JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 19

Answer: (c) the maximum principal quantum number of any element of the period

Explanation:
1. A new period starts when a new principal shell begins to fill.
2. So the period number equals the highest principal quantum number n among the elements of that period.

Teacher's Note:
1. Period number = outermost shell number n.
2. Atomic number and l or m values do not stay constant across a period.

Answer: (a) lanthanoids

Explanation:
1. Filling of 4f orbitals gives the 14 lanthanoids (Ce to Lu).
2. Actinoids fill 5f orbitals.

Teacher's Note:
1. Lanthanoids = 4f series, actinoids = 5f series.
2. Both are inner transition elements.

Answer: (d) It cannot be placed in group 1 of the modern periodic table

Explanation:
1. Hydrogen has one electron in its 1s orbital and forms H⁺, like the alkali metals, so it is placed in group 1.
2. It also forms H^- and diatomic molecules like the halogens.

Teacher's Note:
1. Hydrogen is placed in group 1 although it also shows halogen-like behaviour.
2. Its position is a convention, not a perfect fit.

Answer: (a) a noble gas of the seventh period (Og)

Explanation:
1. Z = 118 completes the seventh period: 7s² 5f¹⁴ 6d¹⁰ 7p⁶.
2. A filled p-subshell means a noble gas, oganesson.

Teacher's Note:
1. The last element of each period is a noble gas.
2. Period 7 ends at Z = 118.

Answer: (d) s > p > d > f

Explanation:
1. Penetration decreases in the order s > p > d > f.
2. Hence the screening power decreases in the same order.

Teacher's Note:
1. More penetrating electrons shield better.
2. d and f electrons shield poorly.

Answer: (a) Sn⁴⁺

Explanation:
1. The cation Sn⁴⁺ has the greatest Z (50) with the fewest electrons (46).
2. The positive charge on the ion raises Z_eff, and tin has a higher Z than indium.

Teacher's Note:
1. Removing electrons raises Z_eff on those left.
2. For the same element, cation > atom.

Answer: (a) 6

Explanation:
1. To reach an octet by gaining two electrons, the atom needs 6 valence electrons.

Teacher's Note:
1. Common oxidation state -2 means two electrons are gained.
2. Group 16 elements have 6 valence electrons.

Answer: (b) -3 to +5

Explanation:
1. Nitrogen is -3 in NH₃ and +5 in HNO₃.
2. It has 5 valence electrons, so -3 is the lowest and +5 the highest.

Teacher's Note:
1. Range = -(8 - group valence electrons) to +group valence electrons.
2. For group 15 that is -3 to +5.

Answer: (d) Fe

Explanation:
1. Fe forms Fe²⁺ and Fe³⁺ compounds because its 4s and 3d electrons have similar energies.
2. Na, Mg and Al show only +1, +2 and +3 respectively.

Teacher's Note:
1. Transition metals show variable oxidation states.
2. Typical s-block metals show one.

Answer: (b) the inert pair of electrons

Explanation:
1. Tl has the configuration 6s² 6p¹. Using only 6p gives +1, while using 6s as well gives +3.
2. The 6s² pair is reluctant to take part, so +1 is the more stable state.

Teacher's Note:
1. Inert pair effect explains stable lower oxidation states of heavy elements.
2. Tl is a p-block, not a transition, element.

Answer: (a) Bi

Explanation:
1. Because of the inert pair effect, the stability of +5 falls down group 15.
2. For Bi the +3 state is clearly more stable than +5.

Teacher's Note:
1. Lower state becomes more stable down groups 13 to 16.
2. Bi(V) compounds are strong oxidising agents.

Answer: (d) Be = 111, C = 77, O = 66, B = 88, N = 74

Explanation:
1. Across period 2 the radius decreases: Be (111) > B (88) > C (77) > N (74) > O (66).

Teacher's Note:
1. Atomic radius falls from left to right in a period.
2. Z_eff rises while the shell stays the same.

Answer: (d) nuclear charge

Explanation:
1. All three have 10 electrons in the same shells, so n is the same.
2. The nuclear charge (8, 9, 11) differs, so the pull on the electrons and hence the size differs.

Teacher's Note:
1. Among isoelectronic species, radius decreases as Z increases.
2. O^2- > F^- > Na⁺.

