JEE Chemistry S Block Elements MCQs Set 08

Answer: (c) Na, \(\ce{H2}\), NaOH and Zn

Explanation:
1. Golden yellow flame means A is Na.
2. \(\ce{2Na + 2H2O -> 2NaOH + H2}\), so B is hydrogen and C is NaOH.
3. Zinc gives hydrogen with NaOH and with dilute \(\ce{H2SO4}\).

Teacher's Note:
1. Zinc is amphoteric and gives hydrogen with both acid and alkali.

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

Explanation:
1. Solubility of group 2 hydroxides increases down the group.
2. So Be(OH)2 is the least soluble and has the lowest Ksp.

Teacher's Note:
1. Hydroxides become more soluble down the group.

Answer: (c) \(\ce{BaS2O3}\)

Explanation:
1. Green flame points to Ba (Cu also gives green-blue).
2. \(\ce{BaS2O3 + 2HCl -> BaCl2 + S + SO2 + H2O}\) gives yellow sulphur (Y) and pungent \(\ce{SO2}\) (Z).
3. \(\ce{Ba^2+ + CrO4^2- -> BaCrO4}\) is the yellow precipitate T.

Teacher's Note:
1. Thiosulphates with acid give sulphur and sulphur dioxide.

Answer: (b) \(\ce{Li2CO3}\)

Explanation:
1. Li+ is the smallest and polarises the carbonate ion most.
2. Li2CO3 decomposes to the oxide; other alkali carbonates do not.

Teacher's Note:
1. Stability of alkali carbonates rises down the group.

Answer: (d) \(\ce{Mg(OH)2}\)

Explanation:
1. A suspension of magnesium hydroxide in water is called milk of magnesia.
2. It neutralises excess stomach acid.

Teacher's Note:
1. A weak base is a gentle antacid.

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

Explanation:
1. A hydrogencarbonate reacts with a hydroxide: \(\ce{HCO3- + OH- -> CO3^2- + H2O}\).
2. The other pairs do coexist.

Teacher's Note:
1. An acid salt and a base react to form a normal salt.

Answer: (d) solvated electrons

Explanation:
1. Dissolved sodium ionises to give \(\ce{Na+}\) and ammoniated electrons.
2. The free electrons readily donate and reduce other substances.

Teacher's Note:
1. Solvated electrons also give the blue colour.

Answer: (b) \(\ce{Na(NH4)HPO4.4H2O}\)

Explanation:
1. Microcosmic salt, \(\ce{Na(NH4)HPO4.4H2O}\), is used in the bead test for metal ions.
2. The others serve other purposes.

Teacher's Note:
1. Microcosmic salt gives coloured phosphate beads with metal oxides.

Answer: (a) \(\ce{H2SO4}\) and \(\ce{NaHCO3}\) solutions

Explanation:
1. When the two solutions mix, \(\ce{2NaHCO3 + H2SO4 -> Na2SO4 + 2H2O + 2CO2}\).
2. The CO2 smothers the fire.

Teacher's Note:
1. The soda-acid extinguisher releases CO2.

Answer: (c) both \(\ce{Ca^2+}\) and \(\ce{Mg^2+}\) ions

Explanation:
1. EDTA forms stable complexes with both ions, so it is used to measure the hardness of water caused by both.

Teacher's Note:
1. EDTA titration is the standard test for hardness.

Answer: (d) \(\ce{NaH2PO2}\)

Explanation:
1. \(\ce{P4 + 3NaOH + 3H2O -> PH3 + 3NaH2PO2}\).
2. It is a disproportionation: P goes from 0 to -3 and +1.

Teacher's Note:
1. The product is sodium hypophosphite.

Answer: (a) atomic size increases

Explanation:
1. Ions that are larger have a lower charge density.
2. So their ability to form complexes falls.

Teacher's Note:
1. Charge density is the deciding factor.

Answer: (b) Be and Mg

Explanation:
1. Be and Mg have very high excitation energy, so their electrons are not excited by the flame.

Teacher's Note:
1. Ca - brick red, Sr - crimson, Ba - apple green.

Answer: (b) barium carbonate

Explanation:
1. A yellow precipitate with chromate in acetic acid and a white sulphate precipitate indicate \(\ce{Ba^2+}\).
2. So B is \(\ce{BaCrO4}\) and C is \(\ce{BaSO4}\).
3. The white precipitate D whose reaction gives such a salt is \(\ce{BaCO3}\).

Teacher's Note:
1. BaCrO4 does not dissolve in acetic acid; SrCrO4 does.

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

Explanation:
1. \(\ce{NaOH + CO -> HCOONa}\).
2. The product is sodium formate.

Teacher's Note:
1. This is an industrial route to formates.

Answer: (c) Y = \(\ce{Mg3N2}\) and T = \(\ce{CuSO4.4NH3}\)

Explanation:
1. X is Mg; it forms \(\ce{Mg3N2}\) in nitrogen.
2. Hydrolysis gives \(\ce{NH3}\).
3. Ammonia with Cu2+ gives the deep blue complex \(\ce{CuSO4.4NH3}\).

Teacher's Note:
1. Ammonia plus Cu2+ gives a deep blue colour.

Answer: (d) X > Y

Explanation:
1. For the alkali metal the second electron is removed from a stable noble-gas core.
2. For the alkaline earth metal it is removed from the outer s-orbital.
3. Hence X is much larger than Y.

Teacher's Note:
1. Second IE is very high for alkali metals.

Answer: (c) \(\ce{KNO3}\)

Explanation:
1. KNO3 decomposes only to \(\ce{KNO2}\) and oxygen.
2. The others give oxide or metal with \(\ce{NO2}\).

Teacher's Note:
1. Nitrates of Na and K give nitrites on heating.

Answer: (a) the anionic part

Explanation:
1. \(\ce{LiAlH4}\) is \(\ce{Li+}\) \(\ce{AlH4-}\).
2. Aluminium is part of the complex anion.

Teacher's Note:
1. Li+ is the cation.

Answer: (a) \(\ce{OCl-}\)

Explanation:
1. The hypochlorite ion in bleaching powder gives oxygen with \(\ce{CoCl2}\) as catalyst: \(\ce{2OCl- -> 2Cl- + O2}\).
2. The bleaching action is due to this ion.

Teacher's Note:
1. Bleaching powder is Ca(OCl)Cl.

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

Explanation:
1. Lattice energy increases as ionic size decreases.
2. Li+ is the smallest alkali cation, so LiF has the highest lattice energy.

Teacher's Note:
1. Same anion, smaller cation means higher lattice energy.

Answer: (b) complexes with alkali metals

Explanation:
1. These ligands have cavities that trap alkali metal ions selectively, forming stable complexes.

Teacher's Note:
1. Selectivity depends on cavity size.

Answer: (b) \(\ce{Cs+ < Rb+ < K+ < Na+ < Li+}\)

Explanation:
1. The degree of hydration is greatest for the smallest ion.
2. Hence Cs+ < Rb+ < K+ < Na+ < Li+.

Teacher's Note:
1. Small size means large hydration.

Answer: (b) \(\ce{BeH2}\)

Explanation:
1. Balanced: \(\ce{2BeCl2 + LiAlH4 -> 2BeH2 + LiCl + AlCl3}\).

Teacher's Note:
1. LiAlH4 is a hydride source.