JEE Chemistry S Block Elements MCQs Set 14

Answer: (d) gives no gas at all

Explanation:
1. \( \ce{K2CO3} \) is thermally stable and melts without decomposing.
2. So no gas is evolved.

Teacher's Note:
1. Only \( \ce{Li2CO3} \) decomposes among alkali metal carbonates.
2. Stability increases with cation size.

Answer: (c) violet

Explanation:
1. The 4s electron of K is excited by the flame and emits light as it falls back.
2. The emitted light is violet.

Teacher's Note:
1. Na golden yellow, K violet, Rb red-violet, Cs blue.
2. Li crimson red.

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

Explanation:
1. Only \( \ce{LiOH} \) loses water on heating: \( \ce{2LiOH -> Li2O + H2O} \).
2. The other hydroxides are thermally stable.

Teacher's Note:
1. Li behaves like Mg in its hydroxide: \( \ce{Mg(OH)2} \) also decomposes.
2. This is an anomalous property of lithium.

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

Explanation:
1. Group 1 metals and Ca, Sr and Ba form ionic (saline) hydrides.
2. \( \ce{BeH2} \) is covalent and polymeric, and Be does not combine directly with hydrogen.

Teacher's Note:
1. Be and Mg hydrides are covalent.
2. \( \ce{BeH2} \) is made from \( \ce{BeCl2} \) and \( \ce{LiAlH4} \).

Answer: (b) 0.2 mole of \( \ce{LiH} \)

Explanation:
1. \( \ce{LiH + H2O -> LiOH + H2} \).
2. One mole of \( \ce{LiH} \) gives one mole of \( \ce{H2} \), so 0.2 mole of \( \ce{H2} \) needs 0.2 mole of \( \ce{LiH} \).

Teacher's Note:
1. Use the mole ratio from the balanced equation.
2. Saline hydrides are a convenient source of hydrogen.

Answer: (c) \( \ce{KO2} \)

Explanation:
1. The oxide and peroxide ions have all electrons paired.
2. The superoxide ion \( \ce{O2^-} \) has one unpaired electron, so \( \ce{KO2} \) is paramagnetic.

Teacher's Note:
1. Count electrons in the molecular orbitals of the oxygen species.
2. Superoxides are coloured and paramagnetic.

Answer: (b) \( \ce{Be^2+} \)

Explanation:
1. The small \( \ce{Be^2+} \) (and \( \ce{Mg^2+} \)) has very high ionisation energy and its electrons are not excited by the flame.
2. Li, Na and K ions give crimson, yellow and violet flames.

Teacher's Note:
1. Be and Mg are the group 2 metals with no flame colour.
2. Excitation energy must be available in the flame.

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

Explanation:
1. The small \( \ce{Li+} \) is strongly hydrated, so the salt crystallises as \( \ce{LiCl.2H2O} \).
2. Larger ions give anhydrous salts.

Teacher's Note:
1. High hydration enthalpy of Li+ explains hydrated salts.
2. Mg salts are hydrated for the same reason.

Answer: (d) Both \( \ce{(1)} \) and \( \ce{(2)} \) are correct

Explanation:
1. Ba dissolves to give ammoniated \( \ce{Ba^2+} \) and ammoniated electrons.
2. The electrons make the solution blue, and the ions and electrons make it conducting.

Teacher's Note:
1. Ca, Sr and Ba behave like the alkali metals in liquid ammonia.
2. Be and Mg do not dissolve.

Answer: (b) \( \ce{Cs} \) > \( \ce{Rb} \) > \( \ce{K} \)

Explanation:
1. In the gaseous state reducing power follows low ionisation energy.
2. Ionisation energy falls from K to Cs, so the order is \( \ce{Cs} \) > \( \ce{Rb} \) > \( \ce{K} \).

Teacher's Note:
1. Gas-phase reactivity follows ionisation energy.
2. Aqueous reactivity is different because hydration enters.

Answer: (d) It is easily oxidised

Explanation:
1. Sodium has low ionisation energy, so it loses its electron easily and is a strong reducing agent.
2. It is extracted from fused NaCl, not aqueous solution, and it is denser than K.

Teacher's Note:
1. Alkali metals are oxidised easily.
2. Hydrogen would be liberated at the cathode in aqueous electrolysis.

Answer: (d) cluster formation of metal ions

Explanation:
1. At high concentration (above about 3 M) metal ions form clusters.
2. The solution takes a bronze colour with a metallic lustre.

Teacher's Note:
1. Dilute solutions are blue, concentrated ones bronze.
2. Amide formation is a slow decomposition, which fades the blue colour.

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

Explanation:
1. \( \ce{Be + 2NaOH + 2H2O -> Na2[Be(OH)4] + H2} \).
2. Be is amphoteric, like Al, but Al is not an s-block element.

Teacher's Note:
1. Only Be among the s-block metals dissolves in alkali.
2. The product is sodium beryllate.

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

Explanation:
1. Melting points: Be 1560 K, Ca 1115 K, Ba 1000 K, Mg 923 K.
2. Mg has the lowest value.

