JEE Chemistry S Block Elements MCQs Set 10

Answer: (a) \(\ce{Rb+}\)

Explanation:
1. Mobility depends on the hydrated size.
2. Rb+ is the least hydrated of the four, so it moves fastest.

Teacher's Note:
1. Least hydration means highest mobility.

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

Explanation:
1. \(\ce{CaCO3 -> CaO + CO2}\).
2. CaO is basic and CO2 is acidic.

Teacher's Note:
1. Metal carbonates often give basic oxide plus CO2.

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

Explanation:
1. \(\ce{2NaHCO3 + H2SO4 -> Na2SO4 + 2H2O + 2CO2}\).
2. The CO2 produced puts out the fire.

Teacher's Note:
1. Soda-acid extinguisher uses bicarbonate.

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

Explanation:
1. Reactivity of alkali metals increases down the group because of the falling ionisation energy.
2. Rb is the lowest among the four in the group.

Teacher's Note:
1. Cs is the most reactive, then Rb.

Answer: (b) \(\ce{Mg3N2}\)

Explanation:
1. Mg heated in N2 gives \(\ce{Mg3N2}\).
2. Hydrolysis gives NH3, which with \(\ce{Cu^2+}\) gives a deep blue complex.

Teacher's Note:
1. Deep blue colour means \(\ce{Cu(NH3)4^2+}\).

Answer: (c) \(\ce{3Ca3(PO4)2.CaF2}\)

Explanation:
1. Fluoride replaces hydroxide in hydroxyapatite, giving fluoroapatite.
2. This is harder and more resistant to acid attack.

Teacher's Note:
1. Hydroxyapatite becomes fluoroapatite.

Answer: (c) \(\ce{Li+ > Na+ > K+ > Rb+ > Cs+}\)

Explanation:
1. Hydration enthalpy falls as the ion becomes bigger down the group.

Teacher's Note:
1. Smallest ion, strongest hydration.

Answer: (c) chain and dimeric

Explanation:
1. In the solid, Be is bridged by chlorine in a polymeric chain.
2. In vapour at about 1000 K it exists as a linear monomer, and near 700 K as a dimer.
3. So the pair is chain (solid) and dimeric (vapour).

Teacher's Note:
1. Be completes its octet through chloride bridges.

Answer: (a) washing soda and soda ash

Explanation:
1. \(\ce{Na2CO3.10H2O -> Na2CO3.H2O -> Na2CO3}\).
2. Washing soda goes to monohydrate (Y) and then to soda ash (Z).

Teacher's Note:
1. Baking soda has no water of crystallisation.

Answer: (a) \(\ce{Be(OH)2}\)

Explanation:
1. Be(OH)2 reacts with both acids and alkalis.
2. The others are basic.

Teacher's Note:
1. Be is the only amphoteric element in group 2.

Answer: (b) i-R, ii-S, iii-Q, iv-P

Explanation:
1. Sodium carbonate is made by the Solvay process (R).
2. Mg(HCO3)2 causes temporary hardness (S).
3. NaOH is made by the Castner-Kellner process (Q).
4. Ca3Al2O6 is a constituent of Portland cement (P).

Teacher's Note:
1. Learn each compound's main use or process.

Answer: (c) \(\ce{Mg(HCO3)2}\) and \(\ce{Mg(OH)2}\)

Explanation:
1. Temporary hardness is due to bicarbonates of Ca and Mg.
2. On boiling, Ca(HCO3)2 gives insoluble CaCO3, while Mg(HCO3)2 gives insoluble Mg(OH)2 because Mg(OH)2 is less soluble than MgCO3.
3. So X = Mg(HCO3)2 and Y = Mg(OH)2.

Teacher's Note:
1. Mg gives hydroxide on boiling; Ca gives carbonate.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. Both statements are true: the blue colour comes from ammoniated electrons.
2. Statement II explains the species formed, but the blue colour itself is due to the electron, so II is not the full explanation.

Teacher's Note:
1. The colour is from the solvated electron, not the solvated cation.

