A Data Booklet may be used.
CommonThe Valence Shell Electron Pair Repulsion theory (VSEPR) is used to predict the bond angles of molecules. Which bond angle is correctly predicted by VSEPR?
| number of bonded electron pairs around central atom | number of lone pairs around central atom | bond angle | |
|---|---|---|---|
| A | 2 | 1 | 105° |
| B | 2 | 2 | 118° |
| C | 3 | 1 | 107° |
| D | 3 | 2 | 90° |
Source: 2024 HCI J2 Prelim P1 Q3
Which bonding type does not correspond to its description of physical properties?
| bonding type | physical properties | |
|---|---|---|
| A | giant covalent | high melting point, conducts electricity when in solution but not when solid |
| B | simple covalent | low melting point, does not conduct electricity in any state |
| C | metallic | variety of melting points, conducts electricity when solid and when molten |
| D | ionic | high melting point, conducts electricity when molten but not when solid |
Source: A Level 2013 P1 Q5, via EJC 2022 JC1 Chemical Bonding Tutorial, A11
Which of the following statements describes a phenomenon which can be explained by intermolecular hydrogen bonding?
Source: A Level 1990 P1 Q4 (modified), via EJC 2022 JC1 Chemical Bonding Tutorial, A10
Predict the most likely bond angle at each nitrogen atom in hydrazine, N2H4.
Source: A Level 2009 P1 Q19 (modified), via EJC 2022 JC1 Chemical Bonding Tutorial, A5
CommonSodium amide, NaNH2, is a salt that dissolves in liquid ammonia to form a solution containing NH2−.
Source: 2023 HCI J2 Prelim P3 Q1(a)
The type of bonding present in a binary compound can be predicted from the electronegativities of the elements involved.
Source: 2023 VJC J2 Prelim P2 Q2(a)(i) and (ii)
Source: 2023 SAJC J2 Prelim P3 Q1(a)(ii)
Source: 2025 NJC J2 Prelim P2 Q3(a)(ii)
Source: 2018 RI J1 Promo Section C Q1(a)
In each pair, which compound has the greater covalent character, and which would you expect to have the higher melting point? Explain.
Source: RJC H2 Chemistry Tutorial 3a (Chemical Bonding I), Q7
Source: RJC H2 Chemistry Tutorial 3b (Chemical Bonding II), Q3(a)
Source: 2023 SAJC J2 Prelim P2 Q2(a)(iii)
Silane, SiH4, is a key raw material in the production of semiconductor-grade silicon.
Source: 2023 YIJC J2 Prelim P2 Q1(b)
Source: 2023 RI J2 Prelim P3 Q1(a)(i)
Source: 2020 YIJC JC1 Chemical Bonding Tutorial, Section C2 Q6, adapted
The bond energies and the boiling points of three halogens are given.
| halogen | bond energy / kJ mol−1 | boiling point / °C |
|---|---|---|
| Cl2 | 244 | −34 |
| Br2 | 193 | 59 |
| I2 | 151 | 184 |
Source: 2020 YIJC JC1 Intermolecular Forces Tutorial, Section E Q6
ChallengeExplain why the first substance in each pair has a higher boiling point than the second.
Source: RJC H2 Chemistry Tutorial 3b (Chemical Bonding II), Q2(a) and (c)
C3 bond pairs, 1 lone pair: trigonal pyramidal, 107°[1]
Based on the number of bond pairs and lone pairs listed in the question, these are the corresponding geometries and bond angles:
Agiant covalent[1]
Giant covalent substances have high melting points and generally do not conduct electricity (except graphite) due to the absence of mobile charge carriers.
DIce has a lower density than water at 0 °C.[1]
Intermolecular hydrogen bonding holds ice molecules in place in an orderly open structure with a large amount of empty space in between. This takes up more space than the more random arrangement of hydrogen bonds in liquid water, resulting in a larger volume and hence lower density.
A107°[1]
Since there are 3 bond pairs and 1 lone pair around each N atom in hydrazine, the shape is trigonal pyramidal and the most likely angle is 107°.
Electron pairs around the central N atom are arranged as far apart as possible to minimise repulsion.[½]
Lone pair–lone pair repulsion is greater than lone pair–bond pair repulsion, which is greater than bond pair–bond pair repulsion.[½]
NH3: trigonal pyramidal about N and 107°, due to 1 lone pair and 3 bond pairs[1]
NH2−: bent about N and 104.5°, due to 2 lone pairs and 2 bond pairs[1]
or NH2− is bent about N with a bond angle of 104.5°, whereas NH3 is trigonal pyramidal about N with a bond angle of 107°, as NH2− has lone pair–lone pair repulsion which is absent in NH3.
More energy is needed to overcome the stronger ionic bonding between Na+ and NH2− ions than the weaker hydrogen bonding between NH3 molecules.[2]
NoteName the particles: "between Na+ and NH2−" for the ionic bonding, and "between molecules" for the hydrogen bonding. Each was worth [½].
