JEE Chemistry S Block Elements MCQs Set 13

Answer: (d) all of these

Explanation:
1. The superoxide ion has one unpaired electron, so the compounds are paramagnetic.
2. They are coloured (\( \ce{KO2} \) orange, \( \ce{RbO2} \) and \( \ce{CsO2} \) dark brown or yellow).
3. Superoxide oxygen is in the -1/2 state and can be reduced to \( \ce{O2} \)- or peroxide, so they act as oxidising agents.
4. All three statements hold.

Teacher's Note:
1. Count the electrons of the superoxide ion by molecular orbital theory to see why it is paramagnetic.
2. Link colour and paramagnetism to the same unpaired electron.

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

Explanation:
1. \( \ce{Li+} \) is very small and has a high polarising power, so \( \ce{LiCl} \) has appreciable covalent character (Fajans' rule).
2. Covalent halides dissolve in organic solvents; the other chlorides are ionic.

Teacher's Note:
1. Smaller cation and larger anion mean greater covalent character.
2. Li shows this anomalous behaviour among alkali metals.

Answer: (a) \( \ce{CsHCO3} \)

Explanation:
1. Solubility of alkali metal hydrogencarbonates increases down the group because lattice energy falls faster than hydration energy.
2. Cs is at the bottom, so \( \ce{CsHCO3} \) is the most soluble.

Teacher's Note:
1. Remember the group trend for large singly charged anions: solubility rises down the group.
2. This is also why \( \ce{KHCO3} \) cannot be made by the Solvay process.

Answer: (b) higher ionisation potential

Explanation:
1. Na lies above K, so it is smaller, with higher ionisation energy, higher electronegativity and a higher melting point (371 K against 336 K).
2. Hence only the second statement is correct.

Teacher's Note:
1. Going up the group, size falls and ionisation energy rises.
2. Melting points of alkali metals fall down the group.

Answer: (c) Be has the most negative standard reduction potential in group 2

Explanation:
1. Standard reduction potentials of group 2 are about Be -1.85 V, Mg -2.37 V, Ca -2.87 V, Ba -2.91 V.
2. Be therefore has the LEAST negative value, so statement 3 is false.
3. Reducing power rises down the group and \( \ce{Mg^2+} \) (less negative E) is reduced more easily than \( \ce{Ba^2+} \), so the other statements are true.

Teacher's Note:
1. Compare actual E values rather than guessing from position.
2. Be is the least reactive and the weakest reducing agent of its group.

Answer: (d) all of these

Explanation:
1. Ca, Sr and Ba are large, with low polarising power, so their halides are ionic.
2. Only the halides of Be (and to some extent Mg) are covalent.

Teacher's Note:
1. Ionic character of halides rises down group 2.
2. Be is the exception because of its small size.

Answer: (d) \( \ce{BaCO3} \)

Explanation:
1. Carbonates of group 2 become more stable down the group as the cation gets larger and its polarising power on the carbonate ion falls.
2. \( \ce{BaCO3} \) decomposes only above about 1630 K.

Teacher's Note:
1. Thermal stability of carbonates, sulphates and nitrates of group 2 rises down the group.
2. \( \ce{BeCO3} \) is so unstable that it is stored under \( \ce{CO2} \).

Answer: (b) potassium hydrogencarbonate is too soluble in water to be precipitated

Explanation:
1. In the Solvay process \( \ce{NaHCO3} \) separates as a precipitate because it is sparingly soluble in cold brine.
2. \( \ce{KHCO3} \) is much more soluble, so it does not precipitate and the process fails.

Teacher's Note:
1. The process depends on precipitation of the hydrogencarbonate.
2. K salts of large anions are generally more soluble than Na salts.

Answer: (c) Setting of cement is an endothermic process

Explanation:
1. Setting involves hydration of calcium silicates and aluminates, and it gives out heat.
2. So it is exothermic, and the statement that it is endothermic is incorrect.

Teacher's Note:
1. Setting is hydration, which releases heat.
2. Concrete is cured with water to control this heat and avoid cracks.

Answer: (b) soluble \( \ce{Ca(HCO3)2} \) is formed

Explanation:
1. First \( \ce{Ca(OH)2 + CO2 -> CaCO3 v + H2O} \) makes the water milky.
2. With excess \( \ce{CO2} \) the precipitate dissolves: \( \ce{CaCO3 + CO2 + H2O -> Ca(HCO3)2} \) (soluble).

Teacher's Note:
1. Milkiness first appears and then disappears with excess \( \ce{CO2} \).
2. On boiling, the hydrogencarbonate gives back \( \ce{CaCO3} \).

Answer: (b) \( \ce{CaOCl2} \)

Explanation:
1. \( \ce{Ca(OH)2 + Cl2 -> CaOCl2 + H2O} \) below 308 K.
2. The product is bleaching powder, \( \ce{CaOCl2} \).

Teacher's Note:
1. Dry slaked lime and chlorine give bleaching powder.
2. Do not confuse it with calcium hypochlorite, which needs different conditions.

Answer: (a) Gypsum contains a lower percentage of Ca than plaster of Paris

Explanation:
1. Calcium is 40 g in 172 g of gypsum (23.3 percent) and 40 g in 145 g of the hemihydrate (27.6 percent).
2. Gypsum therefore has the lower Ca percentage.
3. Plaster of Paris is made by partial DEHYDRATION of gypsum, so the other statements are wrong.

Teacher's Note:
1. Compute mass percentages when asked to compare compounds.
2. Gypsum loses water on heating; plaster of Paris takes water on setting.

