📘 Model Test Questions Class 11 Chemistry Test # 1 for Chapter # 1 (Chemistry of Representative Elements)
Prepared by Inam Jazbi – Learn Chemistry
📘 Short Answer-Questions (SLO based)
Prepared by Inam Jazbi – Learn Chemistry
Q1.
Give reason for the following:
▶ Ionization energy decreases from top to bottom in s-block elements.
▶ Boiling points of halogens increase down the group in the periodic table.
▶ Gallium has smaller atomic radii than aluminium despite being below the aluminium in group IIIA.
▶ Electronegativities of alkali metals decrease from Li to Cs.
▶ Acidity of hydrogen halides increase from HF to HI.
▶ Fluorine is the strongest oxidizing agent.
▶ The ionization energy of Ga is unexpectedly higher than Al in group IIIA.
▶ The ionization energy of thallium is unexpectedly higher than indium in group IIIA.
▶ Electronegativity of Ga is higher than Al.
▶ Electronegativity of In is higher than Ga.
▶ Electronegativity of Tl is higher than In.
▶ IE of Sn is less than Pb.
▶ Chlorine can displace bromide and iodide from their salts but fluorine cannot do this job. Why?
▶ Ionization energy decreases from top to bottom in s-block elements.
▶ Boiling points of halogens increase down the group in the periodic table.
▶ Gallium has smaller atomic radii than aluminium despite being below the aluminium in group IIIA.
▶ Electronegativities of alkali metals decrease from Li to Cs.
▶ Acidity of hydrogen halides increase from HF to HI.
▶ Fluorine is the strongest oxidizing agent.
▶ The ionization energy of Ga is unexpectedly higher than Al in group IIIA.
▶ The ionization energy of thallium is unexpectedly higher than indium in group IIIA.
▶ Electronegativity of Ga is higher than Al.
▶ Electronegativity of In is higher than Ga.
▶ Electronegativity of Tl is higher than In.
▶ IE of Sn is less than Pb.
▶ Chlorine can displace bromide and iodide from their salts but fluorine cannot do this job. Why?
Q2.
What is flame test? Give its basis? Mention the colour flame of alkali metals and alkaline earth metals.
Q3.
What is meant by a diagonal relationship? Mention three pairs of representative elements that show diagonal relationship. Write down three points to show the similarity of diagonal members of group IA and IIA, IIA and IIIA and IIIA and IVA.
Q4.
Discuss the group trend of ionization energy and atomic radii in group IIIA and IVA of the periodic table.
Q5.
Write down four properties of beryllium that show its unique behaviour in group IIA.
Q6.
What is Electrical Conductivity? Briefly explain group trend of electrical conductivity of representative elements.
Q7.
Write down action of oxygen and water on s-block elements with balanced chemical equation.
Q8.
Complete and balance the following chemical equations:
Li(s) + H2(g) →
Mg(s) + H2(g) →
Na(s) + Cl2(g) →
Be(s) + Cl2(g) →
Na(s) + N2(g) →
Ca(s) + N2(g) →
Li(s) + N2(g) →
Li(s) + O2(g) →
Na(s) + O2(g) →
Na(s) + O2(g) — (Excess) →
K(s) + O2(g) →
Rb(s) + O2(g) →
Cs(s) + O2(g) →
Be(s) + O2(g) →
Sr(s) + O2(g) →
Li(s) + H2(g) →
Mg(s) + H2(g) →
Na(s) + Cl2(g) →
Be(s) + Cl2(g) →
Na(s) + N2(g) →
Ca(s) + N2(g) →
Li(s) + N2(g) →
Li(s) + O2(g) →
Na(s) + O2(g) →
Na(s) + O2(g) — (Excess) →
K(s) + O2(g) →
Rb(s) + O2(g) →
Cs(s) + O2(g) →
Be(s) + O2(g) →
Sr(s) + O2(g) →
Q9.
Explain the auto oxidizing and reducing properties of chlorine.
Q10.
Write the balanced equations for the following chemical processes:
▶ A piece of aluminium is dropped into concentrated sulphuric acid.
▶ Ferric chloride is mixed in an aqueous solution of caustic soda.
▶ Magnesium is heated with nitrogen gas.
▶ Potassium is put into ethyl alcohol.
▶ Chlorine gas is passed through an aqueous solution of caustic soda.
▶ A piece of aluminium is dropped into concentrated sulphuric acid.
▶ Ferric chloride is mixed in an aqueous solution of caustic soda.
▶ Magnesium is heated with nitrogen gas.
▶ Potassium is put into ethyl alcohol.
▶ Chlorine gas is passed through an aqueous solution of caustic soda.
Q11.
Write down complete balanced action of following reactions:
(i) Bleaching powder is dissolved in water.
(ii) Fluorine reacts with oxygen.
(iii) Aluminium reacts with water.
(iv) Silicon reacts with steam.
(v) Phosphorus reacts vigorously with water.
(vi) Sulphur reacts at high temperature with water.
(vii) Silicon is heated with nitrogen at high temperatures.
(viii) Phosphorus reacts with nitrogen at high temperature.
(ix) Chlorine reacts with nitrogen.
(x) Nitrogen reacts with oxygen in the presence of catalyst.
(i) Bleaching powder is dissolved in water.
(ii) Fluorine reacts with oxygen.
(iii) Aluminium reacts with water.
(iv) Silicon reacts with steam.
(v) Phosphorus reacts vigorously with water.
(vi) Sulphur reacts at high temperature with water.
(vii) Silicon is heated with nitrogen at high temperatures.
(viii) Phosphorus reacts with nitrogen at high temperature.
(ix) Chlorine reacts with nitrogen.
(x) Nitrogen reacts with oxygen in the presence of catalyst.
📘 Long Questions (SLO based)
Prepared by Inam Jazbi – Learn Chemistry
Q1.
Discuss the group trend of atomic radii, melting and boiling points, oxidation states and electronegativity of representative elements.
📘 MCQs
Prepared by Inam Jazbi – Learn Chemistry
1. Melting and boiling points of which of the following group of representative elements decrease regularly down the group?
🟥 (a) Group IA
🟦 (b) Group IIIA
🟩 (c) Group VIIA
🟨 (d) Group VIIIA
✔ Correct Answer: 🟥 (a) Group IA
Reason: Metallic bonding becomes weaker down Group IA, so melting and boiling points decrease.
Reason: Metallic bonding becomes weaker down Group IA, so melting and boiling points decrease.
2. Which of the following s-block element forms super oxide when burned in air?
🟥 (a) Li
🟦 (b) Na
🟩 (c) K
🟨 (d) Mg
✔ Correct Answer: 🟩 (c) K
Reason: Potassium reacts with oxygen to form potassium superoxide (KO2).
Reason: Potassium reacts with oxygen to form potassium superoxide (KO2).
3. Which of the following formula of nitrides for alkaline earth metal is possible?
🟥 (a) MN3
🟦 (b) M2N3
🟩 (c) M3N
🟨 (d) M3N2
✔ Correct Answer: 🟨 (d) M3N2
Reason: Alkaline earth metals are M2+ and nitride ion is N3−, giving M3N2.
Reason: Alkaline earth metals are M2+ and nitride ion is N3−, giving M3N2.
4. The flame colour of which of the following alkali metal is yellow?
🟥 (a) Na
🟦 (b) K
🟩 (c) Rb
🟨 (d) Cs
✔ Correct Answer: 🟥 (a) Na
Reason: Sodium gives a characteristic bright golden-yellow flame.
Reason: Sodium gives a characteristic bright golden-yellow flame.
5. The chemical used in fireworks is:
🟥 (a) Sodium bicarbonate
🟦 (b) Bleaching powder
🟩 (c) Potassium nitrate
🟨 (d) Potash alum
✔ Correct Answer: 🟩 (c) Potassium nitrate
Reason: Potassium nitrate acts as a strong oxidizing agent in fireworks and gunpowder.
Reason: Potassium nitrate acts as a strong oxidizing agent in fireworks and gunpowder.
6. Cathode in Castner Kellner cell is:
🟥 (a) Titanium blocks
🟦 (b) Carbon rods
🟩 (c) Mercury
🟨 (d) Iron container
✔ Correct Answer: 🟩 (c) Mercury
Reason: Mercury acts as the cathode and forms sodium amalgam during electrolysis.
Reason: Mercury acts as the cathode and forms sodium amalgam during electrolysis.
