JEE Chemistry S Block Elements MCQs Set 17

Answer: (a) A, B and C only

Explanation:
1. \( \ce{BeCl2} \) is covalent because of the small, highly polarising \( \ce{Be^2+} \).
2. Be has only four electrons in its valence shell in the monomer, so it is electron deficient, and it dimerises through chloride bridges to complete its octet.
3. The monomer Cl-Be-Cl is linear and Be is sp hybridised, so D is wrong.

Teacher's Note:
1. Count the valence electrons around Be.
2. Linear geometry means sp hybridisation.

Answer: (a) A and C only

Explanation:
1. \( \ce{4KO2 + 2CO2 -> 2K2CO3 + 3O2} \).
2. It absorbs carbon dioxide and releases oxygen; it does not produce ozone and does not absorb CO.

Teacher's Note:
1. \( \ce{KO2} \) solves two problems at once for life support.
2. Check each fact against the reaction.

Answer: (a) A and B only

Explanation:
1. The pure solution is blue because of ammoniated electrons, and it conducts through both the electrons and the ammoniated ions.
2. It does not decompose quickly in pure form, so no hydrogen or carbonate forms, and it is not red.

Teacher's Note:
1. Use the facts of the dilute solution: blue, conducting, paramagnetic.
2. Colour changes occur only at high concentration.

Answer: (a) A and D

Explanation:
1. Superoxides contain the \( \ce{O2^-} \) ion with one unpaired electron, so \( \ce{KO2} \) and \( \ce{RbO2} \) are paramagnetic.
2. Peroxides contain \( \ce{O2^2-} \) with all electrons paired, so they are diamagnetic.

Teacher's Note:
1. Use molecular orbital theory for the oxygen species.
2. Superoxide has bond order 1.5.

Answer: (b) B and C only

Explanation:
1. \( \ce{CaH2} \) is called hydrolith, so A is true.
2. \( \ce{Be2C} \) gives methane, not acetylene, so B is false.
3. \( \ce{BeO} \) is amphoteric but \( \ce{CaO} \) is basic, so C is false.

Teacher's Note:
1. Match the carbide type with the hydrocarbon.
2. Only BeO is amphoteric in group 2.

Answer: (a) A, B and D

Explanation:
1. \( \ce{CaC2} \), \( \ce{Al4C3} \) and \( \ce{Be2C} \) contain carbide ions (acetylide or methanide type) and are ionic or salt-like.
2. \( \ce{SiC} \) is a giant covalent solid.

Teacher's Note:
1. SiC (carborundum) has a diamond-like network.
2. Ionic carbides give hydrocarbons with water.

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

Explanation:
1. \( \ce{LiCl} \) is partly covalent and melts at 887 K, whereas \( \ce{NaCl} \) melts at 1074 K, \( \ce{KCl} \) at 1043 K and \( \ce{RbCl} \) at 988 K.
2. NaCl has the best combination of high lattice energy and ionic character.

Teacher's Note:
1. Lattice energy alone would favour LiCl, but its covalent character lowers its melting point.
2. Real data: NaCl is highest.

Answer: (a) A, B and C

Explanation:
1. Normal oxides and peroxides have no unpaired electrons.
2. Superoxides have the unpaired electron of \( \ce{O2^-} \), and the small Li and Na do not form them.

Teacher's Note:
1. Combine electronic structure with size arguments.
2. Pure peroxides are essentially white, though impure samples look yellow.

Answer: (a) \( \ce{LiH} \) > \( \ce{NaH} \) > \( \ce{KH} \) > \( \ce{RbH} \)

Explanation:
1. As the cation grows, the lattice energy falls and the hydride becomes less stable.
2. So \( \ce{LiH} \) is the most stable and \( \ce{RbH} \) the least.

Teacher's Note:
1. Small ions have strong ionic bonds.
2. Reactivity with water goes in the opposite order.

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

Explanation:
1. The smaller the bare ion, the more it is hydrated and the larger its hydrated radius, so it moves more slowly.
2. Mobility therefore rises from \( \ce{Li+} \) to \( \ce{Rb+} \).

Teacher's Note:
1. Mobility follows the inverse of the hydrated radius.
2. Compare with the order of bare ionic radii, which is the opposite.

Answer: (c) Chlorophyll

Explanation:
1. Chlorophyll has \( \ce{Mg^2+} \) at the centre of a porphyrin-type ring.
2. Haemoglobin has iron and vitamin \( \ce{B12} \) has cobalt.

Teacher's Note:
1. Mg in chlorophyll, Fe in haemoglobin, Co in vitamin \( \ce{B12} \).
2. ATP uses \( \ce{Mg2} \)+ only as a cofactor.