Answer: (d) (i), (ii), (iii), (iv)

Explanation:
1. Calcium ions take part in nerve impulse transmission, muscle contraction, maintaining cell membranes and blood clotting.
2. So all four are correct.

Teacher's Note:
1. Learn \( \ce{Ca2} \)+ roles as a set: nerves, muscles, membranes, clotting, bones.
2. \( \ce{Mg2} \)+ is linked with chlorophyll and ATP-using enzymes.

Answer: (c) \( \ce{Mg} \) > \( \ce{Ca} \)

Explanation:
1. Densities in g/cm3: Li 0.53, Na 0.97, K 0.86, Rb 1.53, Cs 1.90, Mg 1.74, Ca 1.55.
2. Only \( \ce{Mg} \) > \( \ce{Ca} \) is true.

Teacher's Note:
1. Density generally rises down a group, but K is an exception, lighter than Na.
2. Check each option against the data rather than the trend alone.

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

Explanation:
1. Standard oxidation potentials are Li 3.05 V, K 2.93 V, Cs 2.92 V and Na 2.71 V.
2. Na has the lowest value, so it is the weakest reductant.

Teacher's Note:
1. Li is the best aqueous reductant because of its very high hydration energy.
2. The order is Li greater than Cs, Rb, K greater than Na.

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

Explanation:
1. Cs has the lowest ionisation energy among stable metals, so even visible light can eject its valence electron.
2. It is therefore used in photoelectric cells.

Teacher's Note:
1. Low ionisation energy means easy photoelectron emission.
2. Be and Mg do not show the effect.

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

Explanation:
1. Stability of alkali metal hydrides decreases down the group as the cation grows.
2. \( \ce{LiH} \) is the most stable.

Teacher's Note:
1. Smaller cation gives stronger ionic bonding with the small H- ion.
2. Basic character and reactivity with water rise from LiH to CsH.

Answer: (a) \( \ce{CaH2} \)

Explanation:
1. In \( \ce{CaH2} \), hydrogen is bonded to a much less electronegative metal and exists as H-.
2. In \( \ce{CH4} \) and \( \ce{HCl} \), H is less electronegative than C and Cl.

Teacher's Note:
1. Saline hydrides contain H- ions.
2. Ionic hydrides give \( \ce{H2} \) with water.

Answer: (a) \( \ce{Na} \) > \( \ce{K} \)

Explanation:
1. Na is 0.97 g/cm3 and K is 0.86 g/cm3, so Na is denser.
2. K is anomalous, because its atomic volume is unexpectedly large.

Teacher's Note:
1. The usual rise in density down the group breaks at potassium.
2. Li is the least dense metal.

Answer: (b) diamagnetic and colourless

Explanation:
1. The cations have noble gas configurations with no unpaired electrons, and absorb no visible light.
2. So the salts are diamagnetic and colourless unless the anion is coloured.

Teacher's Note:
1. No unpaired electrons means diamagnetic.
2. Colour in salts needs d-d transitions or coloured anions.

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

Explanation:
1. Melting point follows lattice energy, which is largest for the smallest ions.
2. \( \ce{NaF} \) has the smallest anion and the smallest cation among the options.

Teacher's Note:
1. Lattice energy varies inversely with the sum of ionic radii.
2. NaF melts at about 1266 K against 1074 K for NaCl.

Answer: (d) \( \ce{K2CO3} \)

Explanation:
1. Carbonates of Mg, Li and Zn decompose to the oxide and \( \ce{CO2} \) on heating.
2. \( \ce{K2CO3} \) is very stable and melts without decomposition.

Teacher's Note:
1. Alkali metal carbonates (except Li) are stable to heat.
2. Thermal stability rises with cation size.

Answer: (b) \( \ce{NH3} \)

Explanation:
1. \( \ce{3Mg + N2 -> Mg3N2} \).
2. \( \ce{Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3} \), so ammonia is evolved.

Teacher's Note:
1. Metal nitrides give ammonia on hydrolysis.
2. Be, Mg and Li all form nitrides.

Answer: (b) free ions

Explanation:
1. In the molten state the ionic lattice breaks down and \( \ce{K+} \) and \( \ce{Cl-} \) ions move freely.
2. Their movement carries current.

Teacher's Note:
1. Ionic compounds conduct when molten or dissolved, not when solid.
2. Metals conduct through free electrons instead.

Answer: (d) \( \ce{H2} \)

Explanation:
1. \( \ce{2Na + 2NH3 -> 2NaNH2 + H2} \).
2. The gas evolved is hydrogen, and the solid product is sodamide.

Teacher's Note:
1. Distinguish the gas from the solid product in such questions.
2. Sodium amide is a strong base.

Answer: (c) is more electropositive

Explanation:
1. Na has the lower ionisation energy and is more electropositive.
2. Its heat of reaction melts the metal and the hydrogen can ignite.

Teacher's Note:
1. Reactivity with water increases down the group (apart from the hydration effect that makes Li E-value highest).
2. Li reacts slowly because it has a high ionisation energy and a high melting point.

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

Explanation:
1. Reducing power in group 2 rises with falling ionisation energy.
2. Ba has the most negative reduction potential, so it is the strongest reductant.

Teacher's Note:
1. Be is the weakest reducing agent of the group.
2. Compare standard electrode potentials.

Answer: (c) the formation of an inert oxide layer on the surface of the metal

Explanation:
1. Conc. \( \ce{HNO3} \) oxidises the metal surface and forms a thin, tight oxide film.
2. The film prevents further attack, so the metals become passive.

Teacher's Note:
1. Passivity comes from a protective oxide layer.
2. This is one similarity in the Be and Al diagonal relationship.