Bonding

Question 1

VSEPR shapes of molecules and ions

Answer 1

• Linear (180°)
• Trigonal planar (120°), bent/angular (< 120°)
• Tetrahedral (109°), trigonal pyramidal (< 109°), bent/angular (&laquo_space;109°)
• Trigonal bipyramidal (90°, 120°), sawhorse/seesaw (< 90°, < 120°), t-shape (< 90°), linear (180°)
• Octahedral (90°), square pyramid (< 90°), square planar (90°), t-shaped (< 90°), linear (180°)

Repulsion between electron pairs dictates shape of a molecule/ion

Question 2

Formal charge

Answer 2

• [Of an atom in a molecule,] hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms
• We can double-check calculations by determining the sum of the formal charges for the whole structure (must be zero [if an ion, should equal the charge])
• Not the actual charge; only a useful bookkeeping procedure (does not indicate the presence of actual charges)
• For each atom, FC = # valence electrons - # non-bonding electrons - # bonding electrons/2
• Arrangement of atoms in a molecule/ion is its molecular structure; in many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure (ex. different multiple-bond and lone-pair electron placements or different arrangements of atoms)—the most reasonable one of these will be the one w/ the FC difference (max - min) closest to zero and the one that has the negative charges located on the most electronegative atoms

*Allow us to understand why the less electronegative atom typically occupies the central position

Question 3

Sigma bonds

Answer 3

• When atoms covalently bond to form molecules, atomic orbitals have to combine to form molecular orbitals: Lewis structures are a convenient way to model covalent bonds, but they don’t take into account that valence electrons are in different sub-orbitals
• Covalent bonds (shared pair of electrons)
• Formed by direct overlapping between two adjacent atom’s outermost orbitals along the internuclear axis (i.e., along a line connecting the two bonded atoms)
• Created by combo of the single electrons form each atom’s orbital
• Single bonds are σ bonds
• Rotation around a σ bond is possible

Question 4

Pi bonds

Answer 4

• Formed when two p orbitals overlap “sideways-on” (overlap occurs above and below the line drawn between the two nuclei): has two regions of electron density
• A double bond consists of two parts: a sigma bond caused by the overlapping of atomic orbitals along the internuclear axis between the two atoms and a pi bond made by the overlapping of p orbitals above and below the axis
• Molecular shape only determined by sigma bonds (double bonds treated as one single bond when determining shape of molecule)
• Every single covalent bond is a sigma bond; every double covalent bond has both a sigma and a pi bond in it; every triple covalent bond has a sigma and two pi bonds in it

Question 5

Hybridization (methane)

Answer 5

• B/c of hybridization, carbon forms four identical C-H bonds when it only has two unpaired electrons in 2p w/ which to form bonds
• The electrons rearrange themselves in separate hybrid orbitals w/ equal energy: this happens in two steps for methane
• First, since there is only a small energy gap between the 2s and 2p orbitals, carbon can “promote” one electron from the 2s to the empty 2p to give four unpaired electrons (when bonds are formed, energy is released and the system becomes more stable: if carbon forms four bonds rather than two, twice as much energy is released and so the resulting molecule becomes even more stable)
• But now the electrons are at different energies and won’t give identical bonds, so a further step needs to happen: the electrons rearrange themselves in a process called hybridization (reorganizes the electrons into four identical hybrid orbitals called sp3 hybrids [b/c made from one s orbital and three p orbitals])

Question 6

sp3 hybrid orbital shape

Answer 6

• Looks a bit like half a p orbital
• Arrange themselves in space so that they are as far apart as possible
• Picture the nucleus as being at the centre of a tetrahedron (triangularly based pyramid) w/ the orbitals pointing to the corners