Q 1 – The ability of metals to be drawn into thin wire is known as:
(a) ductility
(b) malleability
(c) sonorousity
(d) conductivity
(a) ductility
Ductility is the property of metals by which they can be drawn into thin wires.
Q 2 – Aluminium is used for making cooking utensils. Which of the following properties of aluminium are responsible for the same?
(i) Good thermal conductivity
(ii) Good electrical conductivity
(iii) Ductility
(iv) High melting point
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i) and (iv)
(d) (i) and (iv)
Aluminium is suitable for cooking utensils because it has:
- Good thermal conductivity
- High melting point
Q 3 – What happens when calcium is treated with water?
(i) It does not react with water
(ii) It reacts violently with water
(iii) It reacts less violently with water
(iv) Bubbles of hydrogen gas formed stick to the surface of calcium
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (iii) and (iv)
(d) (iii) and (iv)
Calcium reacts less violently with water. Hydrogen bubbles formed during the reaction stick to its surface.
Reaction:
Ca + 2H₂O → Ca(OH)₂ + H₂↑
Q 4 – The composition of aqua-regia is
(a) Dil.HCl : Conc. HNO3
3 : 1
(b) Conc.HCl : Dil. HNO3
3 : 1
(c) Conc.HCl : Conc.HNO3
3 : 1
(d) Dil.HCl : Dil.HNO3
3 : 1
(c) Concentrated HCl : Concentrated HNO₃ = 3 : 1
This mixture is called aqua regia.
Q 5 – Silver articles become black on prolonged exposure to air. This is due to the formation of
(a) Ag3N
(b) Ag2O
(c) Ag2S
(d) Ag2S and Ag3N
(c) Ag2S
Silver becomes black due to the formation of silver sulphide (c) Ag2S when it reacts with traces of hydrogen sulphide in air.
Q 6 – Non-metals form covalent chlorides because
(a) they can give electrons to chlorine
(b) they can share electrons with chlorine
(c) they can give electrons to chlorine atoms to form chloride ions
(d) they cannot share electrons with chlorine atoms
(b) They can share electrons with chlorine
Non-metals generally form covalent compounds by sharing electrons.
Q 7 –The electronic configuration of three elements X, Y and Z are as follows:
X = 2, 4, Y = 2, 7, Z = 2,1 Which two elements will combine to form an ionic compound and write the correct formula,
(a) X2Y
(b) YZ
(c) XZ3
(d) Y2Z
(b) YZ
- X = 2,4 → needs 4 electrons
- Y = 2,7 → needs 1 electron
- Z = 2,1 → loses 1 electron
Y and Z form an ionic compound:
YZ
Q 8 – Which of the following non-metal is lustrous?
(a) Sulfur
(b) Oxygen
(c) Nitrogen
(d) Iodine
(d) Iodine
Iodine is a non-metal that has a lustrous appearance.
Q 9 – Which one among the following is an acidic oxide?
(a) Na2O
(b) CO
(c) CO2
(d) Al2O3
(c) CO2
Carbon dioxide is an acidic oxide.
Q 10 –In thermite welding a mixture of ____ and ____ is ignited with a burning magnesium ribbon which produces molten iron metal as large amount of heat is evolved.
(a) iron (III) oxide and aluminium powder
(b) iron (II) oxide and aluminium powder
(c) iron (III) chloride and aluminium powder
(d) iron (III) sulphate and aluminium powder
(a) Iron(III) oxide and aluminium powder
The thermite reaction is:
Fe2O3 + 2Al → Al2O3 + 2Fe + heat
Q 11 – Copper objects lose their shine and form green coating of
(a) Copper oxide
(b) Copper hydroxide and Copper oxide
(c) Basic Copper carbonate
(d) Copper carbonate
(c) Basic copper carbonate
Copper develops a green coating of basic copper carbonate on prolonged exposure to moist air.
Q 12 – Metals a burns in air, on heating to from an oxide A2O3 where another metals burns in air only on heating to from an oxide BO. The two oxide A2O3 and BO can react with hydrochloric acid as well sodium hydroxide solution to from the corresponding salts and water.
(a) What is the nature of Oxide A2O3 ?
(b) What is the nature of Oxide BO?
(c) Name one metal like A.
(d) Name one metal like B.
(a) Nature of A2O3: Amphoteric oxide
(b) Nature of BO: Amphoteric oxide
(c) One metal like A: Aluminium (Al)
(d) One metal like B: Zinc (Zn)
Both aluminium oxide (A2O3) and zinc oxide (ZnO) react with both acids and bases.
Q 13 – A metal A, which is used in thermite process, when heated with oxygen gives an oxide B, which is amphoteric in nature. Identify A and B. Write down the reactions of oxide B with HCl and NaOH.
The metal used in the thermite process is aluminium.
A = Aluminium (Al)
B = Aluminium oxide (A2O3)
Aluminium burns in oxygen:
4Al + 3O2 → 2 A2O3
Al2O3 is amphoteric.
Reaction with HCl:
Al2O3 + 6HCl → 2AlCl3 + 3H2O
Reaction with NaOH:
Al2O3 + 2NaOH → 2NaAlO2 + H₂O
Q 14 – What happens when
(a) ZnCO3 is heated in the absence of oxygen?
(b) a mixture of Cu2O and Cu2S is heated?
(a) Heating ZnCO3 in the absence of oxygen
Zinc carbonate undergoes thermal decomposition to form zinc oxide and carbon dioxide.
ZnCO3 → ZnO + CO2
(b) Heating a mixture of Cu2O and Cu2S
Copper(I) oxide reacts with copper(I) sulphide to form copper metal and sulphur dioxide.