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

Explanation:
1. \( \ce{CaCO3 -> CaO + CO2} \), \( \ce{CaO + H2O -> Ca(OH)2} \) (Y).
2. \( \ce{Ca(OH)2 + 2CO2 -> Ca(HCO3)2} \) (Z), and on boiling \( \ce{Ca(HCO3)2 -> CaCO3 + CO2 + H2O} \).

Teacher's Note:
1. The cycle returns to the starting carbonate.
2. \( \ce{Na2CO3} \) does not decompose on heating.

Answer: (d) A, B, C and D

Explanation:
1. Soda ash is \( \ce{Na2CO3} \).
2. It is used to fuse bauxite in Hall's process, to precipitate Ca and Mg in water softening, as a raw material for soda glass, and in making dichromate from chromite.

Teacher's Note:
1. Sodium carbonate is a major industrial chemical.
2. Learn the four main uses.

Answer: (a) A, B and D

Explanation:
1. Milk of magnesia is an antacid, Be is transparent to X-rays, and EDTA titration estimates hardness.
2. Liquid sodium is used as a COOLANT in fast breeder reactors, not as a moderator, so C is wrong.

Teacher's Note:
1. Be has very low atomic number, hence low X-ray absorption.
2. Liquid sodium transfers heat in breeder reactors.

Answer: (c) hydrated radius: \( \ce{Li+} \)(aq) > \( \ce{Na+} \)(aq) > \( \ce{K+} \)(aq)

Explanation:
1. Lattice energy rises with charge, so it is \( \ce{AlN} \) > \( \ce{Mg3N2} \) > \( \ce{Na3N} \), and radii fall in the isoelectronic series \( \ce{Na+} \) > \( \ce{Mg^2+} \) > \( \ce{Al^3+} \), so the first two are wrong.
2. Mobility of halide ions rises from F- to I- because F- is hydrated most, so option 4 is wrong.
3. Hydrated radius is largest for Li+, so option 3 is correct.

Teacher's Note:
1. Compare isoelectronic ions by nuclear charge.
2. The most hydrated ion moves slowest.

Answer: (b) Boron

Explanation:
1. Second ionisation energies (kJ/mol): Be 1757, B 2427, Mg 1451, Al 1816.
2. Boron has the highest, because removal of the second electron breaks into the stable filled 2s subshell.

Teacher's Note:
1. After losing one electron, boron has a filled 2s orbital.
2. Use data for these anomalous cases.

Answer: (c) \( \ce{Li2CO3(s) -> Li2O(s) + CO2(g)} \) on heating

Explanation:
1. Lithium forms no solid bicarbonate, so option 1 is wrong.
2. \( \ce{LiNO3} \) decomposes to \( \ce{Li2O} \), \( \ce{NO2} \) and \( \ce{O2} \), not \( \ce{N2} \), so option 2 is wrong.
3. \( \ce{Li2CO3} \) decomposes to the oxide and \( \ce{CO2} \).

Teacher's Note:
1. Li behaves like Mg in thermal behaviour.
2. Nitrates of Li and Mg give \( \ce{NO2} \) and \( \ce{O2} \).

Answer: (b) \( \ce{Li} \) > \( \ce{Mg} \): atomic radius

Explanation:
1. Ionic radii are Li+ 76 pm, \( \ce{Mg2} \)+ 72 pm, \( \ce{Be2} \)+ 31 pm and \( \ce{Al3} \)+ 54 pm, so options 1 and 4 are correct, and atomic radii are Be 112 pm and Al 143 pm, so option 3 is correct.
2. Li (152 pm) is SMALLER than Mg (160 pm), so option 2 is incorrect.

Teacher's Note:
1. Diagonal pairs have almost the same size.
2. Check values when two elements are in different periods.

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

Explanation:
1. \( \ce{P4} \) gives \( \ce{PH3} \), \( \ce{NH4Cl} \) gives \( \ce{NH3} \) and \( \ce{PH4I} \) gives \( \ce{PH3} \), all basic gases.
2. Zn gives \( \ce{H2} \) (neutral), so it does not fit.

Teacher's Note:
1. Identify the gas for each reagent.
2. Hydrogen is neutral.

Answer: (b) \( \ce{Mg2C3} \)

Explanation:
1. \( \ce{Mg2C3} \) contains the \( \ce{C3^4-} \) ion and gives propyne.
2. \( \ce{Al4C3} \) and \( \ce{Be2C} \) give methane, and \( \ce{CaC2} \) gives ethyne.

Teacher's Note:
1. Match the carbide anion with the hydrocarbon.
2. \( \ce{Mg2C3} \) is the only common propyne source.

