The melting point of CaS is higher than that of KCl. Which of the following are explanations for the observation?
I Ca2+ is more positively charged than K+.
II S2− is more negatively charged than Cl−.
III The K+ ion is smaller than the Ca2+ ion.
2023 SJChO P1 Q36
Which statement about metals is not correct?
2024 USS O Prelim P2 Q8
Two statements about the structure and properties of metals are given.
statement 1: Metals are malleable and have high melting points.
statement 2: Metals have mobile electrons in their structure.
Which statement is correct?
2024 Woodlands O Prelim P1 Q9
The table shows the sixth ionisation energies of four consecutive elements in the Periodic Table.
| element | W | X | Y | Z |
|---|---|---|---|---|
| sixth ionisation energy / kJ mol−1 | 19 800 | 21 300 | 8500 | 9360 |
What is the formula of the fluoride of W?
Source: 2024 VJC J2 Prelim P1 Q2
Diazene has the molecular formula N2H2. Given that nitrogen and hydrogen show their usual numbers of bonds, which describes the shape and bond angle around each central atom? Each option gives the shape, then the bond angle.
2019 CJC J2 Prelim P1 Q4
Iodine and phosphorus each form a trifluoride. What are the shapes of these two molecules? Each option gives IF3, then PF3.
2020 ACJC J1 Promo Q6
The valence shell electron pair repulsion (VSEPR) theory is used to predict the bond angles of molecules. Which bond angle is correctly predicted by VSEPR?
| bond pairs around the central atom | lone pairs around the central atom | bond angle | |
|---|---|---|---|
| A | 2 | 1 | 105° |
| B | 2 | 2 | 118° |
| C | 3 | 1 | 107° |
| D | 3 | 2 | 90° |
2024 HCI J2 Prelim P1 Q3
Which one of the following pairs of compounds shows the same shape and similar bond angles?
2022 VJC J1 Promo P1 Q4
In which pair of molecules or ions do both species contain a bond angle of 107°?
SJChO Mock 3 P1 Q29
When an iodine atom is covalently bonded to an electron-withdrawing group X, the electron density on the large iodine atom becomes uneven. The tip of the iodine atom directly opposite the X–I bond becomes very electron-deficient, while the region round the middle of the iodine atom is electron-rich. The electron-deficient region can attract an anion A− without forming a covalent bond. What is the most likely X–I···A− bond angle for this attraction? (··· shows the non-covalent attraction.)
2023 SJChO P1 Q6
The table identifies the shape and polarity of four molecules. Which row is correct?
| molecule | molecular shape | polarity | |
|---|---|---|---|
| A | xenon tetrafluoride | tetrahedral | non-polar |
| B | selenium hexafluoride | T-shaped | non-polar |
| C | trichloromethane | tetrahedral | polar |
| D | boron trichloride | trigonal planar | polar |
2024 ASRJC J2 Prelim P1 Q4
The table shows the boiling points of some halogenoalkanes.
| compound | boiling point / °C |
|---|---|
| CH3CH2Cl | 12.3 |
| CH3CH2Br | 34.8 |
| CH3CH2I | 70.0 |
Which of the following correctly explains the difference in the boiling points?
1 The electronegativity difference between the halogen and carbon increases from C–Cl to C–I.
2 The strength of permanent dipole–permanent dipole attraction increases from C–Cl to C–I.
3 The strength of instantaneous dipole–induced dipole attraction increases from CH3CH2Cl to CH3CH2I.
4 The bond energy of the carbon–halogen bond decreases from C–Cl to C–I.
2019 YIJC J1 MYE P1 Q13
Which of the following substances consists of atoms or molecules held together only by instantaneous dipole–induced dipole forces of attraction?
2019 YIJC J1 MYE P1 Q14
Information about some hydrocarbons is given below.
| hydrocarbon | boiling point / °C |
|---|---|
| ethane, CH3CH3 | –89.0 |
| propane, CH3CH2CH3 | –42.0 |
| butane, CH3CH2CH2CH3 | –1.7 |
| 2-methylpropane, (CH3)3CH | –11.7 |
Which statements help to explain the variation in boiling points of the hydrocarbons?
