JEE Chemistry S Block Elements MCQs Set 15

Answer: (c) \( \ce{NaHCO3} \), \( \ce{Na2CO3} \)

Explanation:
1. \( \ce{2NaHCO3 -> Na2CO3 + CO2 + H2O} \), so X is \( \ce{NaHCO3} \) and Y is \( \ce{Na2CO3} \).
2. \( \ce{Na2CO3 + CO2 + H2O -> 2NaHCO3} \), which regenerates X.

Teacher's Note:
1. Heating gives the carbonate, and \( \ce{CO2} \) with water regenerates the hydrogencarbonate.
2. This cycle is used in the Solvay process.

Answer: (c) \( \ce{Na2O2} \)

Explanation:
1. \( \ce{2Na2O2 + 2CO2 -> 2Na2CO3 + O2} \).
2. Sodium peroxide absorbs carbon dioxide and releases oxygen.

Teacher's Note:
1. Superoxides such as \( \ce{KO2} \) are used for the same purpose.
2. Check that the compound does both jobs.

Answer: (d) \( \ce{Mg} \), \( \ce{NH3} \)

Explanation:
1. Mg burns in air to give \( \ce{MgO} \) and \( \ce{Mg3N2} \).
2. \( \ce{Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3} \), so the gas is ammonia.

Teacher's Note:
1. Burning in air produces both oxide and nitride.
2. The nitride is hydrolysed by water.

Answer: (c) both are correct

Explanation:
1. The process in which brine, ammonia and carbon dioxide give sodium hydrogencarbonate is the Solvay (ammonia-soda) process.
2. Both names refer to the same process.

Teacher's Note:
1. Know both names of the process.
2. Limestone, ammonia and brine are the raw materials.

Answer: (a) \( \ce{CaO} \), \( \ce{Ca(OH)2} \)

Explanation:
1. \( \ce{CaCO3 -> CaO + CO2} \), so B is \( \ce{CaO} \).
2. \( \ce{CaO + H2O -> Ca(OH)2} \), so C is \( \ce{Ca(OH)2} \).

Teacher's Note:
1. Quick lime is the oxide and slaked lime is the hydroxide.
2. Follow the sequence step by step.

Answer: (b) \( \ce{NH3} \), \( \ce{NH4HCO3} \), \( \ce{NaHCO3} \)

Explanation:
1. \( \ce{CaO + 2NH4Cl -> CaCl2 + 2NH3 + H2O} \), so D is \( \ce{NH3} \).
2. \( \ce{NH3 + H2O + CO2 -> NH4HCO3} \) gives E, and \( \ce{NH4HCO3 + NaCl -> NaHCO3 + NH4Cl} \) gives F.

Teacher's Note:
1. Ammonia is regenerated from ammonium chloride with lime.
2. F is the precipitate that gives soda ash on heating.

Answer: (d) \( \ce{CaCO3} \)

Explanation:
1. \( \ce{CaCO3 -> CaO + CO2} \), and \( \ce{CaO + H2O -> Ca(OH)2} \) (B).
2. Excess \( \ce{CO2} \) gives \( \ce{Ca(HCO3)2} \) (C), and gentle heating of C gives back \( \ce{CaCO3} \).
3. \( \ce{Na2CO3} \) and \( \ce{K2CO3} \) do not decompose on heating.

Teacher's Note:
1. Follow the cycle carbonate, oxide, hydroxide, bicarbonate, carbonate.
2. Only the carbonate that decomposes on heating fits.

Answer: (d) Aluminium reacts with excess \( \ce{NaOH} \) to give \( \ce{Al(OH)3} \)

Explanation:
1. With excess \( \ce{NaOH} \), aluminium forms soluble sodium aluminate (tetrahydroxoaluminate), not \( \ce{Al(OH)3} \).
2. The other statements are correct.

Teacher's Note:
1. Amphoteric hydroxides redissolve in excess alkali.
2. \( \ce{Al(OH)3} \) forms only with limited alkali.

Answer: (b) \( \ce{Zn} \)

Explanation:
1. \( \ce{Zn + 2NaOH -> Na2ZnO2 + H2} \).
2. Zn is amphoteric, but Mg, Cu and Ag do not react with alkali.

Teacher's Note:
1. Amphoteric metals such as Zn and Al release \( \ce{H2} \) with NaOH.
2. Mg is only basic.

Answer: (a) milk of lime

Explanation:
1. A suspension of \( \ce{Ca(OH)2} \) in water is called milk of lime.
2. The clear saturated solution is limewater.

Teacher's Note:
1. Distinguish suspension from clear solution.
2. Quicklime is the oxide.

Answer: (a) \( \ce{K} \) > \( \ce{Fe} \) > \( \ce{Cu} \)

Explanation:
1. The body contains much more potassium (about 0.25 percent) than iron (about 0.006 percent) or copper (trace).
2. So the order is K greater than Fe greater than Cu.

Teacher's Note:
1. Potassium and sodium are major electrolytes.
2. Fe and Cu are trace elements.

Answer: (d) \( \ce{MgCO3} \)

Explanation:
1. Carbonates of Ba, Sr and Ca are precipitated in ammonium chloride-ammonia buffer (group V in qualitative analysis).
2. \( \ce{MgCO3} \) stays in solution because it is more soluble and \( \ce{NH4+} \) lowers the carbonate concentration.

Teacher's Note:
1. Group V reagent is ammonium carbonate in \( \ce{NH4Cl} \)-\( \ce{NH3} \).
2. Mg is detected separately.

