JEE Chemistry S Block Elements MCQs Set 19

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

Explanation:
1. Sodium burns in excess oxygen to form the peroxide, \(\ce{2Na + O2 -> Na2O2}\).
2. Lithium gives the normal oxide, Na gives the peroxide and K, Rb, Cs give superoxides.

Teacher's Note:
1. Larger cations stabilise larger anions: oxide, peroxide, superoxide.
2. Na2O is formed only in a limited supply of oxygen.

Answer: (c) The lattice enthalpy of s-block compounds increases down the group

Explanation:
1. Down the group the ionic size increases, so hydration enthalpy and hydration number decrease and the hydrated ions become smaller, so conductivity rises.
2. Lattice enthalpy decreases (not increases) with larger ions, so (d) is incorrect.

Teacher's Note:
1. Smaller hydrated radius means higher ionic mobility.
2. Lattice enthalpy depends inversely on the distance between ions.

Answer: (b) IV < II < III < I

Explanation:
1. Thermal stability of carbonates increases down group 2 as the polarising power of the cation decreases: BeCO3 < MgCO3 < CaCO3.
2. K2CO3 does not decompose on heating and is the most stable.
3. The order is IV < II < III < I.

Teacher's Note:
1. A small cation polarises the carbonate ion more and decomposes easily.
2. Alkali metal carbonates (except Li2CO3) are stable to heat.

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

Explanation:
1. KO2 contains the superoxide ion \(\ce{O2-}\) with one unpaired electron in the \(\pi^*\) orbital, so it is paramagnetic.
2. Peroxide \(\ce{O2^{2-}}\), ozone and PbO2 have all electrons paired.

Teacher's Note:
1. Superoxide has 17 electrons, an odd number.
2. Odd electron species are paramagnetic.

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

Explanation:
1. Glauber's salt is sodium sulphate decahydrate.
2. MgSO4.7H2O is Epsom salt, FeSO4.7H2O is green vitriol and Na2CO3.10H2O is washing soda.

Teacher's Note:
1. Learn the common names of hydrated salts.
2. Glauber's salt is used as a laxative.

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

Explanation:
1. Microcosmic salt, \(\ce{Na(NH4)HPO4.4H2O}\), on heating gives a glassy bead of NaPO3 that forms coloured metaphosphates with metal oxides.
2. The other salts are alum, Mohr's salt and blue vitriol.

Teacher's Note:
1. Borax and microcosmic salt are used for bead tests.
2. The colour of the bead helps identify the cation.

Answer: (d) \(\ce{CaCl2}\) and \(\ce{MgCl2}\)

Explanation:
1. Deliquescent chlorides of calcium and magnesium absorb moisture from air.
2. Pure NaCl is not hygroscopic.

Teacher's Note:
1. CaCl2 and MgCl2 are very soluble and deliquescent.
2. Sulphates and carbonates do not take up moisture so readily.

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

Explanation:
1. \(\ce{CaCO3 -> CaO + CO2}\) (colourless gas).
2. CaO dissolves in water to give \(\ce{Ca(OH)2}\) (Y); excess CO2 forms \(\ce{Ca(HCO3)2}\) (Z).
3. Gentle heating of Z gives back CaCO3.

Teacher's Note:
1. Lime water turns milky then clear with excess CO2.
2. Calcium bicarbonate exists only in solution.

Answer: (a) \(\ce{NaCl, NH3, CaCO3}\)

Explanation:
1. Brine (NaCl), ammonia and limestone (CaCO3, which gives CO2 and CaO) are the raw materials.
2. NH3 is recovered and recycled; CaCl2 is the by-product.

Teacher's Note:
1. Ammonia is recycled in the Solvay process.
2. The net reaction is \(\ce{2NaCl + CaCO3 -> Na2CO3 + CaCl2}\).

Answer: (d) Aluminium bronze

Explanation:
1. Electron contains Mg with Zn, Al; magnalium is Al + Mg; duralumin is Al, Cu, Mg, Mn.
2. Aluminium bronze is Cu + Al.

Teacher's Note:
1. Light alloys of aluminium often include magnesium.
2. Bronze alloys are copper-based.

Answer: (b) solvated electrons

Explanation:
1. Alkali metals dissolve in liquid NH3 to give \(\ce{M+}\) and ammoniated electrons \(\ce{e^-(NH3)_x}\).
2. The solvated electrons give the blue colour, conductivity and strong reducing power.

Teacher's Note:
1. The solution is blue and paramagnetic.
2. On standing it slowly forms sodamide and hydrogen.

Answer: (a) Alums form hexagonal crystals

Explanation:
1. Alums crystallise in the cubic (octahedral) system, not hexagonal.
2. The other three statements are correct.

Teacher's Note:
1. Alums have the formula M2SO4.M'2(SO4)3.24H2O.
2. Water softening uses Na2CO3 and lime.

Answer: (d) Li

Explanation:
1. Reactivity with water increases down the group as ionisation enthalpy decreases.
2. Li is the least reactive even though its E° is the most negative, as it melts less easily and its hydroxide coats the surface.

Teacher's Note:
1. Lithium reacts slowly; potassium catches fire; caesium reacts explosively.
2. The trend is Li < Na < K < Rb < Cs.