1 Presence of more hydrogen atoms results in stronger hydrogen bonding.
2 Presence of more covalent bonds results in more energy required to break them.
3 Greater surface area of contact results in stronger instantaneous dipole–induced dipole interactions.
4 Presence of more electrons results in stronger instantaneous dipole–induced dipole interactions.
2020 VJC J1 Promo P1-P2 Q5
Information about the properties of some compounds is given below.
| compound | boiling point / °C |
|---|---|
| HF | 19.5 |
| NH3 | −33.3 |
| H2O | 100 |
Which statement about the hydrogen bonding in these compounds is correct?
2022 MI J2 Prelim P1 Q7
The boiling point of water (100 °C) is greater than that of HF (20 °C). Which statement is a correct explanation of this?
2023 NJC J2 Prelim P1 Q3
The boiling points of H2O, H2S and H2Se are different. Which statements are correct?
I H2S has a lower boiling point than H2Se because H2S molecules have weaker permanent dipole–permanent dipole forces of attraction.
II H2O molecules experience weaker temporary dipole–induced dipole forces of attraction than H2Se molecules.
III H2O molecules experience stronger permanent dipole–permanent dipole forces of attraction than H2S molecules.
SJChO Mock 3 P1 Q17
Use the table of densities and boiling points of five solvents.
| solvent | molecular formula | density / g cm−3 | boiling point / °C | |
|---|---|---|---|---|
| I | chloroform | CHCl3 | 1.49 | 61 |
| II | ethanol | C2H5OH | 0.79 | 78 |
| III | acetone | (CH3)2CO | 0.78 | 56 |
| IV | hexane | C6H14 | 0.66 | 69 |
| V | water | H2O | 1.00 | 100 |
Which pair of liquids can be separated by liquid–liquid extraction?
Source: 2023 SJChO P1 Q28
Which lattice structure is correctly described?
| lattice structure | description | |
|---|---|---|
| A | giant molecular | lattice of molecules held together by strong covalent bonds |
| B | giant ionic | each cation is only bonded to the anion to which electrons are transferred |
| C | giant metallic | all electrons are delocalised within the metal cation lattice |
| D | simple molecular | molecules held together by weak intermolecular forces |
2021 DHS J2 Prelim P1 Q4
Silicon carbide is a shiny, hard and chemically inert material with a very high melting point. It can be used to sharpen knives and make crucibles. Which type of structure explains these properties?
2021 RVHS J1 Common Test P1 Q9
Substances X, Y and Z are all solids. Some of their physical properties are given in the table.
| substance | X | Y | Z |
|---|---|---|---|
| melting point / °C | 772 | 114 | 1610 |
| boiling point / °C | 1407 | 183 | 2205 |
| electrical conductivity when molten | good | poor | poor |
What type of structure could each substance have? Each option gives X, Y and Z in that order.
2021 YIJC J1 MYE P1 Q10
Which row correctly describes the structure and bonding present in the solid lattice of the given substance?
| substance | structure | bonding | |
|---|---|---|---|
| A | ice | giant molecular | covalent and hydrogen bonding |
| B | iodine | simple molecular | covalent bonding and permanent dipoles |
| C | sodium sulfate | giant ionic | ionic and covalent bonding |
| D | copper | giant metallic | ionic bonding |
2023 DHS J2 Prelim P1 Q4
Which property of X2 increases down the group for X = Cl, Br or I?
2024 RI J2 Prelim P1 Q11
Which statement about covalent bonds is true?
2022 EJC J1 MYE P1 Q3, adapted
Consider these chlorine-containing species: ICl2−, ClF3, ClO3−. Which statements about them are correct?