Answer: (d) Liquid ammonia remains diamagnetic

Explanation:
1. Sodium dissolves to form ammoniated \( \ce{Na+} \) and ammoniated electrons.
2. The free electron makes the solution paramagnetic, so it does not remain diamagnetic.

Teacher's Note:
1. Free unpaired electrons give paramagnetism.
2. Blue colour and conductivity come from the same species.

Answer: (a) \( \ce{NaHCO3} \) and \( \ce{NaOH} \)

Explanation:
1. \( \ce{NaHCO3 + NaOH -> Na2CO3 + H2O} \): the acidic salt and the base neutralise each other.
2. The other pairs have no mutual reaction.

Teacher's Note:
1. An acid salt and a base react.
2. Carbonate with hydroxide is a stable alkaline mixture.

Answer: (a) \( \ce{CO2} \) and \( \ce{NH3} \)

Explanation:
1. \( \ce{CO2} \) from heating \( \ce{NaHCO3} \) and from limestone, and \( \ce{NH3} \) recovered from \( \ce{NH4Cl} \) with lime, are reused.
2. \( \ce{CaCl2} \) is discarded as by-product.

Teacher's Note:
1. Recycling of ammonia keeps the process economical.
2. Only \( \ce{CaCl2} \) is a waste product.

Answer: (b) \( \ce{Cl-} \) is oxidised

Explanation:
1. Anions are discharged at the anode by loss of electrons: \( \ce{2Cl- -> Cl2 + 2e-} \).
2. So chloride ions are oxidised.

Teacher's Note:
1. Oxidation occurs at the anode.
2. Sodium is deposited at the cathode.

Answer: (d) metal amide and \( \ce{H2} \)

Explanation:
1. Iron (as \( \ce{Fe3} \)+ compounds) catalyses the reaction \( \ce{2M + 2NH3 -> 2MNH2 + H2} \).
2. The products are metal amide and hydrogen.

Teacher's Note:
1. A catalyst speeds up amide formation.
2. Without it, the blue solution is stable for some time.

Answer: (a) \( \ce{Cs} \)

Explanation:
1. The ionisation energy is lowest for Cs, so it loses electrons most easily to light.
2. It therefore shows the greatest photoelectric effect.

Teacher's Note:
1. Low ionisation energy favours photoemission.
2. Cs is used in photocells.

Answer: (c) Potassium superoxide

Explanation:
1. \( \ce{4KO2 + 2CO2 -> 2K2CO3 + 3O2} \).
2. It absorbs exhaled carbon dioxide and supplies oxygen.

Teacher's Note:
1. A single compound that does both jobs is needed.
2. \( \ce{Na2O2} \) is also used.

Answer: (b) \( \ce{BaO} \) > \( \ce{SrO} \) > \( \ce{CaO} \) > \( \ce{MgO} \)

Explanation:
1. Ionic character rises as the cation gets larger and less polarising.
2. So \( \ce{BaO} \) > \( \ce{SrO} \) > \( \ce{CaO} \) > \( \ce{MgO} \).

Teacher's Note:
1. Fajans' rule links covalency to polarising power.
2. BeO is largely covalent.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true.
2. The blue colour is due to ammoniated electrons, not due to \( \ce{KNH2} \) formation, so R does not explain A.

Teacher's Note:
1. Identify the species responsible for colour.
2. Amide formation is a slow side reaction.

Answer: (d) A is false, but R is true.

Explanation:
1. \( \ce{Na2O2} \) contains the peroxide ion \( \ce{O2^2-} \) with all electrons paired, so it is diamagnetic and pale yellow-white, not paramagnetic.
2. It is a peroxide, not a superoxide, so both statements are false.

Teacher's Note:
1. Peroxide is diamagnetic and superoxide paramagnetic.
2. Check the oxidation state of oxygen.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both statements are true.
2. Burning of Mg in nitrogen is the formation of \( \ce{Mg3N2} \), so R explains A.

Teacher's Note:
1. Mg is a reactive metal that combines with nitrogen at high temperature.
2. Hydrolysis of the nitride gives ammonia.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both statements are true.
2. The small \( \ce{Li+} \) cannot hold the large coordination sphere needed to form alums, so R explains A.

Teacher's Note:
1. Alums need large univalent cations.
2. \( \ce{Li2SO4} \) is the only alkali sulphate that does not form double salts.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true because the high hydration energy of \( \ce{Li+} \) gives it the highest oxidation potential.
2. R is false because the ionisation energy of Li is the HIGHEST in group 1, not the minimum.

Teacher's Note:
1. Aqueous reducing power depends on hydration enthalpy.
2. Cs has the lowest ionisation energy.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both are true.
2. The hardening is due to hydration and formation of an interlocked gel and crystals, not to the heat evolved, so R does not explain A.

Teacher's Note:
1. Heat release is a result of the setting, not its cause.
2. Hydration of silicates and aluminates causes hardening.

Answer: (d) A is false, but R is true.

Explanation:
1. Be has a very HIGH ionisation energy and small size, so it does not show the photoelectric effect.
2. Both statements are false.

Teacher's Note:
1. Low ionisation energy is needed for photoemission.
2. Be has the highest ionisation energy of the group 2 metals.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: small, highly polarising cations make covalent chlorides.
2. R is false because these cations are SMALL.

Teacher's Note:
1. Fajans' rule: small cation, large anion gives covalent character.
2. The covalent chlorides dissolve in organic solvents.