fusing” to produce unique hybrid orbitals. These hybrid orbitals have different energies and “shapes” as compared to the original atomic orbitals. For example, carbon’s valence electrons sit in the 2s and 2p atomic orbitals according to the electron configuration (2s²2p²). When carbon forms four single bonds, these atomic orbitals hybridize into four unique sp³ orbitals. Collectively the four sp³ orbitals produce a tetrahedral orientation. All four sp³ hybrid orbitals are degenerate and their energies are distinct from the energies of the 2s and 2p orbitals.

Question 15

Compare and contrast sigma versus pi bonds.

Answer 15

A sigma bond is the result of a head to head overlap of orbitals. This results in what we normally label as a single bond. A pi bond is the result of a side by side overlap of p orbitals. This side by side overlap occurs both above and below the plane of the sigma bond. A sigma bond and a pi bond between two atoms would constitute a double bond. A sigma bond and two pi bonds would constitute a triple bond.

Question 16

Describe the relationship of electron groups (EG) and hybridization.

Answer 16

For the purpose of the Chem 112 course, the electron groups (which is easily determined by considering the number of sigma bonds and lone pair of electrons about an atom) is a good predictor of the atom’s hybridization. EG = 2 = sp hybridization; EG = 3 = sp² hybridization; EG = 4 = sp³; EG = 5 = sp³d hybridization; EG = 6 = sp³d² hybridization.

Question 17

Describe the relationship of electronic geometry and hybridization.

Answer 17

Given that electronic geometry is determined by an atom’s electron groups (EG) and EG is also a good predictor of hybridization, there is a natural link between electronic geometry and hybridization. For example, all central atoms with EG = 4 will be sp³ hybridized and will have a tetrahedral electronic geometry. Note, however, that the molecular geometry can differ depending on the number of lone pairs.

Question 18

When does electronic geometry = molecular geometry?

Answer 18

Molecular geometry is identical to electronic geometry when there are no lone pairs of electrons. For example the electronic and molecular geometry of CH₄ is tetrahedral (EG = 4, AX₄-type molecule). By contrast the electronic geometry of NH₃ is tetrahedral (EG = 4), but the electronic geometry is trigonal pyramidal (AX₃E-type molecule). In this case NH₃ has one lone pair of electrons.