1 The bond angles increase in the order ClF3, ClO3−, ICl2−.
2 Two of the three species have an overall dipole moment.
3 Only ClO3− is not planar.
2020 VJC J1 Promo P1-P2 Q4
CI and II only[1]
The attraction between ions is stronger when the charges are larger. Ca2+ and S2− carry charges of 2, K+ and Cl− only 1, so I and II both explain it.
III is false: K+ and Ca2+ have the same number of electrons, and Ca2+ has more protons, so Ca2+ is smaller.
Ca giant lattice of oppositely charged ions[1]
A metal is a lattice of positive ions only, held together by the sea of delocalised electrons. There are no negative ions.
The other statements are correct.
BBoth statements are correct but statement 2 does not explain statement 1.[1]
Both statements are true. Mobile electrons explain why metals conduct electricity.
Malleability comes from layers of ions sliding, and the high melting point from the strong metallic bonds, not from the electrons being mobile.
BWF4[1]
The sixth ionisation energy falls sharply from X to Y. For Y it is the last valence electron, so Y has 6 valence electrons. For X the sixth electron comes from an inner shell, so X has 5 valence electrons.
W is the element before X, with 4 valence electrons, and forms WF4.
Abent, 110°[1]
Diazene is H–N=N–H. Each N has a single bond to H, a double bond to N (one region) and one lone pair: 2 bond pairs and 1 lone pair.
Three regions round each N point to the corners of a triangle, and the lone pair squeezes the angle below 120°, so the shape is bent at about 110°.
BT-shaped, trigonal pyramidal[1]
I has 7 valence electrons: 3 bond pairs and 2 lone pairs. The 5 pairs form a trigonal bipyramid with both lone pairs in the middle, so IF3 is T-shaped.
P has 5 valence electrons: 3 bond pairs and 1 lone pair, so PF3 is trigonal pyramidal.
Crow C[1]
3 bond pairs and 1 lone pair give a trigonal pyramid with a bond angle of 107°, as in NH3.
2 bond pairs and 1 lone pair give a bent shape just under 120°. 2 bond pairs and 2 lone pairs give about 104.5°. 3 bond pairs and 2 lone pairs give a T-shape with angles just under 90°.
APOCl3 and CCl4[1]
POCl3 has four regions round P (a P=O double bond and three P–Cl bonds) and no lone pairs, so it is tetrahedral, like CCl4.
AlCl3 is trigonal planar but NCl3 is pyramidal. SO2 is bent but CO2 is linear. ClF3 is T-shaped but SF4 is see-saw.
DNH3 and H3O+[1]
NH3 and H3O+ each have 3 bond pairs and 1 lone pair round the central atom: trigonal pyramidal, 107°.
CH3+ has 3 bond pairs and no lone pair (120°). H2O and NH2− have 2 bond pairs and 2 lone pairs (about 104.5°).
B180°[1]
The anion is attracted to the electron-deficient (slightly positive) region of the iodine atom.
That region lies directly opposite the X–I bond, so X, I and A− lie in a straight line: 180°.
Crow C[1]
Trichloromethane, CHCl3, is tetrahedral round C, and the C–H and C–Cl dipoles do not cancel, so it is polar.
XeF4 is square planar, SeF6 is octahedral, and BCl3 is trigonal planar but non-polar because its dipoles cancel.
C3 only[1]
The number of electrons increases from CH3CH2Cl to CH3CH2I, so the instantaneous dipole–induced dipole attractions get stronger and the boiling point rises (3).
Electronegativity falls from Cl to I, so 1 and 2 are wrong. Statement 4 is true, but boiling does not break covalent bonds, so it does not explain the trend.
ACO2[1]
CO2 molecules are non-polar, so only instantaneous dipole–induced dipole forces act between them.
Cu has metallic bonds, H2O has hydrogen bonds as well, and SiO2 is held by covalent bonds throughout.
C3 and 4 only[1]
From ethane to butane the number of electrons rises, so the instantaneous dipole–induced dipole forces get stronger (4). Butane and 2-methylpropane have the same number of electrons, but the chain-shaped butane has more surface contact (3).