7. The diagonal member of beryllium is:
🟥 (a) Mg
🟦 (b) Al
🟩 (c) Si
🟨 (d) C
✔ Correct Answer: 🟦 (b) Al
Reason: Beryllium shows a diagonal relationship with aluminium due to similar charge density.
Reason: Beryllium shows a diagonal relationship with aluminium due to similar charge density.
8. Purification of sulphur dioxide from arsenic oxide is an essential step in contact process to avoid:
🟥 (a) Catalyst poisoning
🟦 (b) Temperature elevation
🟩 (c) Pressure controlling
🟨 (d) Air mixing
✔ Correct Answer: 🟥 (a) Catalyst poisoning
Reason: Arsenic oxide poisons the V2O5 catalyst used in the contact process.
Reason: Arsenic oxide poisons the V2O5 catalyst used in the contact process.
9. Oil of vitriol refers to:
🟥 (a) Borax
🟦 (b) Sulphuric Acid
🟩 (c) Alum
🟨 (d) Caustic soda
✔ Correct Answer: 🟦 (b) Sulphuric Acid
Reason: Concentrated sulphuric acid is historically known as oil of vitriol.
Reason: Concentrated sulphuric acid is historically known as oil of vitriol.
10. The best oxidizing agent among halogens is:
🟥 (a) F2
🟦 (b) Cl2
🟩 (c) Br2
🟨 (d) I2
✔ Correct Answer: 🟥 (a) F2
Reason: Fluorine has the highest reduction potential, making it the strongest oxidizing agent.
Reason: Fluorine has the highest reduction potential, making it the strongest oxidizing agent.
11. The electronic configuration of s-block metal, M is 1s2 2s2 2p6 3s1. The formula of its oxide would be:
🟥 (a) MO
🟦 (b) M2O
🟩 (c) M2O2
🟨 (d) MO2
✔ Correct Answer: 🟦 (b) M2O
Reason: M belongs to Group IA (+1 oxidation state), so its oxide is M2O.
Reason: M belongs to Group IA (+1 oxidation state), so its oxide is M2O.
12. The element lithium bears resemblance with:
🟥 (a) Al
🟦 (b) Mg
🟩 (c) Si
🟨 (d) None of them
✔ Correct Answer: 🟦 (b) Mg
Reason: Lithium shows a diagonal relationship with magnesium.
Reason: Lithium shows a diagonal relationship with magnesium.
13. The electronic configuration of a metal, M is 1s2 2s2. The diagonal member of this element has electronic configuration:
🟥 (a) 1s2 2s1
🟦 (b) 1s2 2s2 2p6 3s1
🟩 (c) 1s2 2s2 2p6 3s2 3p1
🟨 (d) 1s2
✔ Correct Answer: 🟩 (c) 1s2 2s2 2p6 3s2 3p1
Reason: The given configuration is Be; its diagonal member is Al.
Reason: The given configuration is Be; its diagonal member is Al.
14. Which of the following s-block element only forms normal oxide when burned in air?
🟥 (a) Li
🟦 (b) Be
🟩 (c) Mg
🟨 (d) All of them
✔ Correct Answer: 🟨 (d) All of them
Reason: Li, Be and Mg form only normal oxides on burning in air.
Reason: Li, Be and Mg form only normal oxides on burning in air.
15. Which of the following alkali metal only forms normal oxide when burned in air?
🟥 (a) Li
🟦 (b) Be
🟩 (c) Ca
🟨 (d) Na
✔ Correct Answer: 🟥 (a) Li
Reason: Lithium forms only the normal oxide (Li2O), unlike other alkali metals which form peroxides or superoxides.
Reason: Lithium forms only the normal oxide (Li2O), unlike other alkali metals which form peroxides or superoxides.
16. Which of the following formula of nitrides for alkali metal is possible?
🟥 (a) MN3
🟦 (b) M2N3
🟩 (c) M3N
🟨 (d) MN
✔ Correct Answer: 🟩 (c) M3N
Reason: Alkali metals are monovalent (M+), so their nitride formula is M3N.
Reason: Alkali metals are monovalent (M+), so their nitride formula is M3N.
17. Which of the following formula of nitrides for Group IIIA is possible?
🟥 (a) MN3
🟦 (b) M2N3
🟩 (c) M3N
🟨 (d) MN
✔ Correct Answer: 🟥 (d) MN3
Reason: Group IIIA elements have a +3 oxidation state, giving the nitride formula MN.
Reason: Group IIIA elements have a +3 oxidation state, giving the nitride formula MN.
18. Which group elements form nitrides with general formula M3N4?
🟥 (a) IIIA
🟦 (b) IVA
🟩 (c) VA
🟨 (d) IIA
✔ Correct Answer: 🟦 (b) IVA
Reason: Group IVA elements are tetravalent (+4), so their nitrides have the formula M3N4.
Reason: Group IVA elements are tetravalent (+4), so their nitrides have the formula M3N4.
19. Which group elements form nitrides with general formula M3N5?
🟥 (a) IIIA
🟦 (b) IVA
🟩 (c) VA
🟨 (d) IIA
✔ Correct Answer: 🟩 (c) VA
Reason: Group VA elements have a +5 oxidation state, producing nitrides of formula M3N5.
Reason: Group VA elements have a +5 oxidation state, producing nitrides of formula M3N5.
20. Which group elements form superoxides with general formula MO2?
🟥 (a) IA
🟦 (b) IVA
🟩 (c) IIIA
🟨 (d) IIA
✔ Correct Answer: 🟥 (a) IA
Reason: Heavy alkali metals (K, Rb and Cs) form superoxides with the general formula MO2.
Reason: Heavy alkali metals (K, Rb and Cs) form superoxides with the general formula MO2.
21. Which of the following gas is evolved when phosphorus reacts with water?
🟥 (a) NH3
🟦 (b) PH3
🟩 (c) H2
🟨 (d) H2S
✔ Correct Answer: 🟦 (b) PH3
Reason: Phosphorus reacts with hot water/steam to produce phosphine (PH3).
Reason: Phosphorus reacts with hot water/steam to produce phosphine (PH3).
22. The general formula of alkali metal alkoxide is:
🟥 (a) C2H5OM
🟦 (b) C2H5COONa
🟩 (c) (C2H5)2M
🟨 (d) None of them
✔ Correct Answer: 🟥 (a) C2H5OM
Reason: Alkali metals react with alcohols to form alkoxides having the general formula R–OM.
Reason: Alkali metals react with alcohols to form alkoxides having the general formula R–OM.
23. The chemical formula of lithium ethoxide is:
🟥 (a) C2H5OLi
🟦 (b) C2H5COOLi
🟩 (c) (C2H5)2Li
🟨 (d) None of them
✔ Correct Answer: 🟥 (a) C2H5OLi
Reason: Lithium ethoxide is formed when lithium reacts with ethanol.
Reason: Lithium ethoxide is formed when lithium reacts with ethanol.
24. Alkali metals react with alcohols liberating hydrogen gas along with:
🟥 (a) Alkoxide
🟦 (b) Phenoxide
🟩 (c) Carboxylate
🟨 (d) None of them
✔ Correct Answer: 🟥 (a) Alkoxide
Reason: Alkali metals react with alcohols to form alkoxides (RONa/ROK) and hydrogen gas.
Reason: Alkali metals react with alcohols to form alkoxides (RONa/ROK) and hydrogen gas.
25. The Group IA elements react violently with water making the solution:
🟥 (a) Acidic
🟦 (b) Neutral
🟩 (c) Alkaline
🟨 (d) Amphoteric
✔ Correct Answer: 🟩 (c) Alkaline
Reason: Alkali metals form metal hydroxides (MOH), making the solution strongly alkaline.
Reason: Alkali metals form metal hydroxides (MOH), making the solution strongly alkaline.
26. Which of the following alkali metals imparts golden yellow colour to the flame?
🟥 (a) K
🟦 (b) Li
🟩 (c) Cs
🟨 (d) Na
✔ Correct Answer: 🟨 (d) Na
Reason: Sodium imparts a characteristic bright golden-yellow flame.
Reason: Sodium imparts a characteristic bright golden-yellow flame.
27. Which of the following metals floats over water?
🟥 (a) Cs
🟦 (b) Na
🟩 (c) Rb
🟨 (d) Be
✔ Correct Answer: 🟦 (b) Na
Reason: Sodium has a density (0.97 g cm−3) lower than water, so it floats.
