JEE Chemistry Classification Of Elements and Periodicity In Properties MCQs Set 28

Answer: (d) nuclear charge

Explanation:
1. These species have the same number of electrons (18) and the same shell.
2. The only difference is nuclear charge: larger Z pulls the electrons in more strongly and gives a smaller radius.

Teacher's Note:
1. For isoelectronic species, radius is inversely related to atomic number. Ca²⁺ is the smallest and Cl⁻ the largest here.

Answer: (d) K

Explanation:
1. K has one electron in the outer shell, so IE₁ is low.
2. Removing the second electron breaks the stable 3p⁶ core, so IE₂ is very high and IE₂ − IE₁ is the largest for K.

Teacher's Note:
1. Alkali metals show a huge jump between the first and second ionisation energy because the second electron comes from a noble gas core.

Answer: (b) 15, 5 and 3

Explanation:
1. Z = 15 has configuration 1s² 2s² 2p⁶ 3s² 3p³.
2. It has 5 valence electrons, so it is in group 15, with valency 8 − 5 = 3.

Teacher's Note:
1. Group number and valence electron count are linked: for p-block elements group = 10 + number of valence electrons.

Answer: (a) electronegativity and atomic radius

Explanation:
1. Down a group electronegativity decreases because the atoms get larger and attract bonding electrons less.
2. Atomic radius increases down a group.

Teacher's Note:
1. Electron gain enthalpy generally becomes less negative down a group.

Answer: (d) Ga < Ge

Explanation:
1. Pauling values: Ga 1.6 and Ge 1.8, so Ga < Ge.
2. The others are wrong: S (2.5) > P (2.1), Si (1.8) > Al (1.5), Se (2.4) > Te (2.1).

Teacher's Note:
1. Electronegativity increases across a period and decreases down a group.

Answer: (c) alkaline earth metal

Explanation:
1. Oganesson (Z = 118) is the noble gas of period 7.
2. Z = 119 is an alkali metal and Z = 120 is an alkaline earth metal with configuration [118]8s².

Teacher's Note:
1. Add up the periods: elements 119 and 120 start the eighth period as s-block elements.

Answer: (c) 1.71, 1.40 and 1.36

Explanation:
1. N³⁻, O²⁻ and F⁻ each have 10 electrons.
2. Radius falls as nuclear charge rises: N³⁻ (1.71) > O²⁻ (1.40) > F⁻ (1.36).

Teacher's Note:
1. Among isoelectronic anions, the one with the lowest atomic number is largest.

Answer: (c) Ba < Ca < Se < S < Ar

Explanation:
1. Ionisation enthalpy rises across a period and falls down a group.
2. Ba < Ca (group 2); Se < S (group 16); metals are lower than non-metals; Ar, a noble gas, is highest.

Teacher's Note:
1. Place the elements in the periodic table first and then apply both trends.

Answer: (c) Ca²⁺, K⁺, Cl⁻, S²⁻

Explanation:
1. All have 18 electrons, so radius is inversely related to nuclear charge.
2. Z: Ca 20 > K 19 > Cl 17 > S 16, so the radius increases Ca²⁺ < K⁺ < Cl⁻ < S²⁻.

Teacher's Note:
1. More protons for the same electrons means a smaller ion.

Answer: (a) Al₂O₃ < MgO < Na₂O < K₂O

Explanation:
1. Across a period, oxides change from basic to acidic: Na₂O > MgO > Al₂O₃ (amphoteric).
2. Down a group basic character increases, so K₂O > Na₂O.

Teacher's Note:
1. Metallic character of the element decides basic character of its oxide.

Answer: (d) O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺

Explanation:
1. These five ions are isoelectronic (10 electrons).
2. Radius decreases with increasing nuclear charge: O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺.

Teacher's Note:
1. Order isoelectronic ions by atomic number in reverse.

Answer: (b) Na⁺ > Li⁺ > Mg²⁺ > Be²⁺

Explanation:
1. Radius increases down a group: Na⁺ > Li⁺ and Mg²⁺ > Be²⁺.
2. Across the period the ions with higher charge are smaller, so Li⁺ (0.68 Å) > Mg²⁺ (0.65 Å) > Be²⁺.

Teacher's Note:
1. Compare down a group first, then compare diagonal pairs by charge and size.

Answer: (a) K⁺, Cl⁻, Ca²⁺, Sc³⁺

Explanation:
1. K⁺, Cl⁻, Ca²⁺ and Sc³⁺ each have 18 electrons.
2. The other sets mix different electron counts.

Teacher's Note:
1. Count the electrons: atomic number minus charge.

Answer: (b) Rb⁺

Explanation:
1. Smaller ions are more heavily hydrated, so they are larger when hydrated and move slower.
2. Rb⁺ has the smallest hydrated radius of the four ions listed, so the highest mobility.

Teacher's Note:
1. Hydrated size order: Li⁺ > Na⁺ > K⁺ > Rb⁺.

Answer: (d) In alkali metals the reactivity increases but in halogens it decreases with increase in atomic number down the group

Explanation:
1. Alkali metals react by losing an electron, which becomes easier down the group.
2. Halogens react by gaining an electron, which becomes harder down the group.

Teacher's Note:
1. Metals become more reactive down a group and non-metals less reactive.

Answer: (c) B < C < N < O: increasing first ionisation enthalpy

Explanation:
1. The actual first ionisation order is B < C < O < N, because N has a half-filled 2p subshell.
2. The other arrangements are correct.

Teacher's Note:
1. N is higher than O in ionisation energy, as is P over S.

Answer: (c) NaCl

Explanation:
1. Melting point is governed by lattice energy for ionic halides.
2. LiCl has a covalent character, and the large ions of KCl and RbCl lower the lattice energy; NaCl melts at about 800°C, the highest.

Teacher's Note:
1. LiCl is covalent in part, so its melting point is lower than that of NaCl.

Answer: (c) the O⁻ ion will tend to resist the addition of another electron

Explanation:
1. The second electron must be added to an ion that is already negative.
2. The electron-electron repulsion outweighs the stability gained, so the step is endothermic.

Teacher's Note:
1. Second electron gain enthalpy is positive for all anions with charge 2− or more.

Answer: (c) Hydration enthalpy

Explanation:
1. The oxidising power in water depends on the electrode potential, which includes the hydration of F⁻.
2. F⁻ has a very high hydration enthalpy, which is the dominant factor.

Teacher's Note:
1. Oxidising strength in solution combines bond energy, electron gain and hydration enthalpies.

Answer: (c) atomic numbers

Explanation:
1. The modern periodic law states that properties are periodic functions of atomic number.

Teacher's Note:
1. Mendeleev used atomic mass; the modern law uses atomic number.

Answer: (d) 0.85 Å

Explanation:
1. Across the lanthanides the 4f electrons shield poorly, so the ionic radius shrinks steadily (lanthanide contraction).
2. Lu³⁺ is therefore smaller than La³⁺; only 0.85 Å is smaller.

Teacher's Note:
1. Lanthanide contraction reduces the radii from La³⁺ to Lu³⁺.

Answer: (b) Yb³⁺ < Pm³⁺ < Ce³⁺ < La³⁺

Explanation:
1. Radius decreases steadily along the lanthanides: La (57), Ce (58), Pm (61), Yb (70).
2. So Yb³⁺ < Pm³⁺ < Ce³⁺ < La³⁺.

Teacher's Note:
1. Lanthanide contraction: the radius decreases from La³⁺ to Lu³⁺.