2Cu2O + Cu2S → 6Cu + SO2
Q 15 – An element A reacts with water to form a compound B which is used in white washing. The compound B on heating forms an oxide C which on treatment with water gives back B. Identify A, B and C and give the reactions involved.
The elements/compounds are:
A = Calcium (Ca)
B = Calcium oxide (CaO)
C = Calcium hydroxide [Ca(OH)2]
However, the wording of the question describes the whitewashing cycle in the reverse order. The standard whitewashing reactions are:
Formation of calcium oxide:
CaCO₃ → CaO + CO2
Formation of calcium hydroxide:
CaO + H2O → Ca(OH)2
Calcium hydroxide is used for whitewashing.
On exposure to carbon dioxide, it forms calcium carbonate:
Ca(OH)2 + CO2 → CaCO3 + H2O
Q 16 – A solution of CuSO4 was kept in an iron pot. After few days the iron pot was found to have a number of holes in it. Explain the reason in terms of reactivity. Write the equation of the reaction involved.
Iron is more reactive than copper. Therefore, iron displaces copper from copper sulphate solution.
As a result, the iron pot gradually dissolves and holes are formed in it.
Reaction:
Fe + CuSO4 → FeSO4 + Cu
Thus, the formation of holes is due to the reaction of iron with CuSO4.
Q 17 – An element A burns with golden flame in air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equations for the reactions involved.
Element B has atomic number 17, so:
B = Chlorine (Cl)
An element that burns with a golden/yellow flame and reacts with chlorine is sodium.
Therefore:
A = Sodium (Na)
B = Chlorine (Cl)
C = Sodium chloride (NaCl)
D = Sodium hydroxide (NaOH)
Reaction 1:
2Na + Cl2 → 2NaCl
Electrolysis of aqueous NaCl:
2NaCl + 2H2O → 2NaOH + H2 + Cl2
Thus, the compound formed is NaOH, and hydrogen is liberated.
Q 18 – Iqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in reaction mixture. He made the same observations when this
element was treated with hydrochloric acid. Suggest how can he identify the produced gas. Write chemical equations for both the reactions.
The gas produced is hydrogen (H2).
The gas can be identified by bringing a burning splint near the gas. Hydrogen burns with a characteristic ‘pop’ sound.
Since M is divalent and reacts with both NaOH and HCl, it can be zinc.
M = Zinc (Zn)
Reaction with HCl:
Zn + 2HCl → ZnCl2 + H2↑
Reaction with NaOH:
Zn + 2NaOH → Na2ZnO2 + H2↑
Therefore, the gas evolved in both reactions is hydrogen.
Q 19 – Which of the following reactions will not occur? why not ?
(a) MgSO4 (aq) + Cu (s) → CuSO4 (aq) + Mg (s)
(b) CuSO4 (aq) + Fe (s) → FeSO4 (aq) + Cu (s)
(c) MgSO4 (aq) + Fe (s) → FeSO4 (aq) + Mg (s)
(a) MgSO4 + Cu → CuSO4 + Mg
Will NOT occur.
Copper is less reactive than magnesium. Therefore, Cu cannot displace Mg from MgSO4.
(b) CuSO4 + Fe → FeSO4 + Cu
Will occur.
Iron is more reactive than copper and displaces copper from CuSO4.
(c) MgSO4 + Fe → FeSO4 + Mg
Will NOT occur.
Iron is less reactive than magnesium and therefore cannot displace magnesium from MgSO4.
Final answer:
Reactions (a) and (c) will not occur.
Q 20 – Four elements A, B, C and D have the following electron arrangements in their atoms:
A. 2, 8, 6
B. 2, 8, 8
C. 2, 8, 8, 1
D. 2, 7
(a) What type of bond is formed when element when c & D reacts?
(b) Which element is an inert gas ?
(c) What will be the formula of the compound between A and C ?
Given electron arrangements:
- A = 2, 8, 6
- B = 2, 8, 8
- C = 2, 8, 8, 1
- D = 2, 7
(a) Type of bond formed
A has 6 valence electrons and C has 1 valence electron.
C loses one electron, while A needs two electrons. Therefore, two C atoms combine with one A atom, forming an ionic bond.
(b) Inert gas
B = 2,8,8
It has a completely filled outer shell.
Therefore, B is the inert/noble gas.
(c) Formula of compound between A and C
A needs 2 electrons and each C can donate 1 electron.
Therefore:
Formula = C2A
Q 21 – An elements X of atomic number 12 combines with an elements Y of atomic number 17 to from a compound XY2. State the nature of chemical bond in XY2 and show how the electron configurations of X and Y Change in the formation of this compound.
Atomic number of X = 12
Therefore:
X = Magnesium (Mg)
Electronic configuration:
Mg = 2, 8, 2
Atomic number of Y = 17
Therefore:
Y = Chlorine (Cl)
Electronic configuration:
Cl = 2, 8, 7
The compound is XY2 = MgCl2.
Nature of bond
The bond is an ionic/electrovalent bond.
Formation of Mg²⁺
Magnesium loses two electrons:
Mg → Mg²⁺ + 2e–
Configuration changes:
Mg: 2,8,2 → Mg²⁺: 2,8
Formation of Cl⁻
Each chlorine atom gains one electron:
Cl + e– → Cl –
Configuration:
Cl: 2,8,7 → Cl⁻: 2,8,8
Since one Mg atom loses 2 electrons, two chlorine atoms each gain 1 electron.
Therefore:
Mg²⁺ + 2Cl – → MgCl2
Hence, XY2 is an ionic compound.