JEE Chemistry S Block Elements MCQs Set 06

Answer: (a) only +1 oxidation state

Explanation:
1. Each alkali metal has one electron in its outer s-orbital.
2. Removing it gives a stable noble-gas core.
3. The second ionisation energy is enormous, so only +1 is shown in compounds.

Teacher's Note:
1. Think of the outer electron as the only one easy to lose.
2. Going to +2 would break into a noble-gas core.

Answer: (c) diamagnetic and colourless

Explanation:
1. The ions \(\ce{Na+}\), \(\ce{K+}\) etc. have complete noble-gas shells, so no unpaired electrons.
2. No unpaired electrons means diamagnetic.
3. No d-d or charge-transfer transitions are possible, so the salts are colourless (unless the anion is coloured).

Teacher's Note:
1. Colour in salts needs unpaired electrons or easy electronic transitions.
2. Noble-gas-like ions have neither.

Answer: (c) \(\ce{Li+}\)(aq)

Explanation:
1. The smallest bare ion, \(\ce{Li+}\), has the highest charge density.
2. It attracts the most water molecules.
3. So its hydrated radius is the largest, and the order is Li+ > Na+ > K+ > Rb+ > Cs+ (hydrated).

Teacher's Note:
1. Hydrated size runs opposite to bare ionic size.
2. Smaller ion, more water, bigger hydrated ion.

Answer: (d) violet

Explanation:
1. Heat excites the outer electron of K to a higher level.
2. When it falls back, it emits visible light in the violet region.
3. Na is golden yellow, Li crimson red and Ba apple green.

Teacher's Note:
1. Learn the set: Li crimson, Na golden yellow, K violet (lilac), Rb red-violet, Cs blue.

Answer: (c) low ionisation potential

Explanation:
1. The flame supplies only a small amount of energy.
2. Because the ionisation (excitation) energy of alkali metals is low, the valence electron is excited by this energy.
3. On returning to the ground state it emits visible light.

Teacher's Note:
1. Low excitation energy is the real reason; having one electron alone is not enough (Li and Be also differ).

Answer: (a) low melting point

Explanation:
1. Li has the most negative electrode potential, but it reacts less vigorously.
2. Its melting point is higher, so the heat of reaction does not melt it easily.
3. Sodium melts, exposes fresh surface and reacts violently.

Teacher's Note:
1. Reaction rate depends on exposed surface as well as on thermodynamics.

Answer: (b) \(\ce{Na2SO4.10H2O}\)

Explanation:
1. Glauber's salt is sodium sulphate decahydrate.
2. Na2S2O3.5H2O is hypo, and Na2CO3.10H2O is washing soda.

Teacher's Note:
1. Match common names: Glauber's salt = sodium sulphate decahydrate, hypo = sodium thiosulphate pentahydrate.

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

Explanation:
1. Photoelectric cells need a metal that loses electrons with visible light.
2. Cs has the lowest ionisation energy and work function.
3. Hence Cs (and K) are used.

Teacher's Note:
1. Lowest ionisation energy means electrons are ejected by the weakest light.

Answer: (a) solvated (ammoniated) electrons

Explanation:
1. \(\ce{Na + (x+y)NH3 -> [Na(NH3)x]+ + [e(NH3)y]-}\).
2. The ammoniated electron absorbs light in the red region.
3. So the transmitted light is blue.

Teacher's Note:
1. The blue colour and the strong reducing power both come from the ammoniated electron.

Answer: (a) \(\ce{Na2CO3.10H2O}\)

Explanation:
1. Washing soda is sodium carbonate decahydrate.
2. The anhydrous form is soda ash and the monohydrate is a different crystalline form.

Teacher's Note:
1. Washing soda has 10 water molecules; soda ash has none.

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

Explanation:
1. Baking soda is sodium hydrogencarbonate.
2. On heating it gives \(\ce{Na2CO3}\), water and carbon dioxide, which makes dough rise.

Teacher's Note:
1. Baking soda = bicarbonate, washing soda = carbonate.

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

Explanation:
1. Electropositive character increases down the group as ionisation energy falls.
2. Cs has the lowest ionisation energy among the choices.

