CBSE Class 11 Chemistry Classification of Elements and Periodicity in Properties MCQs Set 03

Answer: (c) electronic configuration

Explanation:
1. The long form of the periodic table is based on the electronic configuration of elements (Bohr–Bury concept).
2. Elements with similar outer electronic configurations fall in the same group, and each period begins with a new principal shell.

Teacher's Note:
1. The modern periodic law uses atomic number, and atomic number decides the electronic configuration.
2. Atomic mass was the basis of Mendeleev’s table, not the long form.

Answer: (d) All of these

Explanation:
1. Elements with similar outer electronic configurations are arranged in vertical columns.
2. These vertical columns are called groups or families, so all three terms are correct.

Teacher's Note:
1. Group, family and vertical column all mean the same thing in the periodic table.
2. Horizontal rows are periods, not groups.

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

Explanation:
1. Each period starts with the filling of electrons in a new principal quantum number (shell).
2. Therefore, period number = maximum principal quantum number (n) of any element in that period.

Teacher's Note:
1. For example, every element of period 4 has electrons up to n = 4.
2. Do not confuse the azimuthal quantum number (l) with the principal quantum number (n).

Answer: (d) 8

Explanation:
1. The 3s orbital holds 2 electrons and the three 3p orbitals hold 6 electrons.
2. So 2 + 6 = 8 elements, from sodium (Z = 11) to argon (Z = 18), make up the third period.

Teacher's Note:
1. 3d orbitals are not filled in period 3 because 3d is higher in energy than 4s.
2. Number of elements in a period = maximum electrons in the orbitals filled in that period.

Answer: (a) lanthanoids, actinoids

Explanation:
1. The 14 elements of the 6th period that follow lanthanum (Z = 58 to 71) are called lanthanoids.
2. The 14 elements of the 7th period that follow actinium (Z = 90 to 103) are called actinoids.

Teacher's Note:
1. Lanthanoids fill 4f orbitals; actinoids fill 5f orbitals.
2. Both series are placed separately at the bottom of the periodic table.

Answer: (c) lanthanoids

Explanation:
1. The elements in which electrons are progressively filled in the 4f orbitals are called lanthanoids.
2. Lanthanoids consist of elements from Z = 58 (cerium) to Z = 71 (lutetium).

Teacher's Note:
1. 4f filling gives lanthanoids, 5f filling gives actinoids.
2. Transition elements fill d-orbitals, not f-orbitals.

Answer: (a) Period 4, group 6

Explanation:
1. The highest principal quantum number is n = 4, so the element lies in the 4th period.
2. It is a d-block element, so group number = ns electrons + (n − 1)d electrons = 1 + 5 = 6.
3. The element is chromium (Z = 24): period 4, group 6.

Teacher's Note:
1. For d-block elements, add the (n − 1)d and ns electrons to get the group number.
2. Cr has 3d⁵4s¹ because a half-filled d subshell is extra stable.

Answer: (a) f-block elements

Explanation:
1. The anti-penultimate shell is the (n − 2) shell, i.e. the shell inside the penultimate shell.
2. Elements in which the differentiating electron enters the (n − 2)f subshell are f-block or inner-transition elements.

Teacher's Note:
1. Outermost shell = n (s, p-block), penultimate = n − 1 (d-block), anti-penultimate = n − 2 (f-block).
2. This is why f-block elements are called inner transition elements.

Answer: (c) d-block

Explanation:
1. Electronic configuration of Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.
2. The last (differentiating) electron enters a 3d orbital, so Cu is a d-block element.

Teacher's Note:
1. Cu has 3d¹⁰4s¹, not 3d⁹4s², because a full d subshell is extra stable.
2. An outer 4s¹ does not make Cu an s-block element; the block is decided by the subshell being filled.

Answer: (a) fifth group

Explanation:
1. The last electron enters the 3d subshell (3d³), so it is a d-block element.
2. Group number of a d-block element = ns electrons + (n − 1)d electrons = 2 + 3 = 5.
3. Hence, it is placed in the fifth group (it is vanadium, Z = 23).

Teacher's Note:
1. Group 15 would need ns²np³, which is a p-block configuration.
2. Always check which subshell is being filled before counting electrons for the group.