Answer: (b) 3

Explanation:
1. Metals above hydrogen in the electrochemical series liberate \( \ce{H2} \) with dilute HCl.
2. Zn, Na and K do so, while Cu, Ag and Au lie below hydrogen.

Teacher's Note:
1. Use the reactivity series.
2. Noble-type metals do not displace hydrogen from acids.

Answer: (b) 4

Explanation:
1. \( \ce{AgNO3} \) gives yellow \( \ce{Ag2CO3} \), \( \ce{CuSO4} \) gives a blue-green basic carbonate, \( \ce{HgCl2} \) gives a reddish-brown precipitate and \( \ce{Hg2(NO3)2} \) gives yellow \( \ce{Hg2CO3} \).
2. The remaining salts give white precipitates.

Teacher's Note:
1. Learn colours of carbonates of Ag, Cu and Hg.
2. Alkaline earth, Zn and Pb carbonates are white.

Answer: (c) 5

Explanation:
1. Iodide is oxidised to iodine by \( \ce{K2Cr2O7} \), \( \ce{HNO3} \), \( \ce{H2O2} \), \( \ce{KMnO4} \) and \( \ce{CuSO4} \) (which gives \( \ce{CuI} \) and \( \ce{I2} \)).
2. \( \ce{AgNO3} \) and \( \ce{Pb(CH3COO)2} \) give precipitates of \( \ce{AgI} \) and \( \ce{PbI2} \), and \( \ce{CaSO4} \) does not react.

Teacher's Note:
1. I- is a reducing agent.
2. Precipitation of AgI or \( \ce{PbI2} \) is not a redox reaction.

Answer: (b) 6

Explanation:
1. Cement contains CaO (about 61.5 percent), \( \ce{SiO2} \), \( \ce{Al2O3} \), \( \ce{Fe2O3} \), \( \ce{MgO} \) and \( \ce{SO3} \) among the main oxides.
2. ZnO, NiO and CuO are not constituents.

Teacher's Note:
1. Remember the major oxides of cement.
2. Lime and silica make up most of it.

Answer: (b) 5

Explanation:
1. Lattice energy rises with ionic charge and falls with ionic size.
2. \( \ce{CaCl2} \), \( \ce{LiF} \), \( \ce{MgO} \), \( \ce{Al2O3} \) and \( \ce{TiO2} \) are higher, while \( \ce{NaI} \) and \( \ce{CsBr} \) have larger ions and lower lattice energies.

Teacher's Note:
1. Compare both charge and radius.
2. Divalent and trivalent compounds have much higher lattice energy.

Answer: (c) 6

Explanation:
1. Calgon is the cyclic sodium hexametaphosphate with a six-membered ring of alternating P and O atoms.
2. Each ring oxygen links two phosphorus atoms, so there are six P-O-P linkages.

Teacher's Note:
1. The ring has six P and six bridging O atoms.
2. Calgon softens water by complexing \( \ce{Ca2} \)+.

Answer: (a) A-S, B-R, C-Q, D-T

Explanation:
1. Dead burnt plaster is anhydrous \( \ce{CaSO4} \) (S), quicklime is \( \ce{CaO} \) (R), slaked lime is \( \ce{Ca(OH)2} \) (Q) and gypsum is \( \ce{CaSO4.2H2O} \) (T).
2. Hence A-S, B-R, C-Q, D-T.

Teacher's Note:
1. Quick lime is the oxide and slaked lime the hydroxide.
2. Dead burnt plaster has no water of crystallisation.

Answer: (d) P, Q, R and S

Explanation:
1. Solubility of alkali metal hydroxides and carbonates rises down the group because lattice energy falls faster than hydration energy.
2. The same rise is seen for the oxalates from Ca to Ba, and for the oxides (through their hydroxides) from Mg to Ba.
3. So all four sets are in increasing order.

Teacher's Note:
1. These series follow the group trend of solubility of hydroxides.
2. Ca, Sr and Ba oxalates increase in solubility.

Answer: (b) P, Q and R only

Explanation:
1. Thermal stability of hydroxides, carbonates and oxalates rises down the group as the cation gets larger.
2. Oxides are all very stable, and their melting points actually fall from MgO to BaO, so S is not in increasing order.

Teacher's Note:
1. Larger cations stabilise larger anions.
2. Check the oxide data separately.

Answer: (a) A-S, B-P, C-R, D-Q

Explanation:
1. Quicklime is used in making dyestuffs, plaster of Paris sets fractured bones, slaked lime is used for bleaching powder, and limestone is a constituent of chewing gum.
2. So A-S, B-P, C-R, D-Q.

Teacher's Note:
1. Learn the main use of each calcium compound.
2. Slaked lime with chlorine gives bleaching powder.