Hydrocarbons have no H on F, O or N, so there is no hydrogen bonding, and boiling breaks no covalent bonds.
DNH3 forms few hydrogen bonds on average and has the smallest bond dipole.[1]
NH3 has only one lone pair, so it forms 1 hydrogen bond per molecule on average, and N–H is the least polar of the three bonds, so each hydrogen bond is the weakest.
HF and NH3 both form 1 per molecule. H–F is more polar than H–O. Water boils highest because it forms 2 per molecule, not because it dimerises.
CThere are, on average, more hydrogen bonds between water molecules.[1]
Water has 2 H atoms and 2 lone pairs, so it forms 2 hydrogen bonds per molecule on average. HF has only 1 H atom, so it forms 1.
H–F is more polar than H–O, so each hydrogen bond between HF molecules is actually stronger: the extent, not the strength, explains the result.
CII and III only[1]
H2O has far fewer electrons than H2Se, so its instantaneous dipole–induced dipole (temporary dipole) forces are weaker (II). O is more electronegative than S, so H2O is more polar and has stronger permanent dipole–permanent dipole forces than H2S (III).
I is wrong: H2S is the more polar molecule. H2Se boils higher because of its stronger instantaneous dipole–induced dipole forces from its larger number of electrons.
BIV and V[1]
Liquid–liquid extraction separates liquids that are immiscible, because they form two layers.
Hexane (IV) is non-polar and water (V) is polar, so they do not mix. The pairs in the other options mix: ethanol and hexane (A), acetone and hexane (C), and chloroform and ethanol (D).
Drow D[1]
Simple molecular substances are separate molecules held to one another by weak intermolecular forces.
A giant molecular lattice is atoms, not molecules, joined by covalent bonds. Each ion attracts all the oppositely charged ions round it. Only the valence electrons of a metal are delocalised.
Ba giant tetrahedral network of covalent bonds[1]
Hardness and a very high melting point mean many strong covalent bonds throughout the solid, as in diamond: each Si and C bonded to 4 others in a tetrahedral network.
Si and C are both non-metals, so they do not form ions. A layer structure would be soft, and a simple molecular structure would melt at a low temperature.
Cionic, simple molecular, giant molecular[1]
X melts high and conducts when molten: ionic. Y melts low and does not conduct: simple molecular.
Z melts very high but does not conduct when molten: giant molecular.
Crow C[1]
Sodium sulfate is a lattice of Na+ and SO42− ions (ionic bonding), and the S and O atoms inside each sulfate ion are joined by covalent bonds.
Ice is simple molecular. Iodine is non-polar, so it has instantaneous dipole–induced dipole forces, not permanent dipoles. Copper has metallic bonding.
Bbond length[1]
The atoms get larger down the group, so the distance between the two nuclei increases. Their larger, more diffuse orbitals overlap less effectively, so the bond energy falls: Cl–Cl 244, Br–Br 193 and I–I 151 kJ mol−1.
The number of electrons increases, so the instantaneous dipole–induced dipole forces get stronger and the halogens get less volatile.
DA dative bond can be formed from the head-on overlap of orbitals.[1]
Once formed, a dative bond is identical to any other covalent bond, so it can be a σ bond.
A is false: π bonds are weaker than σ bonds, as sideways overlap is less effective than head-on overlap. B is false: when the number of shared pairs increases, the bond length usually decreases. C is false: some non-polar bonds are stronger than polar bonds, because the number of shared pairs and how well the orbitals overlap matter more.
A1, 2 and 3[1]
ICl2− is linear (2 bond pairs, 3 lone pairs), planar, 180°. ClF3 is T-shaped (3 bond pairs, 2 lone pairs), planar, 90°. ClO3− is trigonal pyramidal (3 bond pairs, 1 lone pair), 107°.
1 is correct: the bond angles are ClF3 < ClO3− < ICl2−. 2 is correct: ClF3 and ClO3− have a net dipole moment, and only ICl2− is non-polar. 3 is correct: only ClO3− is not planar.