Answer: (b) d-block

Explanation:
1. Metals usually used as catalysts, e.g. Fe, Ni, Pt, V, belong to the d-block of the periodic table.
2. Their catalytic activity is due to variable oxidation states and their ability to provide a large surface for adsorption of reactants.

Teacher's Note:
1. Fe in the Haber process and V₂O₅ in the Contact process are common examples.
2. Variable oxidation state is the key reason transition metals are good catalysts.

Answer: (c) Either [(a) or (b)]

Explanation:
1. Elements such as silicon, germanium, arsenic, antimony and tellurium show characteristics of both metals and non-metals.
2. They are called semi-metals or metalloids, so either name is correct.

Teacher's Note:
1. Metalloids lie along the zig-zag line in the p-block.
2. ‘Amphoteric’ describes oxides or hydroxides that react with both acids and bases, not elements.

Answer: (b) II is a d-block element

Explanation:
1. I ([Xe] 6s¹) is caesium. It loses one electron easily to attain a stable configuration, so it is a strong reducing agent.
2. II ([Xe] 4f¹⁴ 5d¹ 6s²) is lutetium, the last member of the lanthanoid series; it is placed with the f-block elements, so calling it a d-block element is not correct.
3. III ([Ar] 4s² 4p⁵, actually [Ar] 3d¹⁰ 4s² 4p⁵) is bromine. It gains one electron to attain a noble gas configuration, so it has high electron affinity.
4. IV ([Ar] 3d⁷ 4s²) is cobalt, a d-block element that shows variable oxidation states.
5. Note: The general electronic configuration of f-block elements is (n − 2)f^1–14 (n − 1)d^0–1 ns².

Teacher's Note:
1. A filled 4f¹⁴ with 5d¹ 6s² is lutetium, which is counted among the lanthanoids.
2. Read ‘not correct’ questions carefully and test each statement separately.

Answer: (c) [Xe] 4f⁷ 5d¹ 6s²

Explanation:
1. The electronic configuration of La (Z = 57) is [Xe] 5d¹ 6s². Further electrons go into the lower-energy 4f orbitals until they are exactly half-filled at Eu (Z = 63): [Xe] 4f⁷ 6s².
2. The next electron does not enter the stable half-filled 4f⁷ subshell but goes into the slightly higher-energy 5d orbital.
3. Thus, the electronic configuration of Gd (Z = 64) is [Xe] 4f⁷ 5d¹ 6s².

Teacher's Note:
1. Check the electron count first: 54 + 7 + 1 + 2 = 64.
2. Gd keeps a half-filled 4f⁷ subshell, which is extra stable.

Answer: (d) Terbium (Z = 65)

Explanation:
1. Elements with atomic numbers Z = 90 to 103 are called actinoids.
2. Terbium (Z = 65) lies in the range 58–71, so it is a lanthanoid, not an actinoid.

Teacher's Note:
1. Remember the ranges: lanthanoids 58–71, actinoids 90–103.
2. Elements with names ending in ‘-ium’ are not necessarily actinoids, so check the atomic number.

Answer: (b) A-2, B-3, C-1, D-1

Explanation:
1. (A) The first period contains 2 elements → 2.
2. (B) The third period contains 8 elements → 3.
3. (C) and (D) The lanthanoids (Z = 58 to 71) and the actinoids (Z = 90 to 103) each contain 14 elements → 1.
4. Correct match: A-2, B-3, C-1, D-1.

Teacher's Note:
1. An option of Column II can be used more than once, as 14 is here.
2. Period lengths are 2, 8, 8, 18, 18, 32, 32.

Answer: (b) dubnium

Explanation:
1. The element with atomic number 105 is dubnium (Db).
2. Its IUPAC temporary name was unnilpentium (Unp).
3. Db (105) = [Rn] 5f¹⁴ 6d³ 7s².

Teacher's Note:
1. Nobelium is Z = 102 and holmium is Z = 67.
2. Temporary names are built from digit roots: 1 = un, 0 = nil, 5 = pent.

Answer: (b) a transition metal

Explanation:
1. The electronic configuration of element 112 is [Rn] 5f¹⁴ 6d¹⁰ 7s².
2. Its differentiating electron enters the 6d subshell, so it belongs to the d-block (group 12), i.e. it is a transition metal.

Teacher's Note:
1. Element 112 is copernicium (Cn), placed below Zn, Cd and Hg.
2. Actinoids end at Z = 103; after that the 6d series begins.