Reason: Sodium has a density (0.97 g cm−3) lower than water, so it floats.
28. The basic strength of alkaline earth metals oxides in water _____ from Be to Ba.
🟥 (a) Decreases
🟦 (b) Increases
🟩 (c) Remains same
🟨 (d) Not predictable
✔ Correct Answer: 🟦 (b) Increases
Reason: Down Group IIA, metallic character increases, making the oxides more basic.
Reason: Down Group IIA, metallic character increases, making the oxides more basic.
29. Which of the following sulphates is not soluble in water?
🟥 (a) BaSO4
🟦 (b) ZnSO4
🟩 (c) K2SO4
🟨 (d) Na2SO4
✔ Correct Answer: 🟥 (a) BaSO4
Reason: Barium sulphate is highly insoluble in water.
Reason: Barium sulphate is highly insoluble in water.
30. The element cesium bears resemblance with:
🟥 (a) Cr
🟦 (b) Ca
🟩 (c) Both (a) and (b)
🟨 (d) None of them
✔ Correct Answer: 🟨 (d) None of them
Reason: Cesium does not show a diagonal relationship with Cr or Ca. Its closest diagonal analogue would be francium (not commonly considered due to its rarity).
Reason: Cesium does not show a diagonal relationship with Cr or Ca. Its closest diagonal analogue would be francium (not commonly considered due to its rarity).
31. The formula of carnallite is:
🟥 (a) KCl.MgCl2.2H2O
🟦 (b) KCl.MgCl2.6H2O
🟩 (c) CaF2
🟨 (d) MgSO4.7H2O
✔ Correct Answer: 🟦 (b) KCl.MgCl2.6H2O
Reason: Carnallite is the double salt of potassium chloride and magnesium chloride with six water molecules.
Reason: Carnallite is the double salt of potassium chloride and magnesium chloride with six water molecules.
32. The difference of water molecules in gypsum and plaster of Paris is:
🟥 (a) 5/2
🟦 (b) 2
🟩 (c) ½
🟨 (d) 1½
✔ Correct Answer: 🟥 (d) 5/2
Reason: Gypsum = CaSO4.2H2O, while Plaster of Paris = CaSO4.½H2O; difference = 2 − ½ = 3/2or 1½
Reason: Gypsum = CaSO4.2H2O, while Plaster of Paris = CaSO4.½H2O; difference = 2 − ½ = 3/2or 1½
33. Which of the following salts are composed of isoelectronic cations and anions?
🟥 (a) NaCl
🟦 (b) MgF2
🟩 (c) CaS
🟨 (d) Both (b) and (c)
✔ Correct Answer: 🟨 (d) Both (b) and (c)
Reason: Mg2+ and F− are isoelectronic (10 e−), while Ca2+ and S2− are isoelectronic (18 e−).
Reason: Mg2+ and F− are isoelectronic (10 e−), while Ca2+ and S2− are isoelectronic (18 e−).
34. The name "Blue John" is given to which of the following compounds?
🟥 (a) CaH2
🟦 (b) CaF2
🟩 (c) Ca3(PO4)2
🟨 (d) CaO
✔ Correct Answer: 🟦 (b) CaF2
Reason: Blue John is a purple-blue variety of fluorite (CaF2).
Reason: Blue John is a purple-blue variety of fluorite (CaF2).
35. The formula of washing soda is:
🟥 (a) Na2CO3.10H2O
🟦 (b) 2KCl.MgCl2.6H2O
🟩 (c) NaHCO3
🟨 (d) MgSO4.7H2O
✔ Correct Answer: 🟥 (a) Na2CO3.10H2O
Reason: Washing soda is sodium carbonate decahydrate.
Reason: Washing soda is sodium carbonate decahydrate.
36. The formula of Baking Soda is:
🟥 (a) Na2CO3.10H2O
🟦 (b) 2KCl.MgCl2.6H2O
🟩 (c) NaHCO3
🟨 (d) MgSO4.7H2O
✔ Correct Answer: 🟩 (c) NaHCO3
Reason: Baking soda is sodium hydrogen carbonate (sodium bicarbonate).
Reason: Baking soda is sodium hydrogen carbonate (sodium bicarbonate).
37. Which element is involved in muscle contraction and bone health?
🟥 (a) Magnesium
🟦 (b) Calcium
🟩 (c) Potassium
🟨 (d) Sodium
✔ Correct Answer: 🟦 (b) Calcium
Reason: Calcium is essential for muscle contraction and maintaining healthy bones.
Reason: Calcium is essential for muscle contraction and maintaining healthy bones.
38. Which element is essential for the growth of bones and teeth?
🟥 (a) Magnesium
🟦 (b) Calcium
🟩 (c) Potassium
🟨 (d) Sodium
✔ Correct Answer: 🟦 (b) Calcium
Reason: Calcium is the major mineral responsible for the formation of bones and teeth.
Reason: Calcium is the major mineral responsible for the formation of bones and teeth.
39. Which element regulates the fluid balance inside and outside our tissues and facilitates the absorption of various materials?
🟥 (a) Magnesium
🟦 (b) Calcium
🟩 (c) Potassium
🟨 (d) Sodium
✔ Correct Answer: 🟨 (d) Sodium
Reason: Sodium regulates extracellular fluid balance and helps in the absorption of nutrients.
Reason: Sodium regulates extracellular fluid balance and helps in the absorption of nutrients.
40. Which element helps balance the pH level in the body?
🟥 (a) Magnesium
🟦 (b) Calcium
🟩 (c) Potassium
🟨 (d) Sodium
✔ Correct Answer: 🟩 (c) Potassium
Reason: Potassium helps maintain acid-base (pH) balance and proper cellular function.
Reason: Potassium helps maintain acid-base (pH) balance and proper cellular function.
41. The chemical formula of Chile saltpeter is:
🟥 (a) KNO3
🟦 (b) NaNO3
🟩 (c) AgNO3
🟨 (d) None of them
✔ Correct Answer: 🟦 (b) NaNO3
Reason: Chile saltpeter is the common name for sodium nitrate (NaNO3).
Reason: Chile saltpeter is the common name for sodium nitrate (NaNO3).
42. Which compound is used in fireworks?
🟥 (a) KNO3
🟦 (b) NaNO3
🟩 (c) AgNO3
🟨 (d) None of them
✔ Correct Answer: 🟥 (a) KNO3
Reason: Potassium nitrate is a strong oxidizing agent widely used in fireworks and gunpowder.
Reason: Potassium nitrate is a strong oxidizing agent widely used in fireworks and gunpowder.
43. Which one of the following metals is a self-protected metal?
🟥 (a) Magnesium
🟦 (b) Beryllium
🟩 (c) Aluminium
🟨 (d) All of them
✔ Correct Answer: 🟨 (d) All of them
Reason: Mg, Be and Al form a thin protective oxide layer that prevents further corrosion.
Reason: Mg, Be and Al form a thin protective oxide layer that prevents further corrosion.
44. Which alkaline earth metal does not react with either cold water or steam?
🟥 (a) Magnesium
🟦 (b) Beryllium
🟩 (c) Aluminium
🟨 (d) All of them
✔ Correct Answer: 🟦 (b) Beryllium
Reason: Beryllium is protected by a strong oxide film and does not react with either cold water or steam.
Reason: Beryllium is protected by a strong oxide film and does not react with either cold water or steam.
45. Which alkaline earth metal reacts with boiling water and steam?
🟥 (a) Magnesium
🟦 (b) Beryllium
🟩 (c) Aluminium
🟨 (d) All of them
✔ Correct Answer: 🟥 (a) Magnesium
Reason: Magnesium reacts slowly with boiling water and readily with steam to produce MgO and hydrogen gas.
Reason: Magnesium reacts slowly with boiling water and readily with steam to produce MgO and hydrogen gas.
46. Most alkali and alkaline earth metals react violently with water liberating H2 gas along with respective alkalis or metal hydroxides with general formula ............... respectively.
🟥 (a) MOH, M'(OH)2
🟦 (b) M'(OH)2, MOH
🟩 (c) M'(OH)3, MOH
🟨 (d) MOH, M'(OH)3
✔ Correct Answer: 🟥 (a) MOH, M'(OH)2
Reason: Alkali metals form MOH, while alkaline earth metals form M(OH)2.
Reason: Alkali metals form MOH, while alkaline earth metals form M(OH)2.
