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9701 · 3.4

Covalent bonding and coordinate (dative covalent) bonding — practice questions

Practice and worked examples for 9701 Covalent bonding and coordinate (dative covalent) bonding. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Draw a 'dot-and-cross' diagram for a molecule of methane, CH4CH_4. Show outer shell electrons only.

Show solution outline
  1. Identify valence electrons: Carbon is in Group 14, so it has 4 valence electrons (we'll use ×). Hydrogen is in Group 1, so it has 1 valence electron (we'll use •).
  2. Determine bonding: Carbon needs 4 more electrons to achieve a stable octet. Each hydrogen needs 1 more electron to complete its first shell. Therefore, the central carbon atom will form four single covalent bonds, one with each of the four hydrogen atoms.
  3. Draw the diagram: Place the carbon atom in the centre. Arrange the four hydrogen atoms around it. For each C-H bond, draw one cross from the carbon and one dot from the hydrogen in the overlapping region. The final diagram shows the carbon atom sharing four pairs of electrons, and each hydrogen atom sharing one pair.

Worked example 2

A molecule of carbon dioxide, CO2CO_2, contains two carbon-oxygen double bonds. Draw its 'dot-and-cross' diagram.

Show solution outline
  1. Valence electrons: Carbon (Group 14) has 4 valence electrons (×). Oxygen (Group 16) has 6 valence electrons (•).
  2. Determine bonding: The central carbon atom needs to form bonds to satisfy its octet. Each oxygen atom needs two electrons. To achieve this, the carbon atom forms a double bond with each of the two oxygen atoms.
  3. Draw the diagram: Place C in the centre with an O on either side. For each C=O double bond, show two electrons from carbon (2 ×) and two electrons from oxygen (2 •) being shared. This gives carbon a share in 8 electrons. Each oxygen atom also has a share in 8 electrons (4 shared + 4 non-bonding/lone pair electrons). The non-bonding electrons on each oxygen should be shown as two lone pairs.

Worked example 3

The reaction between boron trifluoride, BF3BF_3, and ammonia, NH3NH_3, forms an addition compound, F3BNH3F_3BNH_3. Explain how a coordinate bond is formed and draw the dot-and-cross diagram for the product.

Show solution outline
  1. Identify donor and acceptor: Ammonia (NH3NH_3) has a lone pair on the nitrogen atom, so it can act as an electron-pair donor. Boron trifluoride (BF3BF_3) is electron deficient; the boron atom only has 6 electrons in its outer shell and has a vacant p-orbital. It can act as an electron-pair acceptor.
  2. Bond formation: The nitrogen atom in NH3NH_3 donates its lone pair into the vacant orbital of the boron atom in BF3BF_3. This forms a coordinate (dative covalent) bond between N and B.
  3. Draw the diagram: Draw the NH3NH_3 molecule and the BF3BF_3 molecule. Let's use dots (•) for N and H electrons, and crosses (×) for B and F electrons. Show the N atom with its lone pair (two dots). Show the B atom bonded to three F atoms. In the product, draw a bond between N and B, where the shared pair consists of the two dots from the nitrogen. Both N and B now have a full octet of electrons. The overall molecule is neutral, so no brackets or charge are needed.
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