Electronegativity increases across a period. Nuclear charge increases (or number of protons increases) while shielding effect is relatively constant (or number of inner shell electrons is the same).[1]
Hence there is an increase in effective nuclear charge, leading to stronger attraction between the nucleus and the electrons, and so the ability of the atom to attract electrons to itself increases.[1]
Type of bonding in NaF: ionic
Type of bonding in SiF4: covalent
Type of bonding in Mg-Al: metallic[2]
The carbonate ion has 3 bond pairs and 0 lone pairs around the central C.[1]
To minimise (electrostatic) repulsion (between the electron pairs), the shape of the carbonate ion will be trigonal planar.[1]
For CH3OCH3, there are 2 bond pairs and 2 lone pairs of electrons around the O atom.[1]
By VSEPR theory, electron pairs around the central O atom are arranged as far apart as possible to minimise repulsion. Since lone pair–lone pair repulsion is greater than lone pair–bond pair repulsion, which is greater than bond pair–bond pair repulsion, the bond angle is 105° or 104.5°.[1]
NoteCommon mistakes: not stating the order of the three repulsions, and leaving out "arranged as far apart as possible to minimise repulsion".
Both have simple molecular structures. CH3CH2CH2CH3 molecules form weaker instantaneous dipole–induced dipole interactions with one another, while CH3CH2CH2CH2Cl molecules form stronger permanent dipole–permanent dipole interactions.[1]
Less energy is needed to overcome the weaker instantaneous dipole–induced dipole interactions than that needed to overcome the stronger permanent dipole–permanent dipole interactions. Hence, CH3CH2CH2CH3 has a lower boiling point than CH3CH2CH2CH2Cl.[1]
NoteBoiling overcomes the intermolecular forces, not the covalent bonds. Write the names of the interactions in full.
NaI has greater covalent character as its anion is larger and more easily polarisable than that of NaCl.[1]
NaCl: Cl⁻ is distorted slightlyNaI: the larger I⁻ is distorted more
NaCl has the higher melting point as its lattice energy is more exothermic than that of NaI, since there is greater attraction between Na+ and the smaller Cl−.[1]
MgO has greater covalent character as its cation is more highly charged than Na+, so it polarises O2− to a greater extent.[1]
Na₂O: Na⁺ distorts O²⁻ slightlyMgO: the smaller, more highly charged Mg²⁺ distorts O²⁻ more
MgO has the higher melting point as Mg2+, being more highly charged than Na+, attracts the O2− more strongly. Also, the ionic radius of Mg2+ is smaller than that of Na+, so Mg2+ is closer to O2−, causing the electrostatic attraction to be stronger.[1]
In HF, each fluorine atom has three lone pairs and is bonded to only one hydrogen atom. In water, the oxygen atom has two lone pairs and is bonded to two hydrogen atoms. Since a hydrogen bond is formed between a hydrogen atom bonded to electronegative N, O or F and a lone pair of electrons on N, O or F, hydrogen bonding in H2O is more extensive than between HF molecules.[1]
Thus, more energy is required to overcome the more extensive hydrogen bonds between water molecules than the hydrogen bonds between HF molecules, and water has a higher boiling point.[1]
Calcium has a giant metallic structure.[1]
Electrostatic forces of attraction (metallic bonding) exist between the Ca2+ ions and the sea of delocalised valence electrons.[1]
[1]
The molecular shape about Si is tetrahedral.[1]
Silane is non-polar, as the dipole moments of the four polar Si–H bonds, arranged tetrahedrally, cancel each other out, resulting in a net zero dipole moment.[1]
Both diamond and silicon have giant molecular structures. During melting, the strong Si–Si and C–C covalent bonds between the atoms are broken in silicon and diamond respectively.[1]
Si–Si covalent bonds are weaker than C–C covalent bonds, as the valence orbitals of Si are more diffuse than those of C, so the overlap of the orbitals of Si is less effective. Hence less energy is required to break the weaker Si–Si bonds during melting than the C–C bonds in diamond.[1]
A σ bond is formed when two orbitals overlap head-on. In an oxygen molecule, the head-on overlap is between two p orbitals.[1]
A π bond is formed when the p orbitals of the two atoms overlap sideways.[1]
(i) C2H6: 7 σ, 0 π. (ii) C2H4: 5 σ, 1 π. (iii) C2H2: 3 σ, 2 π. (iv) CO2: 2 σ, 2 π.[2]
The size of the valence orbital of the atoms increases down the group from Cl to Br to I, so the overlap between the orbitals is less effective.[1] The covalent bond formed is weaker and has a lower bond energy.[1]
The halogens have a simple molecular structure and are non-polar, with instantaneous dipole–induced dipole interactions between the molecules. The electron cloud gets larger and more polarisable from Cl2 to Br2 to I2, so these interactions get stronger.[1] More energy is required to overcome them, so the boiling point increases.[1]
SiCl4 has a tetrahedral molecular shape, so it is non-polar and its dominant intermolecular forces of attraction are instantaneous dipole–induced dipole interactions. PCl3 has a trigonal pyramidal molecular shape and is polar. It has both permanent dipole–permanent dipole and instantaneous dipole–induced dipole interactions.[1]
Even though PCl3 (66 electrons) has fewer electrons than SiCl4 (82 electrons), the additional, stronger permanent dipole–permanent dipole interactions cause the boiling point of PCl3 to be higher than that of SiCl4.[1]
The greater number of electrons in iodine (106 electrons) causes the electron cloud to be easily polarisable.[1]
This gives rise to relatively strong instantaneous dipole–induced dipole interactions, which are stronger than the hydrogen bonding in water (10 electrons). This results in a higher boiling point for iodine.[1]