Teacher's Note:
1. Group 2 melting points are irregular.
2. Be is the highest.

Answer: (b) \( \ce{BaO2} \)

Explanation:
1. \( \ce{BaO2} \) is a peroxide containing \( \ce{O2^2-} \).
2. \( \ce{RbO2} \) and \( \ce{KO2} \) are superoxides, and \( \ce{Na2O} \) is an oxide.

Teacher's Note:
1. Read the formula to find the oxidation state of oxygen.
2. Superoxide has -1/2 per oxygen.

Answer: (b) a, b, d

Explanation:
1. Metallic character, ionic radius and density increase down the group (K is a slight exception for density).
2. Melting point and ionisation potential decrease.

Teacher's Note:
1. Separate the trends into increasing and decreasing sets.
2. Melting point decreases because metallic bonds weaken.

Answer: (c) \( \ce{Mg} \) + \( \ce{NaOH} \)

Explanation:
1. Al and Zn are amphoteric and dissolve in NaOH with evolution of hydrogen, and \( \ce{LiH} \) gives hydrogen with water.
2. Mg does not react with NaOH, since it is not amphoteric.

Teacher's Note:
1. Only amphoteric metals react with alkali.
2. Be is the amphoteric member of group 2.

Answer: (d) low ionisation potential

Explanation:
1. Superoxide formation needs a large, strongly electropositive cation, i.e. a low ionisation potential.
2. These are K, Rb and Cs.

Teacher's Note:
1. Large, weakly polarising cations stabilise large anions.
2. The lattice energy gain compensates for the formation of \( \ce{O2^-} \).

Answer: (c) It becomes diamagnetic

Explanation:
1. Dilute solutions are paramagnetic because of free solvated electrons, so the statement that it becomes diamagnetic is wrong.
2. The blue colour and conductivity are correct.

Teacher's Note:
1. Concentrated solutions become diamagnetic as electrons pair, but dilute ones are paramagnetic.
2. Hence the statement is false for the usual case.

Answer: (d) low ionisation energy

Explanation:
1. The flame supplies enough energy to excite the outer electron because ionisation energy is low.
2. The electron then emits visible light on returning.

Teacher's Note:
1. Low ionisation energy lets a small energy input excite the electron.
2. Be and Mg have high ionisation energy and give no colour.

Answer: (a) quick lime

Explanation:
1. \( \ce{CaO + H2O -> Ca(OH)2} \) with evolution of heat.
2. The product is slaked lime.

Teacher's Note:
1. Slaking is the reaction of quick lime with water.
2. The reaction is highly exothermic.

Answer: (c) combining with water

Explanation:
1. On mixing with water it takes up water and forms interlocking crystals of gypsum.
2. This is the setting reaction.

Teacher's Note:
1. \( \ce{2CaSO4.H2O + 3H2O -> 2CaSO4.2H2O} \).
2. Do not confuse setting with the dehydration used to make it.

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

Explanation:
1. K, Rb and Cs form superoxides in excess oxygen.
2. Li and Na do not, and Ba forms the peroxide.

Teacher's Note:
1. Large alkali metals form superoxides.
2. Link the metal to the oxide type.

Answer: (d) all of these

Explanation:
1. All three statements show anomalous behaviour of lithium.
2. So the answer is all of these.

Teacher's Note:
1. Lithium is smaller and more polarising.
2. It resembles Mg through the diagonal relationship.

Answer: (c) (b) and (c)

Explanation:
1. (b) is true: only Ca, Sr and Ba of group 2 form peroxides.
2. (c) is true: K is less dense than Na.
3. (a) is false in aqueous solution, where Li is the strongest reductant, and (d) is false because Li has the HIGHEST melting point of the four.

Teacher's Note:
1. Specify whether reducing power is aqueous or gaseous.
2. Li has the highest melting point in group 1.

Answer: (b) \( \ce{Mg^2+} \) is relatively smaller

Explanation:
1. A small, highly charged cation polarises large peroxide and superoxide ions and decomposes them to the oxide.
2. \( \ce{Mg^2+} \) is small, so only \( \ce{MgO} \) is stable.

Teacher's Note:
1. Large anions need large cations for stability.
2. Only Ca, Sr and Ba form peroxides in group 2.

Answer: (b) All form ionic salts such as hydrides \( \ce{MH} \)

Explanation:
1. All alkali metals form ionic hydrides \( \ce{MH} \) when heated with hydrogen.
2. Li does not form a solid bicarbonate, only K, Rb and Cs form superoxides, and only Li forms a nitride directly.

Teacher's Note:
1. Check each general claim for exceptions.
2. Lithium is the exception in most cases.

Answer: (a) \( \ce{LiOH} \) is amphoteric in nature

Explanation:
1. \( \ce{LiOH} \) is a strong base, like other alkali hydroxides, though weaker than the rest.
2. The other statements are correct.

Teacher's Note:
1. Only \( \ce{BeO} \) and \( \ce{Be(OH)2} \) are amphoteric among s-block oxides.
2. LiCl is covalent enough to dissolve in pyridine.