47. The chemical formula of peroxide and normal oxide of strontium is respectively:
🟥 (a) SrO2, Sr2O
🟦 (b) SrO2, SrO
🟩 (c) SrO, SrO2
🟨 (d) Sr2O2, SrO2
✔ Correct Answer: 🟦 (b) SrO2, SrO
Reason: Strontium forms the normal oxide SrO and the peroxide SrO2.
Reason: Strontium forms the normal oxide SrO and the peroxide SrO2.
48. The chemical formula of normal oxide and peroxide of sodium is respectively:
🟥 (a) Na2O, Na2O2
🟦 (b) Na2O2, Na2O
🟩 (c) Na2O, NaO2
🟨 (d) NaO2, Na2O
✔ Correct Answer: 🟥 (a) Na2O, Na2O2
Reason: Sodium forms the normal oxide Na2O and the peroxide Na2O2.
Reason: Sodium forms the normal oxide Na2O and the peroxide Na2O2.
49. The chemical formula of normal oxide and superoxide of potassium is respectively:
🟥 (a) K2O, KO2
🟦 (b) K2O2, K2O
🟩 (c) KO, KO2
🟨 (d) KO2, K2O
✔ Correct Answer: 🟥 (a) K2O, KO2
Reason: Potassium forms the normal oxide K2O and the superoxide KO2.
Reason: Potassium forms the normal oxide K2O and the superoxide KO2.
50. Which one of the following elements will not form peroxide?
🟥 (a) Na
🟦 (b) Sr
🟩 (c) Li
🟨 (d) Ba
✔ Correct Answer: 🟩 (c) Li
Reason: Lithium forms only the normal oxide (Li2O) and does not form peroxide under ordinary conditions.
Reason: Lithium forms only the normal oxide (Li2O) and does not form peroxide under ordinary conditions.
51. Which one of the following will form normal oxide?
🟥 (a) K
🟦 (b) Cs
🟩 (c) Be
🟨 (d) Rb
✔ Correct Answer: 🟩 (c) Be
Reason: Beryllium forms only the normal oxide (BeO), unlike heavier alkali metals.
Reason: Beryllium forms only the normal oxide (BeO), unlike heavier alkali metals.
52. Which one of the following elements will not form superoxide?
🟥 (a) K
🟦 (b) Cs
🟩 (c) Li
🟨 (d) Rb
✔ Correct Answer: 🟩 (c) Li
Reason: Lithium forms only the normal oxide (Li2O) and does not form superoxide.
Reason: Lithium forms only the normal oxide (Li2O) and does not form superoxide.
53. Which of the following alkali metals imparts lilic colour to the flame?
🟥 (a) K
🟦 (b) Li
🟩 (c) Cs
🟨 (d) Na
✔ Correct Answer: 🟨 (a) K
Reason: Sodium gives a characteristic lilic flame.
Reason: Sodium gives a characteristic lilic flame.
54. On heating sodium carbonate (Na2CO3), ............. is evolved:
🟥 (a) CO2 gas
🟦 (b) CO gas
🟩 (c) NO gas
🟨 (d) Water vapours
✔ Correct Answer: 🟨 (d) Water vapours
Reason: Washing soda (Na2CO3.10H2O) loses water of crystallization on heating, while anhydrous Na2CO3 itself is thermally stable.
Reason: Washing soda (Na2CO3.10H2O) loses water of crystallization on heating, while anhydrous Na2CO3 itself is thermally stable.
55. Which one of the following is the formula of peroxide of alkali metals (M)?
🟥 (a) MO2
🟦 (b) M2O2
🟩 (c) MO
🟨 (d) M2O
✔ Correct Answer: 🟦 (b) M2O2
Reason: Alkali metal peroxides have the general formula M2O2 (e.g., Na2O2).
Reason: Alkali metal peroxides have the general formula M2O2 (e.g., Na2O2).
56. The oxide of beryllium is:
🟥 (a) Basic
🟦 (b) Acidic
🟩 (c) Neutral
🟨 (d) Amphoteric
✔ Correct Answer: 🟨 (d) Amphoteric
Reason: BeO reacts with both acids and bases, so it is amphoteric.
Reason: BeO reacts with both acids and bases, so it is amphoteric.
57. Which is the most abundant alkali metal in the earth’s crust?
🟥 (a) Li
🟦 (b) Na
🟩 (c) K
🟨 (d) Cs
✔ Correct Answer: 🟩 (b) Na
Reason: Sodium is the most abundant alkali metal in the Earth's crust comprising of 2.3–2.8% of the Earth's crust by mass.
Reason: Sodium is the most abundant alkali metal in the Earth's crust comprising of 2.3–2.8% of the Earth's crust by mass.
58. Which of the following alkali metals is the most electropositive with the largest atomic radius?
🟥 (a) Li
🟦 (b) Cs
🟩 (c) K
🟨 (d) Na
✔ Correct Answer: 🟦 (b) Cs
Reason: Cesium has the largest atomic radius and is the most electropositive stable alkali metal.
Reason: Cesium has the largest atomic radius and is the most electropositive stable alkali metal.
59. Which of the following hydroxides is not amphoteric?
🟥 (a) Al(OH)3
🟦 (b) Mg(OH)2
🟩 (c) Be(OH)2
🟨 (d) Zn(OH)2
✔ Correct Answer: 🟦 (b) Mg(OH)2
Reason: Mg(OH)2 is a basic hydroxide, whereas Al(OH)3, Be(OH)2, and Zn(OH)2 are amphoteric.
Reason: Mg(OH)2 is a basic hydroxide, whereas Al(OH)3, Be(OH)2, and Zn(OH)2 are amphoteric.
60. The lowest melting point (29°C) among alkali metals is for:
🟥 (a) Cesium
🟦 (b) Rubidium
🟩 (c) Potassium
🟨 (d) Lithium
✔ Correct Answer: 🟥 (a) Cesium
Reason: Cesium has a melting point of about 28.5°C (≈29°C), the lowest among the common stable alkali metals.
Reason: Cesium has a melting point of about 28.5°C (≈29°C), the lowest among the common stable alkali metals.
61. Which of the following is the crystal carbonate or natron?
🟥 (a) Na2CO3
🟦 (b) Na2CO3.10H2O
🟩 (c) NaHCO3
🟨 (d) Na2CO3.H2O
✔ Correct Answer: 🟦 (b) Na2CO3.10H2O
Reason: Crystal carbonate (natron) is sodium carbonate decahydrate.
Reason: Crystal carbonate (natron) is sodium carbonate decahydrate.
62. The formula of bleaching powder is:
🟥 (a) HOCl
🟦 (b) Ca(OCl)2
🟩 (c) CaOCl2
🟨 (d) None of above
✔ Correct Answer: 🟩 (c) CaOCl2
Reason: Bleaching powder is commonly represented by the formula CaOCl2.
Reason: Bleaching powder is commonly represented by the formula CaOCl2.
63. Sodium hydroxide is used as a precipitating agent in qualitative salt analysis to precipitate heavy metal cations from their salt solutions as hydroxides. Which type of precipitated hydroxides redissolve in excess sodium hydroxide solution?
🟥 (a) Basic
🟦 (b) Acidic
🟩 (c) Amphoteric
🟨 (d) None of them
✔ Correct Answer: 🟩 (c) Amphoteric
Reason: Amphoteric hydroxides (e.g., Al(OH)3, Zn(OH)2) dissolve in excess NaOH to form complex ions.
Reason: Amphoteric hydroxides (e.g., Al(OH)3, Zn(OH)2) dissolve in excess NaOH to form complex ions.
64. In Castner-Kellner’s process, which ions are easily discharged over the moving mercury cathode?
🟥 (a) H+
🟦 (b) Na+
🟩 (c) Both of them
🟨 (d) None of them
✔ Correct Answer: 🟦 (b) Na+
Reason: Sodium ions are discharged on the mercury cathode to form sodium amalgam.
Reason: Sodium ions are discharged on the mercury cathode to form sodium amalgam.
65. The flame colour of which of the following alkali metal is violet?
🟥 (a) Na
🟦 (b) K
🟩 (c) Rb
🟨 (d) Cs
✔ Correct Answer: 🟦 (b) K
Reason: Potassium imparts a characteristic lilac (violet) flame colour.
Reason: Potassium imparts a characteristic lilac (violet) flame colour.