Answer: (b) hydration enthalpy

Explanation:
1. The Li+ ion is very small and has the highest hydration enthalpy.
2. This large energy release outweighs its higher ionisation enthalpy, giving the most negative \(E^\circ\) for lithium.

Teacher's Note:
1. E° (Li+/Li) = -3.05 V is the lowest in the series.
2. Aqueous reducing power involves sublimation, ionisation and hydration.

Answer: (c) \(\ce{BeCl2}\)

Explanation:
1. BeCl2 is largely covalent because of the high polarising power of the small Be2+ ion.
2. Covalent halides dissolve in organic solvents such as ethanol and ether.

Teacher's Note:
1. Fajans' rule: small, highly charged cations give covalent compounds.
2. The halides of Mg, Ca, Sr and Ba are ionic.

Answer: (a) high lattice enthalpy

Explanation:
1. The small Li+ and F- ions pack closely, giving a very high lattice enthalpy.
2. This exceeds the hydration enthalpy, so LiF is sparingly soluble.

Teacher's Note:
1. Solubility depends on hydration enthalpy versus lattice enthalpy.
2. LiF is the exception among the alkali metal halides.

Answer: (a) Be

Explanation:
1. Mg, Ca, Sr and Ba form hydrides on heating with H2.
2. BeH2 is covalent and polymeric and must be prepared indirectly, e.g. from BeCl2 and LiAlH4.

Teacher's Note:
1. Be is the exception in group 2.
2. BeH2 is made by reduction of BeCl2.

Answer: (c) hydration energies of cations

Explanation:
1. The lattice energy changes only a little down the group, because the large carbonate ion dominates.
2. The hydration energy of the cation falls markedly as the size increases, so solubility decreases.

Teacher's Note:
1. The same reasoning applies to the hydroxides' trend (opposite).
2. Compare hydration energy and lattice energy.

Answer: (d) Mg > Ca > Sr > Ba

Explanation:
1. The hydration energy falls more rapidly than the lattice energy down the group, so the solubility of sulphates decreases.
2. The order is Mg > Ca > Sr > Ba.

Teacher's Note:
1. BeSO4 and MgSO4 are quite soluble; BaSO4 is practically insoluble.
2. BaSO4 precipitation is used to test for sulphate.

Answer: (d) \(\ce{BeSO4}\)

Explanation:
1. Be2+ is so small that its hydration enthalpy is very large and exceeds the lattice enthalpy.
2. BeSO4 is therefore readily soluble; the other sulphates are sparingly soluble.

Teacher's Note:
1. Solubility requires hydration enthalpy to exceed lattice enthalpy.
2. Be2+ has the highest charge density.

Answer: (c) MI > MBr > MCl > MF

Explanation:
1. Large anions are more polarisable, so covalency increases from F- to I-.
2. Covalent character decreases in the order MI > MBr > MCl > MF.

Teacher's Note:
1. Fajans' rule: large anions are polarised more.
2. LiI is the most covalent of the lithium halides.

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

Explanation:
1. Smaller ions are heavily hydrated and so have a large hydrated radius, which lowers mobility.
2. Li+ is the most hydrated and the least mobile, so the order is Rb+ > K+ > Na+ > Li+.

Teacher's Note:
1. Hydrated size is opposite to crystal size.
2. Cs+ is slightly less mobile than Rb+ in some data, but Rb+ is taken as maximum here.

Answer: (d) Na, H2, NaOH, Zn

Explanation:
1. A gives a golden yellow flame, so it is sodium. \(\ce{2Na + 2H2O -> 2NaOH + H2}\), so B = H2 and C = NaOH.
2. Zn reacts with warm NaOH to give H2 and with dilute H2SO4 at room temperature also gives H2.

Teacher's Note:
1. Na gives golden yellow flame, K lilac, Ca brick red.
2. Amphoteric metals (Zn, Al) react with NaOH.

Answer: (c) Ca and \(\ce{CaH2}\)

Explanation:
1. Ca + H2O gives H2 and \(\ce{CaH2 + 2H2O -> Ca(OH)2 + 2H2}\) also gives H2.
2. KO2 gives O2 (with H2O2), Na2O2 gives H2O2/O2 and BaO2 gives H2O2, so these pairs differ.

Teacher's Note:
1. Metals give H2; peroxides give H2O2.
2. Hydrides and metals both release hydrogen with water.

Answer: (c) lower ionisation energies

Explanation:
1. Alkali metals have larger atomic and ionic radii, are softer and have lower melting and boiling points.
2. They have lower first ionisation energies (one outer electron, larger size).

Teacher's Note:
1. Group 2 metals are harder and denser.
2. Each alkali metal has the lowest IE in its period.

Answer: (b) 42

Explanation:
1. Be2C is a methanide and gives methane, X = CH4 (16).
2. CaC2 gives acetylene, Y = C2H2 (26).
3. Sum = 16 + 26 = 42.

Teacher's Note:
1. Carbides: Be2C and Al4C3 give CH4; CaC2 gives C2H2.
2. Calcium carbide contains the C2 2- ion.

Answer: (d) 3

Explanation:
1. K, Rb and Cs form superoxides MO2 as the large cations stabilise the large anion.
2. Li forms the normal oxide and Na the peroxide, so the number is 3.

Teacher's Note:
1. Larger cation means more stable larger oxygen anion.
2. Superoxides are paramagnetic and coloured.