(Solutions from Past Exam Questions)
1.
(i) metallic bond
strong electrostatic attraction between metallic cations & delocalised electron cloud
(ii) ionic bond
strong electrostatic attraction between Ca2+ and F- ions
2.
(i)
3.
4.
5.
½
1
½
1
(ii) co-ordination no. = 8
½
d-block elements have low-lying d-orbitals which can accept electron pairs from ligands
ions of d-block elements have a high charge-to-radius ratio
∴ can exert stronger attraction on the lone pairs of the ligands
1
1
K has electronic configuration of [Ar] 4s1
formation of K+ leads to a stable octet electronic structure
1
SO2 exists as simple molecules with weak van der Waals’ forces between them
Na2O and Al2O3 are ionic compounds with ionic bond stronger than van der Waals’ forces
½,½
½,½
Al3+ has higher charge : radius ratio than that of Na+
Al2O3(s) has a much higher/negative lattice energy than Na2O(s)
½
½
(ii) (I) Mg2+ and O2- are doubly charged, while Na+ and F- are singly charged
Ions in MgO(s) are more strongly binded than those in NaF(s) ∴MgO(s) has a much higher m.p.
1
1
6.
(II) The solubility of an ionic compound, MX(s), depends on the ∆Hlattice of MX(s) and
∆Hhydration of Mn+(aq) & Xn-(aq)
1
(1)
For compounds with small cations & anions (as in MgO & NaF), ∆Hlattice is the dominant factor
1
MgO(s) has a very negative ∆Hlattice
∴ MgO has a lower solubility than NaF(s)
1
7.
1×3
(i) atomic size: Li < Na < K
∴ attraction of nucleus on the delocalised electron decrease
strength of metallic bond decrease m.p. decrease
½
½
(ii) the atomic radii decreases and the no. of valence electrons involved in metallic bond increase
∴ attraction of nucleus on the delocalised electron increases b.p. increase
½
½
1
(iii) Both the magnitudes of the hydration enthalpy of the X2+ cation and the lattice enthalpy of XSO4 decreases
as the group is descended
With a large anion SO42-, the variation in ∆Hlattice is smaller than that in ∆Hhydration
∴ solubility of Group II sulphates(VI) decreases down the group
½,½
½,½
8.
LiNO3(s) decomposes to give Li2O, NO2 and O2 while KNO3(s) decomposes to give KNO2 and O2
Li+ has a much smaller size than K+
the more polarizing Li+ forms more stable ionic compounds with smaller (less polarizable) anions (O2-)
while K+ forms more stable ionic compounds with larger anions (NO2-)
( lattice made up of ions with comparable size is more stable)
1
½
½
9.
The solubility of an ionic compound depends on ∆Hsolution = ∆Hhydration - ∆Hlattice
OH- is a small anion
The difference in ∆Hlattice between Mg(OH)2 & Ba(OH)2 is much greater than the difference in ∆Hhydration between
Mg2+(aq) & Ba2+(aq) ions
∴ Ba(OH)2 is more soluble
½
½
10. (a) The solubility of Group II sulphate depends on the ∆Hlattice of sulphate and ∆Hhydration of M2+(aq) & SO42-(aq)
1
1
(1)
For ionic compounds with a large anion like SO42-, the solubility will mainly depend on the ∆Hhydration of M2+(aq)
1
The size of Mg2+ is smaller than that of Ba2+
the magnitude of ∆Hhydration[Mg2+(aq)] is large
∴ MgSO4 is more soluble in water
½
½
(b) MgSO4 as drying agent / laxative; BaSO4 as paint additive
11. True
12. (1) no observable change
(2) turns milky, then becomes clear again
CO2 + Ca(OH)2 CaCO3 + H2O
CaCO3 + CO2 + H2O Ca(HCO3)2
13.
14.