66. The flame colour of which of the following alkali metal is red violet?
🟥 (a) Na
🟦 (b) K
🟩 (c) Rb
🟨 (d) Cs
✔ Correct Answer: 🟩 (c) Rb
Reason: Rubidium imparts a characteristic red-violet flame colour.
Reason: Rubidium imparts a characteristic red-violet flame colour.
67. The flame colour of cesium is:
🟥 (a) Blue Violet
🟦 (b) Red Violet
🟩 (c) Violet
🟨 (d) Crimson red
✔ Correct Answer: 🟥 (a) Blue Violet
Reason: Cesium gives a characteristic blue-violet flame colour.
Reason: Cesium gives a characteristic blue-violet flame colour.
68. The flame colour of lithium is:
🟥 (a) Blue Violet
🟦 (b) Red Violet
🟩 (c) Violet
🟨 (d) Crimson red
✔ Correct Answer: 🟨 (d) Crimson red
Reason: Lithium produces a characteristic crimson-red flame.
Reason: Lithium produces a characteristic crimson-red flame.
69. Which elements have the largest atomic radii of all the elements in their respective periods?
🟥 (a) Alkaline earth metals
🟦 (b) Alkali metals
🟩 (c) Halogens
🟨 (d) Chalcogens
✔ Correct Answer: 🟦 (b) Alkali metals
Reason: Alkali metals are the first elements of each period and possess the largest atomic radii.
Reason: Alkali metals are the first elements of each period and possess the largest atomic radii.
70. Which element in group IIIA does not follow the normal group trend of atomic radii?
🟥 (a) B
🟦 (b) Al
🟩 (c) Ga
🟨 (d) In
✔ Correct Answer: 🟩 (c) Ga
Reason: Gallium has a smaller atomic radius than expected due to poor shielding by 3d electrons (d-block contraction).
Reason: Gallium has a smaller atomic radius than expected due to poor shielding by 3d electrons (d-block contraction).
71. Gallium has slightly smaller atomic radii than aluminium despite being placed below it in the group. This is because of the poor shielding effect caused by electrons of:
🟥 (a) d-orbitals
🟦 (b) f-orbitals
🟩 (c) p-orbitals
🟨 (d) s-orbitals
✔ Correct Answer: 🟥 (a) d-orbitals
Reason: Poor shielding by 3d electrons increases the effective nuclear charge, making Ga smaller than expected.
Reason: Poor shielding by 3d electrons increases the effective nuclear charge, making Ga smaller than expected.
72. The ionization energy (IE) of the elements of Boron family (Group IIIA) shows:
🟥 (a) Regular trend
🟦 (b) Irregular trend
🟩 (c) Normal trend
🟨 (d) No trend
✔ Correct Answer: 🟦 (b) Irregular trend
Reason: The IE trend is irregular due to poor shielding by d- and f-electrons.
Reason: The IE trend is irregular due to poor shielding by d- and f-electrons.
73. The anomalous IE trend in Group IIIA has been observed between:
🟥 (a) Al and Ga
🟦 (b) Indium and Thallium
🟩 (c) Both of them
🟨 (d) None of them
✔ Correct Answer: 🟩 (c) Both of them
Reason: Irregularities occur between Al–Ga and In–Tl because of poor shielding by d- and f-electrons.
Reason: Irregularities occur between Al–Ga and In–Tl because of poor shielding by d- and f-electrons.
74. The anomalous IE trend in Group IIIA has been observed between Al and Ga and between Indium and Thallium. The reason for these irregularities is due to the:
🟥 (a) Poor shielding effect
🟦 (b) Good shielding effect
🟩 (c) Both of them
🟨 (d) None of them
✔ Correct Answer: 🟥 (a) Poor shielding effect
Reason: d- and f-electrons shield the nucleus poorly, increasing the effective nuclear charge.
Reason: d- and f-electrons shield the nucleus poorly, increasing the effective nuclear charge.
75. Ga shows an unexpectedly higher ionization energy compared to Al due to the poor shielding of nuclear charge by:
🟥 (a) 3d electrons
🟦 (b) 4d electrons
🟩 (c) 4f electrons
🟨 (d) 5f electrons
✔ Correct Answer: 🟥 (a) 3d electrons
Reason: The 3d electrons shield the nucleus poorly, increasing the effective nuclear charge and hence the ionization energy of Ga.
Reason: The 3d electrons shield the nucleus poorly, increasing the effective nuclear charge and hence the ionization energy of Ga.
76. Thallium exhibits an unexpectedly higher ionization energy than indium due to the poor shielding of nuclear charge by:
🟥 (a) 3d electrons
🟦 (b) 4d electrons
🟩 (c) 4f electrons
🟨 (d) 5f electrons
✔ Correct Answer: 🟩 (c) 4f electrons
Reason: Poor shielding by 4f electrons (lanthanide contraction) increases the effective nuclear charge, raising the ionization energy of Tl.
Reason: Poor shielding by 4f electrons (lanthanide contraction) increases the effective nuclear charge, raising the ionization energy of Tl.
77. The anomalous or irregular ionization energy trend in Group IVA has been observed between:
🟥 (a) Tin and Lead
🟦 (b) Ge and Sn
🟩 (c) Si and Ge
🟨 (d) All of them
✔ Correct Answer: 🟥 (a) Tin and Lead
Reason: Sn and Pb have nearly similar atomic radii because of lanthanide contraction, causing an irregular IE trend.
Reason: Sn and Pb have nearly similar atomic radii because of lanthanide contraction, causing an irregular IE trend.
78. The anomalous ionization energy trend in Group IVA has been observed between tin and lead due to their almost same atomic radii because of:
🟥 (a) Lanthanide contraction
🟦 (b) Actinide contraction
🟩 (c) Both of them
🟨 (d) None of them
✔ Correct Answer: 🟥 (a) Lanthanide contraction
Reason: Lanthanide contraction reduces the expected increase in atomic size from Sn to Pb.
Reason: Lanthanide contraction reduces the expected increase in atomic size from Sn to Pb.
79. Which one of the following elements is/are metalloids?
🟥 (a) B
🟦 (b) Si
🟩 (c) Ge
🟨 (d) All of them
✔ Correct Answer: 🟨 (d) All of them
Reason: Boron, silicon, and germanium are all classified as metalloids.
Reason: Boron, silicon, and germanium are all classified as metalloids.
80. Which one of the following elements is/are metalloids?
🟥 (a) As
🟦 (b) Sb
🟩 (c) Te
🟨 (d) All of them
✔ Correct Answer: 🟨 (d) All of them
Reason: Arsenic, antimony, and tellurium are all metalloids with properties intermediate between metals and non-metals.
Reason: Arsenic, antimony, and tellurium are all metalloids with properties intermediate between metals and non-metals.
81. Which element of Group VA is a semiconductor?
🟥 (a) As
🟦 (b) Sb
🟩 (c) Te
🟨 (d) All of them
✔ Correct Answer: 🟨 (a) As
Reason: Arsenic exhibits semiconducting properties.
Reason: Arsenic exhibits semiconducting properties.
82. Which element of Group IVA is a semiconductor?
🟥 (a) C
🟦 (b) Si
🟩 (c) Ge
🟨 (d) Both (b) and (c)
✔ Correct Answer: 🟨 (d) Both (b) and (c)
Reason: Silicon and germanium are well-known semiconductor materials used in electronic devices.
Reason: Silicon and germanium are well-known semiconductor materials used in electronic devices.
83. The element with the highest boiling and melting points in Group IA is:
🟥 (a) Cs
🟦 (b) Na
🟩 (c) Li
🟨 (d) K
✔ Correct Answer: 🟩 (c) Li
Reason: Lithium has the strongest metallic bonding among alkali metals, giving it the highest melting and boiling points.
Reason: Lithium has the strongest metallic bonding among alkali metals, giving it the highest melting and boiling points.
84. The element with the highest boiling and melting points in Group IIA is:
🟥 (a) Ba
🟦 (b) Be
🟩 (c) Ca
🟨 (d) Mg
✔ Correct Answer: 🟦 (b) Be
Reason: Beryllium has the strongest metallic bonding in Group IIA, resulting in the highest melting and boiling points.
Reason: Beryllium has the strongest metallic bonding in Group IIA, resulting in the highest melting and boiling points.