Answer: (c) \(\ce{NH4NO3}\) and \(\ce{NH4NO2}\)

Explanation:
1. NH4+ with NaOH gives NH3 on boiling, and the evolution stops when NH4+ is used up.
2. Zn in alkali reduces nitrate or nitrite to NH3, so evolution restarts.
3. Both NH4NO3 and NH4NO2 fit; chloride and sulphate have no reducible anion.

Teacher's Note:
1. Zn in alkaline solution reduces nitrate and nitrite to ammonia.

Answer: (c) disproportionation reaction

Explanation:
1. \(\ce{P4 + 3NaOH + 3H2O -> PH3 + 3NaH2PO2}\).
2. P goes from 0 to -3 and +1, so P is both oxidised and reduced.

Teacher's Note:
1. One species forming both a higher and lower oxidation state is disproportionation.

Answer: (a) \(\ce{NaHSO4}\)

Explanation:
1. Chlorine oxidises thiosulphate to sulphate: \(\ce{Na2S2O3 + 4Cl2 + 5H2O -> 2NaHSO4 + 8HCl}\).
2. Iodine instead gives tetrathionate.

Teacher's Note:
1. Cl2 is a stronger oxidant than I2, so it takes S further to sulphate.

Answer: (b) \(\ce{Na2O2}\) and \(\ce{Na2O}\)

Explanation:
1. Na burns in excess air to give mainly \(\ce{Na2O2}\) with a little \(\ce{Na2O}\).
2. NaOH needs water.

Teacher's Note:
1. Excess air gives the peroxide for Na.

Answer: (b) 3

Explanation:
1. A solution turns red litmus blue if it is basic.
2. Basic by hydrolysis: KCN, \(\ce{K2CO3}\) and LiCN (salts of strong base and weak acid).
3. \(\ce{(NH4)2C2O4}\) is nearly neutral; the other salts are neutral or acidic.
4. So the count is 3.

Teacher's Note:
1. Salts of strong base and weak acid give basic solutions.

Answer: (a) \(\ce{Na2S4O6}\)

Explanation:
1. A = NaHSO3.
2. B = Na2SO3.
3. C = Na2S2O3 (sodium thiosulphate, from sulphur).
4. D = Na2S4O6 (sodium tetrathionate, average oxidation state of S = +2.5).

Teacher's Note:
1. Hypo plus iodine gives tetrathionate, the basis of iodometry.

Answer: (c) the M2+ ions are smaller and have higher charge density, so they hydrate more strongly

Explanation:
1. Group 2 ions have double charge and are smaller than the alkali ions in the same period.
2. So they hold water molecules strongly and give hydrated crystals.

Teacher's Note:
1. Charge density controls hydration.

Answer: (c) \(\ce{SrSO4 > CaSO4 > MgSO4 > BeSO4}\)

Explanation:
1. Thermal stability of sulphates increases as the cation gets larger.
2. So the decreasing order is SrSO4 > CaSO4 > MgSO4 > BeSO4.

Teacher's Note:
1. Same trend as carbonates and nitrates.

Answer: (a) dissolution of calcium acetate is exothermic and that of lead nitrate is endothermic

Explanation:
1. By Le Chatelier's principle, heating shifts an exothermic dissolution towards the solid (lower solubility).
2. Calcium acetate dissolves with release of heat, so its solubility falls on heating.
3. Lead nitrate dissolves by absorbing heat, so its solubility rises.

Teacher's Note:
1. The sign of enthalpy of solution decides the effect of temperature.

Answer: (b) the common ion effect of \(\ce{NH4+}\) reduces \(\ce{OH-}\) below the level needed for \(\ce{Mg(OH)2}\) to precipitate

Explanation:
1. \(\ce{NH4Cl}\) supplies \(\ce{NH4+}\), which suppresses ionisation of \(\ce{NH4OH}\).
2. The \(\ce{OH-}\) concentration falls, so the ionic product of \(\ce{Mg(OH)2}\) stays below Ksp.

Teacher's Note:
1. This is the reason group IV reagent is used with NH4Cl in qualitative analysis.