2
1+1
15. (a) Lithium carbonate decomposes on heating to give carbon dioxide
Li2CO3 → Li2O + CO2
but other alkali metal carbonates do not
1
(1)
1
Lithium hydroxide decomposes on heating to give Li2O and water
2LiOH → Li2O + H2O
but other alkali metal hydroxides do not
1
(1)
1
(b) (i) NaHCO3 decomposes on heating
to give CO2 gas which can turn limewater milky
but Na2CO3 does not decompose
½
1
½
or: dissolve the solid in water and add MgSO4(aq) to the solutions
Na2CO3(aq) gives a white ppt; but NaHCO3(aq) does not
½,½
½,½
(ii) carry out a flame test
KCl(s) gives a lilac flame while MgCl2(s) gives no characteristic flame colour
1
½,½
or: add NaOH(aq) / Na2CO3(aq) to aqueous solution of the substance
MgCl2 gives white ppt while KCl does not
(1)
(1)
(c) The solubility of an ionic compound depends on ∆Hsolution = ∆Hhydration - ∆Hlattice
SO42- is a large anion
The difference in ∆Hhydration between Mg2+(aq) & Ba2+(aq) ions is much greater than the difference in ∆Hlattice
between MgSO4 & BaSO4
∆Hhydration of Mg2+(aq) is more exothermic than that of Ba2+(aq)
due to the very small size of Mg2+ cation
16. (i) NaH is an ionic compound. Na atom donates electron to H atom to form Na+ & H- ions
Electrostatic attraction between ions is ionic bond
HCl is a covalent compound. H & Cl atom share a pair of electrons with each other to form a covalent bond
As Cl is more electronegative than H, HCl molecule is polar
Attraction between HCl molecules is dipole-dipole attraction
(ii) NaH(s) reacts with water to give NaOH(aq) & H2(g)
HCl ionizes in water to give H3O+(aq) ions
1
1
½
½
½
½
1
1
1
17. At temperature of a Bunsen flame, a compound will decompose to give gaseous atoms of its constituent elements.
1
When an electron absorbs energy from the flame, it will move from an orbital with lower energy to an
orbital with higher energy.
1
When an electron returns from an orbital of higher energy to an orbital of lower energy, a photon with energy equal
to the difference in energy of the two orbitals will be emitted. An emission line will be resulted.
1
Each element has a unique emission spectrum. A metal can be identified if it has a strong emission line in the
visible region of the electromagnetic radiation.
1
18. (i) (I) clean a platinum wire (dip Pt wire into conc. HCl & heat the wire until the flame has no noticeable color)
stick a sample of the salt onto the Pt wire with conc. HCl
heat the wire with the sample in a non-luminous Bunsen flame
1
1
1
(II) electron transition from a higher energy level to a lower energy level leads to the emission of a photon
with a wavelength in the visible (red in the case of Ca) region
(ii) heat a known mass (m1) of a sample of blackboard chalk in a crucible
until there is no further reduction in mass and record the final mass (m2)
m2
m
= 2
no of moles of CaSO4 =
40 + 32 + 16 × 4 136
m1 − m2
no of moles of H2O =
18
m − m2
m
no of moles of water of crystallization per formula unit of CuSO4 = 1
/ 2
18
136
19. (i) Ca : [Ar]
1
1
1
½
½
1
1
4s
(ii) limestone/marble/calcite etc.
1
3
(iii) amount of H+(aq) exchanged = 0.02 × 15 × 10-3 = 3.0 × 10-4 mol
2+
M (aq)
1
+
2 H (aq)
total hardness of water sample =
1
× 3.0 ×10 −4
2
−3
50 ×10
= 3 × 10-3 mol dm-3
1
20. Electrolysis of brine using steel as cathode and carbon as anode
At cathode: H+(aq) is discharged to give hydrogen gas [ 2H+(aq) + 2e- → H2(g) ]
At anode: Cl-(aq) is discharged to give chlorine gas [ 2 Cl-(aq) → Cl2(g) + 2e- ]
Na+(aq) and OH-(aq) ions remaining in the solution becomes sodium hydroxide
1
½
½
1
21. NaOH(aq) is produced by electrolysis of brine / concentrated NaCl(aq)
Other products: H2 / Cl2 / NaOCl (bleach solution) [any TWO]
1
½,½
22.
4