85. In Group IVA, which element shows the lowest melting point?
🟥 (a) Tin
🟦 (b) C
🟩 (c) Si
🟨 (d) Pb
✔ Correct Answer: 🟥 (a) Tin
Reason: Tin has the lowest melting point in Group IVA (≈232°C), lower than lead, silicon, and carbon.
Reason: Tin has the lowest melting point in Group IVA (≈232°C), lower than lead, silicon, and carbon.
86. The flame colour of barium is:
🟥 (a) Apple green
🟦 (b) Brick red
🟩 (c) Violet
🟨 (d) Crimson red
✔ Correct Answer: 🟥 (a) Apple green
Reason: Barium salts impart a characteristic apple-green flame colour in the flame test.
Reason: Barium salts impart a characteristic apple-green flame colour in the flame test.
📘 Smart Answers Short Questions of Model Test Questions Class 10 Chemistry Test # 1 for Chapter # 1 (Chemical Equilibrium)
Prepared by Inam Jazbi – Learn Chemistry
Q1.Give reason for the following:
▶ Ionization energy decreases from top to bottom in s-block elements.
▶ Boiling points of halogens increase down the group in the periodic table.
▶ Gallium has smaller atomic radii than aluminium despite being below the aluminium in group IIIA.
▶ Electronegativities of alkali metals decrease from Li to Cs.
▶ Acidity of hydrogen halides increase from HF to HI.
▶ Fluorine is the strongest oxidizing agent.
▶ The ionization energy of Ga is unexpectedly higher than Al in group IIIA
▶ The ionization energy of thallium is unexpectedly higher than indium in group IIIA
▶ Electronegativity of Ga is higher than Al.
▶ Electronegativity of In is higher than Ga.
▶ Electronegativity of Tl is higher than In.
▶ IE of Sn is less than Pb ▶ Chlorine can displace bromide and iodide from their salts but fluorine cannot do this job why?
▶ Ionization energy decreases from top to bottom in s-block elements.
▶ Boiling points of halogens increase down the group in the periodic table.
▶ Gallium has smaller atomic radii than aluminium despite being below the aluminium in group IIIA.
▶ Electronegativities of alkali metals decrease from Li to Cs.
▶ Acidity of hydrogen halides increase from HF to HI.
▶ Fluorine is the strongest oxidizing agent.
▶ The ionization energy of Ga is unexpectedly higher than Al in group IIIA
▶ The ionization energy of thallium is unexpectedly higher than indium in group IIIA
▶ Electronegativity of Ga is higher than Al.
▶ Electronegativity of In is higher than Ga.
▶ Electronegativity of Tl is higher than In.
▶ IE of Sn is less than Pb ▶ Chlorine can displace bromide and iodide from their salts but fluorine cannot do this job why?
Answer
⚗️ Ionization Energy (IE) Trend
Ionization energy decreases from top to bottom in s-block elements due to the increase in atomic size and shielding effect.
📌 Reason
Larger atomic size and greater shielding reduce the attraction between the nucleus and outer electrons, making them easier to remove.
🌡️ Boiling Point of Halogens
Boiling points of halogens increase from F₂ to I₂ due to stronger intermolecular (van der Waals) forces.
📌 Reason
Atomic size increases down the group, strengthening intermolecular attractions.
📏 Gallium has Smaller Atomic Radius than Aluminium
Gallium has a slightly smaller atomic radius than aluminium despite being below it in Group IIIA.
📌 Reason
Poor shielding by 3d-electrons increases the effective nuclear charge, pulling the outer electrons closer.
⚡ Electronegativity of Alkali Metals
Electronegativity decreases from Li to Cs.
📌 Reason
Increasing atomic size and shielding effect reduce the attraction for shared electrons.
🧪 Acidity of Hydrogen Halides
Acidity increases from HF to HI.
📌 Order
HF < HCl < HBr < HI
📖 Reason
Bond strength decreases from HF to HI, making proton release easier.
🔥 Fluorine – Strongest Oxidizing Agent
Fluorine is the strongest oxidizing agent.
📌 Reason
It has the highest electronegativity, smallest atomic size, and greatest tendency to gain electrons.
⚡ IE of Ga is Higher than Al
Gallium has slightly higher ionization energy than aluminium.
📌 Reason
Poor shielding by 3d-electrons increases the effective nuclear charge.
⚡ IE of Tl is Higher than In
Thallium has higher ionization energy than indium.
📌 Reason
Poor shielding by 4f-electrons increases the effective nuclear charge.
⚡ EN of Ga is Higher than Al
Gallium has slightly higher electronegativity than aluminium.
📌 Reason
Poor shielding by 3d-electrons increases the effective nuclear charge.
⚡ EN of In is Higher than Ga
Indium has slightly higher electronegativity than gallium.
📌 Reason
Greater effective nuclear charge increases attraction for shared electrons.
⚡ EN of Tl is Higher than In
Thallium has slightly higher electronegativity than indium.
📌 Reason
Poor shielding by 4f-electrons increases the effective nuclear charge.
⚗️ IE of Sn is Less than Pb
Lead has slightly higher ionization energy than tin.
📌 Reason
Due to lanthanide contraction, both have nearly the same atomic size, resulting in stronger nuclear attraction in Pb.
🧪 Why Fluorine Cannot Displace Br⁻ and I⁻?
Fluorine does not undergo simple aqueous displacement reactions.
📌 Reason
It reacts vigorously with water, producing HF and oxygen instead of displacing bromide or iodide ions.
📖 Reaction
2F₂ + 2H₂O → 4HF + O₂
Q2. What is flame test? Give its basis? Mention the colour flame of alkali metals and alkaline earth metals.
Answer
🔥 Flame Test
Flame test is a preliminary qualitative test used to identify s-block metal ions (except Be and Mg) by their characteristic flame colours.
📌 Principle
On heating, electrons absorb energy and jump to higher energy levels. When they return to lower levels, they emit light of a characteristic colour.
📖 Flame Colours
| Element | Flame Colour |
|---|---|
| Li | Crimson Red |
| Na | Golden Yellow |
| K | Violet (Lilac) |
| Rb | Red Violet |
| Cs | Blue Violet |
| Be | No Flame Colour |
| Mg | No Flame Colour |
| Ca | Brick Red |
| Sr | Crimson Red |
| Ba | Apple Green |
⚡ Important Points
👉 Used for identification of alkali and alkaline earth metals.👉 Be and Mg do not produce a characteristic flame colour.
👉 Each metal emits light of a specific wavelength, producing a unique flame colour.
Q3. What is meant by a diagonal relationship? Mention three pairs of representative elements that show diagonal relationship. Write down three points to show the similarity of diagonal members of group IA and IIA, IIA and IIIA and IIIA and IVA.
Answer
📘 Definition of Diagonal Relationship
The close resemblance in the properties exhibited by certain pairs of elements that are located diagonally to each other in two adjacent periods and two adjacent groups within the periodic table is referred to as Diagonal Relationship.
📌 Important Diagonal Members
✔ Lithium – Magnesium (Li – Mg) ✔ Beryllium – Aluminium (Be – Al)
✔ Boron – Silicon (B – Si)
⚗️ Lithium–Magnesium (Li–Mg) Diagonal Relationship
1. Both have almost similar atomic radii (Li = 152 pm and Mg = 160 pm).2. Both have nearly identical electronegativities (Li = 1.0 and Mg = 1.2).
3. Both are the lightest elements in their respective groups.
4. Oxides of both elements are less soluble in water than those of other members of their respective groups.
⚗️ Beryllium–Aluminium (Be–Al) Diagonal Relationship
1. Both have nearly the same electronegativity (Be = 1.5 and Al = 1.5).2. Both become passive towards concentrated nitric acid.
3. BeCl₂ and AlCl₃ both behave as Lewis acids.
⚗️ Boron–Silicon (B–Si) Diagonal Relationship
1. Both have almost the same electronegativity (B = 2.0 and Si = 1.8).2. Both have nearly the same density (B = 2.35 g/cm³ and Si = 2.34 g/cm³).
3. Both are metalloids and neither forms simple cations.
Q4. Discuss the group trend of ionization energy and atomic radii in Group IIIA and IVA of the periodic table.
Answer
⚡ Group Trend of Ionization Energy in Group IIIA
The ionization energy (IE) of the elements of the Boron family (Group IIIA) shows an irregular trend down the group.
📌 Important Exceptions
• Gallium (Ga) has an unexpectedly higher ionization energy than Aluminium (Al).• Thallium (Tl) has a higher ionization energy than Indium (In).