Answer: (b) A-2, B-1, C-3, D-4

Explanation:
1. 109 = un + nil + enn + ium = Unnilennium → 2.
2. 112 = un + un + bi + ium = Ununbium → 1.
3. 115 = un + un + pent + ium = Ununpentium → 3.
4. 118 = un + un + oct + ium = Ununoctium → 4.
5. Correct match: A-2, B-1, C-3, D-4.

Teacher's Note:
1. Write the digit roots in order: 0 nil, 1 un, 2 bi, 3 tri, 4 quad, 5 pent, 6 hex, 7 sept, 8 oct, 9 enn.
2. When ‘enn’ is followed by ‘ium’, the double n is kept but one i is dropped: unnilennium.

Answer: (b) 1.86 Å

Explanation:
1. Radius of Na atom = bond distance ÷ 2
2. = 3.72 Å ÷ 2 = 1.86 Å

Teacher's Note:
1. The radius is half the distance between the two nuclei.
2. Multiplying by 2 (7.44 Å) is a common slip.

Answer: (a) H⁻ > H > H⁺

Explanation:
1. (a) A cation is always smaller than its neutral atom (fewer electrons, same nuclear charge), and an anion is always larger (more electrons, lower effective nuclear charge per electron). So H⁻ > H > H⁺ is correct.
2. (b) and (c) Na⁺, F⁻ and O²⁻ are isoelectronic (10 electrons); radius decreases as atomic number increases (O = 8, F = 9, Na = 11). Correct order: O²⁻ > F⁻ > Na⁺.
3. (d) Al³⁺, Mg²⁺ and N³⁻ are isoelectronic (atomic numbers 13, 12, 7). Correct order: N³⁻ > Mg²⁺ > Al³⁺.

Teacher's Note:
1. For isoelectronic ions, the greater the nuclear charge, the smaller the ion.
2. Anion > neutral atom > cation for the same element.

Answer: (a) 128 pm

Explanation:
1. Metallic radius = ½ × distance between two adjacent metal atoms
2. = ½ × 256 pm = 128 pm

Teacher's Note:
1. Metallic radius is half the internuclear distance between neighbouring atoms in the metal lattice.
2. Watch the units: 128 pm = 1.28 Å, not 12.87 Å.

Answer: (a) A-3, B-4, C-5, D-2, E-1

Explanation:
1. All the given elements are in the second period, and atomic radius decreases along a period because effective nuclear charge increases.
2. So the order of atomic radii is Be > B > C > N > O.
3. Be = 111 pm (3), C = 77 pm (4), O = 66 pm (5), B = 88 pm (2), N = 74 pm (1).
4. Correct match: A-3, B-4, C-5, D-2, E-1.

Teacher's Note:
1. Arrange the elements by position in the period first, then assign radii from largest to smallest.
2. Second-period radii to remember: Be 111, B 88, C 77, N 74, O 66 pm.

Answer: (b) (C) < (B) < (A) < (D) < (E)

Explanation:
1. Atomic radius decreases from left to right in a period, so in period 2: C > O > F.
2. Elements of later periods are larger, and radius increases down a group: Br > Cl > F.
3. So Br > Cl > C > O > F, i.e. the increasing order is F < O < C < Cl < Br = (C) < (B) < (A) < (D) < (E).

Teacher's Note:
1. Cl (99 pm) in period 3 is larger than C (77 pm) in period 2.
2. Note that the letters here label elements, so translate carefully back to the codes.

Answer: (b) Mg²⁺ < Na⁺ < F⁻ < O²⁻

Explanation:
1. For isoelectronic species, ionic radius ∝ 1/atomic number (nuclear charge).
2. Radius increases as positive charge decreases or negative charge increases.
3. Mg²⁺ (12) < Na⁺ (11) < F⁻ (9) < O²⁻ (8)

Teacher's Note:
1. All four species have 10 electrons, so only the nuclear charge decides the size.
2. The ion with the most protons pulls the same 10 electrons in most strongly.

Answer: (c) X = alkali metal, Y = Noble gas, Z = alkaline earth metal

Explanation:
1. First ionisation enthalpies of noble gases are the highest and those of alkali metals are the lowest.
2. IE of alkaline earth metals is higher than that of alkali metals but lower than that of the elements that follow.
3. So X (520 kJ/mol, Li) is an alkali metal, Y (2080 kJ/mol, Ne) is a noble gas and Z (899 kJ/mol, Be) is an alkaline earth metal.

