JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 08

Answer: (b) The Lothar Meyer curve is plotted between atomic mass and atomic density

Explanation:
1. Lothar Meyer plotted atomic volume against atomic mass, not atomic density.
2. The other statements are correct: Mendeleev used atomic mass, the modern law uses atomic number and the long form follows electronic configuration.

Teacher's Note:
1. The Lothar Meyer curve shows periodic rise and fall of atomic volume.
2. Moseley's work led to the atomic number basis.

Answer: (b) 62

Explanation:
1. 4d electrons first appear in yttrium (Z = 39).
2. Every element from Z = 39 onwards has at least one 4d electron (4d is completely filled after cadmium and stays filled).
3. Number of elements from 39 to 100 is \( 100 - 38 = 62 \).

Teacher's Note:
1. Count atomic numbers from the first element with a 4d electron to the last element.
2. Palladium (4d¹⁰ 5s⁰) still counts.

Answer: (b) P³⁻

Explanation:
1. S²⁻, P³⁻, Ca²⁺ and Sc³⁺ are isoelectronic (18 electrons).
2. In an isoelectronic series the radius decreases as the nuclear charge increases, so P³⁻ (Z = 15) is the largest.

Teacher's Note:
1. Isoelectronic species: larger Z means a smaller radius.
2. Anions are larger than the corresponding cations.

Answer: (d) F⁻ < Cl⁻ < Cl < F

Explanation:
1. Anions lose an electron more easily than neutral atoms, so F⁻ and Cl⁻ have lower ionisation energies than Cl and F.
2. The ionisation energy of an anion equals the electron affinity of the neutral atom: for F⁻ it is about 328 kJ/mol and for Cl⁻ about 349 kJ/mol, so F⁻ < Cl⁻.
3. For the atoms, Cl (1251 kJ/mol) < F (1681 kJ/mol). The order is F⁻ < Cl⁻ < Cl < F.

Teacher's Note:
1. Electron affinity of Cl is higher than that of F.
2. Ionisation energy of an anion = electron affinity of the atom.

Answer: (d) HClO₄ < HBrO₄ < HIO₄

Explanation:
1. Oxoacids of the same halogen become stronger as the number of oxygen atoms rises: HOCl < HClO₂ < HClO₃, so (a) is correct.
2. With the same oxidation state, acidity falls as the halogen gets larger (less electronegative): HOCl > HOBr > HOI, so (b) is correct. Hydrogen halides get stronger down the group: HCl < HBr < HI, so (c) is correct.
3. For the perhalic acids the order is HClO₄ > HBrO₄ > HIO₄, so the order written in (d) is incorrect.

Teacher's Note:
1. Larger halogen means weaker oxoacid but stronger hydrohalic acid.
2. HClO₄ is the strongest known common oxoacid.

Answer: (c) The second electron affinity is an endothermic process

Explanation:
1. Adding a second electron to an anion has to overcome electrostatic repulsion, so the second electron affinity is endothermic (positive).
2. Nitrogen has a low (nearly zero) electron affinity because of its half-filled p subshell; F has a lower electron affinity than Cl because of its compact size; electronegativity does not increase for half-filled configurations.

Teacher's Note:
1. EA values: Cl > F, and N is close to zero or slightly positive.
2. Second EA of O is +744 kJ/mol.

Answer: (a) 2X - EA - IP = 0

Explanation:
1. Mulliken's electronegativity is the mean of ionisation potential and electron affinity: \( X = \frac{IP + EA}{2} \).
2. So \( 2X - EA - IP = 0 \).

Teacher's Note:
1. Mulliken: X = (IP + EA)/2.
2. Pauling's scale is based on bond energies instead.