📖 Reason
These irregularities are due to the poor shielding effect of:
• 3d-electrons in gallium.
• 4f-electrons in thallium.
Poor shielding increases the effective nuclear charge, making the outermost electrons more strongly attracted to the nucleus. Therefore, more energy is required to remove them.
⚡ Group Trend of Ionization Energy in Group IVA
The ionization energy (IE) of the elements of the Carbon family (Group IVA) generally decreases from top to bottom in the group.
📌 Exception
An irregular trend is observed between Tin (Sn) and Lead (Pb).
📖 Reason
Tin and lead have nearly the same atomic radii due to lanthanide contraction. As a result, lead experiences stronger nuclear attraction, requiring more energy to remove an electron. Therefore, the ionization energy of lead is slightly higher than that of tin.
📏 Group Trend of Atomic Radii in Group IIIA
The atomic radii of the elements of the Boron family (Group IIIA) generally increase from top to bottom due to the addition of a new electron shell in each successive period.
📌 Exception
Gallium has a slightly smaller atomic radius than aluminium despite being placed below it in the group.
📖 Reason
This exception is due to the poor shielding effect of 3d-electrons, which increases the effective nuclear charge and pulls the outer electrons closer to the nucleus.
⚡ Summary
✔ Group IIIA shows an irregular trend in ionization energy due to poor shielding by d- and f-electrons.✔ Group IVA generally shows a decrease in ionization energy down the group, with an exception between Sn and Pb due to lanthanide contraction.
✔ Atomic radii in Group IIIA generally increase down the group, except that Ga has a slightly smaller atomic radius than Al because of poor shielding by 3d-electrons.
Q5. Write down four properties of beryllium that show its unique behaviour in Group IIA.
Answer
🧪 Unique Behaviour of Beryllium
The first member of Group IIA (Alkaline Earth Metals), Beryllium (Be), differs markedly from its heavier congeners because of its very small atomic size, high ionization energy, and relatively high electronegativity. Therefore, it exhibits several unique properties compared to the other members of the group.
📌 Important Properties of Beryllium
1. Beryllium preferably forms covalent compounds because of its very small atomic size and high polarizing power.2. It is harder and more rigid than the other members of Group IIA.
3. It has a low density and a high melting point compared to the other alkaline earth metals.
4. It is chemically stable because a thin protective oxide layer (BeO) forms on its surface, preventing further oxidation and corrosion.
5. It is the smallest atom among all the alkaline earth metals.
6. It possesses the highest ionization energy in Group IIA.
7. It has the highest electronegativity among all the alkaline earth metals.
8. It has no vacant d-orbitals in its valence shell.
⚗️ Exceptional Chemical Behaviour
9. Unlike the other alkaline earth metals, beryllium does not react with water because of the protective oxide film on its surface.10. Beryllium oxide (BeO) and beryllium hydroxide [Be(OH)₂] are amphoteric, whereas the oxides and hydroxides of the other alkaline earth metals are basic.
11. Beryllium carbide (BeC) is covalent, whereas the carbides of the other Group IIA elements are ionic.
12. Beryllium carbide reacts with water to produce methane (CH₄), whereas the carbides of the other alkaline earth metals produce acetylene (C₂H₂).
Reaction:
BeC + 4H₂O → Be(OH)₂ + CH₄
CaC₂ + 2H₂O → Ca(OH)₂ + HC≡CH
⚡ Coordination Behaviour
13. Beryllium does not exhibit a coordination number greater than four because its valence shell contains only one s and three p orbitals. It has no vacant d-orbitals. In contrast, the heavier members of Group IIA can show a coordination number of six by utilizing their vacant d-orbitals.
📖 Summary
✔ Smallest atomic size in Group IIA.✔ Highest ionization energy and electronegativity.
✔ Forms covalent compounds.
✔ Does not react with water.
✔ BeO and Be(OH)₂ are amphoteric.
✔ BeC is covalent and produces methane on hydrolysis.
✔ Shows a maximum coordination number of four.
Q6. What is Electrical Conductivity? Briefly explain the group trend of electrical conductivity of representative elements.
Answer
⚡ Electrical Conductivity
Electrical conductivity is the measure of the ability of a material to conduct electric current.
📌 Group Trends of Electrical Conductivity
1. Group IA (Alkali Metals)Alkali metals are good conductors of electricity due to the presence of mobile electrons in their metallic lattices. Their electrical conductivity generally increases from Li → Cs because ionization energy decreases down the group, making the valence electron easier to remove.
2. Group IIA (Alkaline Earth Metals)
Alkaline earth metals are also good conductors of electricity. They are generally more conductive than alkali metals because each atom contributes two loosely held valence electrons (ns²) for electrical conduction.
3. Group IIIA (Boron Family)
The boron family shows moderate electrical conductivity. Boron behaves as a metalloid (poor conductor), whereas aluminium, gallium, indium and thallium are metallic and conduct electricity.
4. Group IVA (Carbon Family)
The electrical conductivity varies considerably in this group. Carbon is a poor conductor (except graphite), silicon and germanium are semiconductors, while tin and lead are metallic conductors.
5. Group VA (Pnictogens)
Electrical conductivity changes gradually across the group. Nitrogen is a non-conductor, phosphorus is a poor conductor, arsenic is a semiconductor, whereas antimony and bismuth are good conductors.
6. Groups VIA, VIIA and VIIIA
Most elements of Groups VIA (Chalcogens), VIIA (Halogens) and VIIIA (Noble Gases) are poor or non-conductors of electricity.
📖 Summary of Electrical Conductivity Trends
| Group | Elements | Electrical Conductivity |
|---|---|---|
| Alkali Metals (IA) | Li, Na, K, Rb, Cs | High; generally increases down the group |
| Alkaline Earth Metals (IIA) | Be, Mg, Ca, Sr, Ba | High; generally increases down the group |
| Boron Family (IIIA) | B, Al, Ga, In, Tl | Moderate; generally increases |
| Carbon Family (IVA) | C, Si, Ge, Sn, Pb |
C = Poor conductor (except graphite) Si & Ge = Semiconductors Sn & Pb = Good conductors |
| Pnictogens (VA) | N, P, As, Sb, Bi |
N = Non-conductor P = Poor conductor As = Semiconductor Sb & Bi = Good conductors |
| Chalcogens (VIA) | O, S, Se, Te, Po | Poor electrical conductivity |
| Halogens (VIIA) | F, Cl, Br, I, At | Poor electrical conductivity |
| Noble Gases (VIIIA) | He, Ne, Ar, Kr, Xe, Rn | Extremely low electrical conductivity |
⚡ Key Points
✔ Electrical conductivity depends on the availability of free or mobile electrons.✔ Metals are generally good conductors, metalloids are semiconductors, and non-metals are poor conductors or insulators.
✔ Semiconductors such as silicon, germanium and arsenic are widely used in electronic devices.
Q7. Write down action of oxygen and water on s-block elements with balanced chemical equation.
Answer
Addition Reaction with Oxygen / Action of Oxygen and Formation of Variety of Ionic Basic Oxides
All alkali and alkaline earth metals directly combine with oxygen to form a variety of oxides namely normal oxides, M2O (O2−) e.g. Li2O, Na2O, peroxides, M2O2 (O22−) e.g. Na2O2 and superoxides, MO2 (O2−) e.g. KO2, RbO2, CsO2.
All alkali metals except Li readily react with oxygen and are quickly tarnished to form a variety of oxides i.e. normal oxide (O2−), peroxide (O22−) and superoxide (O2−). Lithium forms normal oxide, sodium forms peroxide in excess of air while the rest of the metals of Group IA form superoxides.
Balanced Chemical Equations
4Na(s) + O2(g) → 2Na2O(s) (Normal oxide)
2Na(s) + O2(g) (Excess) → Na2O2(s) (Peroxide)
K(s) + O2(g) → KO2(s) (Superoxide)
Rb(s) + O2(g) → RbO2(s) (Superoxide)
Cs(s) + O2(g) → CsO2(s) (Superoxide)
All the alkaline earth metals like Be, Mg, Ca react readily on heating with oxygen to form corresponding bivalent normal mono oxides except Sr and Ba which form peroxides.