Teacher's Note:
1. Down the group, the outer electron is held more loosely.

Answer: (a) \(\ce{O2 + NaNO2}\)

Explanation:
1. Nitrates of Na and K decompose only up to the nitrite.
2. \(\ce{2NaNO3 -> 2NaNO2 + O2}\).
3. Only Li and the nitrates of heavier metals give the oxide and \(\ce{NO2}\).

Teacher's Note:
1. Li nitrate behaves differently from the other alkali nitrates.

Answer: (d) more ionisation potential

Explanation:
1. Na is smaller than K, so its outer electron is held more strongly and the ionisation energy is higher.
2. Na also has higher electronegativity and a higher melting point.

Teacher's Note:
1. Going up the group, size falls and ionisation energy rises.

Answer: (b) \(\ce{NaNO3}\)

Explanation:
1. Large deposits of sodium nitrate occur in Chile, hence the name.
2. KNO3 is Indian saltpetre.

Teacher's Note:
1. Chile saltpetre = sodium nitrate; Indian saltpetre = potassium nitrate.

Answer: (c) a basic oxide

Explanation:
1. The configuration (19 electrons) is that of potassium.
2. The oxides of alkali metals are strongly basic.

Teacher's Note:
1. Count electrons first to identify the element.

Answer: (b) \(\ce{LiHCO3}\)

Explanation:
1. Bicarbonates are stabilised by hydrogen bonding in the solid for Na and K.
2. Lithium bicarbonate does not exist in the solid state and is known only in solution.

Teacher's Note:
1. Li is the exception among alkali metals in many compounds.

Answer: (a) caustic soda

Explanation:
1. Causticisation is the treatment of sodium carbonate solution with lime.
2. \(\ce{Na2CO3 + Ca(OH)2 -> 2NaOH + CaCO3}\).

Teacher's Note:
1. Insoluble \(\ce{CaCO3}\) precipitates and leaves NaOH solution.

Answer: (b) sodium peroxide

Explanation:
1. Sodium peroxide, \(\ce{Na2O2}\), is called oxone as it supplies oxygen (with water or acids) and is a strong oxidiser.

Teacher's Note:
1. Remember \(\ce{Na2O2}\) as an oxidising agent and an air purifier along with \(\ce{KO2}\).

Answer: (b) nickel

Explanation:
1. Castner's process electrolyses fused NaOH.
2. The cathode is iron and the anode is nickel.
3. Graphite is used in Down's process for fused NaCl.

Teacher's Note:
1. Different processes use different electrodes; do not mix Castner and Down's cells.

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

Explanation:
1. Li is very small and prefers the small oxide ion, giving only \(\ce{Li2O}\) (with a little nitride).
2. Na, Ba and Ca form peroxides on heating in air or oxygen.

Teacher's Note:
1. A small cation stabilises a small anion (oxide), a large cation stabilises larger anions (peroxide, superoxide).

Answer: (a) \(\ce{Li2O + NO2}\) (and \(\ce{O2}\))

Explanation:
1. \(\ce{4LiNO3 -> 2Li2O + 4NO2 + O2}\).
2. Li nitrate resembles the nitrates of Mg, not those of Na and K.

Teacher's Note:
1. Lithium follows its diagonal neighbour Mg here.

Answer: (b) \(\ce{MgCl2}\)

Explanation:
1. Ca, Sr and Ba are excited by flame energy and give brick red, crimson and apple green.
2. Be and Mg have very high excitation energy and give no colour.

Teacher's Note:
1. Remember: Be and Mg give no flame colour.

Answer: (c) \(\ce{Na+}\)

Explanation:
1. Hydration energy rises with charge and falls with size.
2. \(\ce{Al^3+}\) and \(\ce{Be^2+}\) have higher hydration energy than \(\ce{Mg^2+}\).
3. \(\ce{Na+}\) has lower charge, so its hydration energy is lower.
4. \(\ce{Mg^3+}\) does not exist under ordinary conditions.

Teacher's Note:
1. Compare charge and size together when ranking hydration energies.