Home / AS & A Level Chemistry 3.4 Covalent bonding and coordinate (dative covalent) bonding Exam Style Practice Questions Paper 1
The diagram shows the bonding in a molecule of propyne.
Which types of hybridisation are shown by the carbon atoms in propyne?
Correct Answer: \( \boxed{\mathrm{B}} \)
Propyne has the structure:
\(\mathrm{CH\equiv CCH_3}\)
No carbon atom is \(\mathrm{sp^2}\)-hybridised.
Therefore, the correct answer is (B).
The bonding between two atoms of nitrogen in an \(\mathrm{N_2}\) molecule involves the hybridisation of atomic orbitals to form sp orbitals.
Which row is correct?
| Formation of the \(\sigma\) bond between the nitrogen atoms in \(\mathrm{N_2}\) | Type of orbital which contains the lone pair of electrons on each nitrogen atom in \(\mathrm{N_2}\) | |
|---|---|---|
| A | an sp orbital from one atom overlaps with an sp orbital of the other atom | p |
| B | an sp orbital from one atom overlaps with an sp orbital of the other atom | sp |
| C | an s orbital from one atom overlaps with a p orbital of the other atom | p |
| D | an s orbital from one atom overlaps with a p orbital of the other atom | sp |
Correct Answer: \( \boxed{\mathrm{B}} \)
In \(\mathrm{N_2}\), each nitrogen atom is sp hybridised.
The \(\sigma\) bond is formed by overlap of one sp orbital from each nitrogen atom.
The remaining sp orbital on each nitrogen contains the lone pair, while the two unhybridised p orbitals form the two \(\pi\) bonds.
Therefore, the correct answer is (B).
\(\mathrm{NH_3}\) and \(\mathrm{HCN}\) react together to form \(\mathrm{NH_4CN}\), an ionic compound.
Which row states the number of coordinate bonds and the number of \(\pi\) bonds in one formula unit of \(\mathrm{NH_4CN}\)?
| Number of coordinate bonds | Number of \(\pi\) bonds | |
|---|---|---|
| A | 0 | 2 |
| B | 1 | 2 |
| C | 0 | 3 |
| D | 1 | 3 |
Correct Answer: \( \boxed{\mathrm{B}} \)
In \(\mathrm{NH_4CN}\), the ions present are \(\mathrm{NH_4^+}\) and \(\mathrm{CN^-}\).
The ammonium ion, \(\mathrm{NH_4^+}\), contains one coordinate (dative covalent) bond formed when the lone pair on nitrogen is donated to a hydrogen ion.
The cyanide ion, \(\mathrm{CN^-}\), contains a carbon-nitrogen triple bond consisting of one \(\sigma\) bond and two \(\pi\) bonds.
Therefore, one formula unit of \(\mathrm{NH_4CN}\) contains:
Therefore, the correct answer is (B).