Balanced Chemical Equations
2M(s) + O2(g) → 2MO(s) (Normal oxide) [M = Be, Mg, Ca]
M′(s) + O2(g) → M′O2(s) (Peroxide) [M′ = Sr, Ba]
Displacement (Redox) Reaction with Water
Most alkali and alkaline earth metals react violently with water liberating H2 gas along with respective alkalis or metal hydroxides with general formula MOH and M′(OH)2 respectively.
Alkali metals react very vigorously and violently with water to produce respective metal hydroxides and liberate hydrogen gas with the evolution of a very large amount of heat due to their low melting points except Li. The oxidation of cesium by water is often accompanied by violent explosion.
Alkali metals are highly reactive, therefore they are stored under an inert solvent like kerosene oil to avoid reaction with atmospheric water vapours, oxygen and CO2.
Balanced Chemical Equation
2M(s) + 2H2O(l) → 2MOH(aq) + H2↑ (Alkali metals)
The alkaline earth metals react with water slowly and less violently to form respective hydroxides and H2 gas. Among alkaline earth metals, Be does not react with either cold water or steam, whereas magnesium reacts with boiling water and steam. Be and Mg are protected due to formation of a stable oxide layer. The remaining members react with water to form hydroxides and hydrogen gas.
Balanced Chemical Equation
M′(s) + 2H2O(l) → M′(OH)2 + H2↑ [M′ = Mg, Ca, Sr and Ba]
Q8. Complete and balance the following chemical equations.
Answer
🧪 Reaction with Hydrogen (Formation of Hydrides)
| 2Li(s) + H2(g) | → | 2LiH(s) |
| Mg(s) + H2(g) | → | MgH2(s) |
🧪 Reaction with Chlorine
| 2Na(s) + Cl2(g) | → | 2NaCl(s) |
| Be(s) + Cl2(g) | → | BeCl2(s) |
🧪 Reaction with Nitrogen
| 6Na(s) + N2(g) | → | 2Na3N(s) |
| 3Ca(s) + N2(g) | → | Ca3N2(s) |
| 6Li(s) + N2(g) | → | 2Li3N(s) |
🧪 Reaction with Oxygen (Alkali Metals)
| 4Li(s) + O2(g) | → | 2Li2O(s) |
| 4Na(s) + O2(g) | → | 2Na2O(s) |
| 2Na(s) + O2(g) (Excess) | → | Na2O2(s) |
| K(s) + O2(g) | → | KO2(s) |
| Rb(s) + O2(g) | → | RbO2(s) |
| Cs(s) + O2(g) | → | CsO2(s) |
🧪 Reaction with Oxygen (Alkaline Earth Metals)
| 2Be(s) + O2(g) | → | 2BeO(s) |
| Sr(s) + O2(g) | → | SrO2(s) |
Q9. Explain the auto oxidizing and reducing properties of chlorine.
Answer
⚗️ Auto Oxidation-Reduction (Disproportionation) Reaction
The special type of redox reaction in which a single substance (species) undergoes simultaneous oxidation and reduction is called auto-redox, self-redox, auto-oxidation-reduction or disproportionation reaction. Auto-redox reaction occurs when an element is both oxidized and reduced at the same time.
📌 Auto Oxidizing and Reducing Property of Chlorine
Chlorine undergoes auto-oxidation-reduction (auto-redox) reaction with water in which chlorine simultaneously reduces to hydrochloric acid (Cl−1) and oxidizes to hypochlorous acid (HOCl) (Cl+1), thereby acting as both an oxidizing agent and a reducing agent.
📖 Balanced Chemical Equation
Cl2(g)
+
H2O(l)
→
HCl(aq)
+
HOCl(aq)
| Species | Oxidation State of Cl | Process |
|---|---|---|
| Cl₂ | 0 | Initial State |
| HCl | −1 | Reduction |
| HOCl | +1 | Oxidation |
⚡ Key Points
✔ Chlorine undergoes disproportionation (auto-redox) reaction with water.✔ One chlorine atom is reduced from 0 → −1 forming HCl.
✔ The other chlorine atom is oxidized from 0 → +1 forming HOCl.
✔ Therefore, chlorine acts as both an oxidizing agent and a reducing agent simultaneously.
Q10. Write the balanced equations for the following chemical processes:
▶ A piece of aluminium is dropped into concentrated sulphuric acid.
▶ Ferric chloride is mixed in an aqueous solution of caustic soda.
▶ Magnesium is heated with nitrogen gas.
▶ Potassium is put into ethyl alcohol.
▶ Chlorine gas is passed through an aqueous solution of caustic soda.
▶ A piece of aluminium is dropped into concentrated sulphuric acid.
▶ Ferric chloride is mixed in an aqueous solution of caustic soda.
▶ Magnesium is heated with nitrogen gas.
▶ Potassium is put into ethyl alcohol.
▶ Chlorine gas is passed through an aqueous solution of caustic soda.
Answer
🧪 Balanced Chemical Equations
| ▶ A piece of aluminium is dropped into concentrated sulphuric acid. | 2Al(s) + 6H2O(l) ⟶ 2Al(OH)3(s) + 3H2↑ |
| ▶ Ferric chloride is mixed in an aqueous solution of caustic soda. | FeCl3(aq) + 3NaOH(aq) ⟶ 3NaCl(aq) + Fe(OH)3↓ |
| ▶ Magnesium is heated with nitrogen gas. | 3Mg(s) + N2(g) ⟶ Mg3N2(s) |
| ▶ Potassium is put into ethyl alcohol. | 2K(s) + 2C2H5OH(aq) ⟶ 2C2H5OK(aq) + H2(g)↑ |
| ▶ Chlorine gas is passed through an aqueous solution of caustic soda. |
Hot / Conc. NaOH (Excess Cl₂)
6NaOH(aq) + 3Cl2(g) ⟶ NaClO3 + 5NaCl + 3H2O Cold / Dilute NaOH 2NaOH(aq) + Cl2(g) ⟶ NaOCl + NaCl + H2O |
Q11. Write down complete balanced action of following reactions:
(i) Bleaching powder is dissolved in water.
(ii) Fluorine reacts with oxygen.
(iii) Aluminium reacts with water.
(iv) Silicon reacts with steam.
(v) Phosphorus reacts vigorously with water.
(vi) Sulphur reacts at high temperature with water.
(vii) Silicon is heated with nitrogen at high temperatures.
(viii) Phosphorus reacts with nitrogen at high temperature.
(ix) Chlorine reacts with nitrogen.
(x) Nitrogen reacts with oxygen in the presence of catalyst.
(i) Bleaching powder is dissolved in water.
(ii) Fluorine reacts with oxygen.
(iii) Aluminium reacts with water.
(iv) Silicon reacts with steam.
(v) Phosphorus reacts vigorously with water.
(vi) Sulphur reacts at high temperature with water.
(vii) Silicon is heated with nitrogen at high temperatures.
(viii) Phosphorus reacts with nitrogen at high temperature.
(ix) Chlorine reacts with nitrogen.
(x) Nitrogen reacts with oxygen in the presence of catalyst.
Answer
🧪 Complete Balanced Chemical Equations
| (i) Bleaching powder is dissolved in water. | CaOCl2 + H2O ⟶ Ca(OH)2 + Cl2↑ |
| (ii) Fluorine reacts with oxygen. | 2F2(g) + O2(g) ⟶ 2OF2(g) |
| (iii) Aluminium reacts with water. | 2Al(s) + 6H2O(l) ⟶ 2Al(OH)3(s) + 3H2↑ |
| (iv) Silicon reacts with steam. | Si(s) + 2H2O(l) ⟶ SiO2(s) + 2H2↑ |
| (v) Phosphorus reacts vigorously with water. | 2P4(s) + 12H2O(l) ⟶ 3H3PO4(aq) + 5PH3↑ |
| (vi) Sulphur reacts at high temperature with water. | S(s) + 2H2O(l) ⟶ SO2(aq) + 2H2↑ |
| (vii) Silicon is heated with nitrogen at high temperatures. | 3Si(s) + 2N2(g) ⟶ Si3N4(s) |
| (viii) Phosphorus reacts with nitrogen at high temperature. | 6P(s) + 5N2(g) ⟶ 2P3N5(s) |
| (ix) Chlorine reacts with nitrogen. | 3Cl2(g) + N2(g) ⟶ 2NCl3(g) |
| (x) Nitrogen reacts with oxygen in the presence of catalyst. |
Catalyst 2N2(g) + O2(g) ⟶ 2N2O(g) |