Answer: (d) 14.6 and 13.6

Explanation:
1. N has a stable half-filled 2p³ configuration, so it is harder to ionise than O (2p⁴).
2. Hence 14.6 eV for N and 13.6 eV for O.

Teacher's Note:
1. IE of N is higher than O.
2. Pairing repulsion in 2p⁴ lowers the IE of O.

Answer: (d) Na < Mg > Al < Si

Explanation:
1. Values (kJ/mol): Na 496, Mg 738, Al 578, Si 786.
2. Mg is higher than Al because Mg has a filled 3s orbital and Al removes a 3p electron.

Teacher's Note:
1. The 3s² to 3p¹ dip occurs at group 13.
2. Overall IE rises across a period.

Answer: (a) IE2 of K is greater than IE2 of Ca

Explanation:
1. K⁺ has an argon core, so IE2 of K (3052 kJ/mol) is very large.
2. Ca⁺ still has a 4s electron, so IE2 of Ca (1145 kJ/mol) is much smaller.

Teacher's Note:
1. Removing an electron from a noble gas core is very difficult.
2. IE1 of Ca (590) is greater than IE1 of K (419).

Answer: (a) F < Cl > Br > I

Explanation:
1. Cl (349 kJ/mol) is highest.
2. F (328) is lower because the small 2p subshell has strong repulsion; then Br (325) and I (295).

Teacher's Note:
1. F is an exception, as in the first member of a group.
2. Order: Cl > F > Br > I.

Answer: (c) S > Se > O

Explanation:
1. Values (kJ/mol): S 200, Se 195, O 141.
2. O is lower due to repulsion in its compact 2p subshell.

Teacher's Note:
1. Second period elements have anomalously low affinity.
2. Order: S > Se > O.

Answer: (c) (i) < (iii) < (ii) < (iv)

Explanation:
1. (i) is neon (filled shell, positive egH), (iii) is sodium (small negative), (ii) is oxygen and (iv) is fluorine.
2. Fluorine, one electron short of an octet, gains most readily.

Teacher's Note:
1. Noble gases do not gain electrons.
2. Halogens gain most readily.

Answer: (a) increase from carbon to fluorine

Explanation:
1. Pauling values: C 2.55, N 3.04, O 3.44, F 3.98.
2. They rise steadily across the period.

Teacher's Note:
1. Electronegativity rises across a period.
2. F is the most electronegative element.

Answer: (d) N > C > P > Si

Explanation:
1. Values: N 3.04, C 2.55, P 2.19, Si 1.90.
2. So N > C > P > Si.

Teacher's Note:
1. Electronegativity falls down a group and rises across a period.
2. Compare diagonally using values.

Answer: (a) H

Explanation:
1. Pauling values: H 2.20, Li 0.98, Mg 1.31, Na 0.93.
2. So hydrogen is the most electronegative.

Teacher's Note:
1. Metals have low electronegativity.
2. Hydrogen is intermediate.

Answer: (d) For transition elements the 3d orbitals are filled after 3p and before 4s

Explanation:
1. The 4s orbital is filled before 3d, so 3d is filled after 4s (from Sc onwards).
2. The other three statements are correct.

Teacher's Note:
1. Aufbau order: 3p, 4s, 3d.
2. Non-metals are only about 22 of the elements.

Answer: (d) 3, 20, 55, 87

Explanation:
1. Z = 3 (Li), 20 (Ca), 55 (Cs) and 87 (Fr) all have ns¹ or ns² valence configuration.
2. The other sets contain p- or d-block elements.

Teacher's Note:
1. Alkali and alkaline earth metals are s-block.
2. Check each Z against the configuration.

Answer: (a) all of these: He⁺, Li²⁺ and Be³⁺

Explanation:
1. Each of these ions is hydrogen-like with only one electron.
2. Screening needs at least two electrons, so it is absent in all three.

Teacher's Note:
1. Single-electron species have no screening.
2. Z_eff = Z for them.

Answer: (c) Mn

Explanation:
1. Mn has seven valence electrons (3d⁵ 4s²) and is +7 in KMnO₄.
2. F is only -1, In is +3 or +1 and N is at most +5.

Teacher's Note:
1. Highest state of a d-block element can equal the sum of its ns and (n-1)d electrons.
2. Mn(VII) in permanganate.