Teacher's Note:
1. Very low IE (about 400–520 kJ/mol) points to an alkali metal; above 2000 kJ/mol points to a noble gas.
2. These three values belong to Li, Ne and Be of period 2.

Answer: (a) He

Explanation:
1. Helium has the stable configuration 1s², with a completely filled K-shell and a very small size.
2. So the ionisation energy of helium (2372 kJ/mol) is the highest of all the elements.

Teacher's Note:
1. IE increases across a period and decreases down a group, so the top-right element (He) has the highest IE.
2. F has the highest electronegativity, not the highest ionisation energy.

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

Explanation:
1. Electronic configurations: Na = [Ne] 3s¹, Mg = [Ne] 3s², Al = [Ne] 3s² 3p¹, Si = [Ne] 3s² 3p².
2. IE generally increases from left to right along a period.
3. However, IE of Mg is higher than that of Al because Mg has a completely filled 3s orbital, while the 3p electron of Al is easier to remove.
4. So the order is Na < Mg > Al < Si (Na 496, Mg 737, Al 577, Si 786 kJ/mol).

Teacher's Note:
1. Group 2 > group 13 and group 15 > group 16 are the two exceptions to the period trend.
2. A p electron is shielded more than an s electron, so it is removed more easily.

Answer: (b) second ionisation enthalpy

Explanation:
1. Second ionisation enthalpy is the energy required to remove the second electron, i.e. an electron from the unipositive gaseous ion.
2. The given equation removes an electron from X⁺(g), so it represents the second ionisation enthalpy.

Teacher's Note:
1. Second IE is always greater than first IE because the electron is removed from a positively charged ion.
2. Count the charge on the starting species: X → X⁺ is first, X⁺ → X²⁺ is second.

Answer: (c) Beryllium

Explanation:
1. Ionisation enthalpy increases on moving from left to right in a period and decreases on moving down a group.
2. So the order of ionisation enthalpy is Be > Mg > Na > K, and beryllium has the highest value.

Teacher's Note:
1. Be is the smallest of the four atoms and has a fully filled 2s subshell.
2. Use both trends together: higher up and further right means higher IE.

Answer: (a) N < O < S < Cl

Explanation:
1. Generally, electron affinity increases from left to right in a period and decreases down a group.
2. Second-period elements (N, O) have less negative electron gain enthalpies than the corresponding third-period elements because of their small size and greater electron–electron repulsion.
3. N has a stable half-filled 2p³ configuration and almost zero electron affinity.
4. So the correct order is N < O < S < Cl.

Teacher's Note:
1. S has higher electron affinity than O, and Cl higher than F, because of the small size of O and F.
2. Half-filled p³ atoms like N have very low electron affinity.

Answer: (d) Cl and P

Explanation:
1. Electron gain enthalpy generally becomes more negative across a period from left to right and less negative down a group.
2. However, adding an electron to the small 2p subshell of F causes greater repulsion than adding one to the larger 3p subshell of Cl, so Cl has a more negative value than F.
3. Chlorine (−349 kJ/mol) has the most negative and phosphorus (−74 kJ/mol) the least negative electron gain enthalpy.

Teacher's Note:
1. Chlorine, not fluorine, has the most negative electron gain enthalpy of all elements.
2. P has a half-filled 3p³ subshell, so it accepts an electron less readily.

Answer: (b) A-4, B-1, C-2, D-3

Explanation:
1. A. 1s² 2s² 2p⁶ is neon, a noble gas. Noble gases have positive Δ_egH values, so A → +48 kJ mol⁻¹ (4).
2. B. 1s² 2s² 2p⁶ 3s¹ is sodium, an alkali metal. Alkali metals have small negative Δ_egH values, so B → −53 kJ mol⁻¹ (1).
3. C. 1s² 2s² 2p⁵ is fluorine, a halogen. Halogens have large negative Δ_egH values, so C → −328 kJ mol⁻¹ (2).
4. D. 1s² 2s² 2p⁴ is oxygen, a chalcogen, whose Δ_egH is less negative than that of the halogens, so D → −141 kJ mol⁻¹ (3).
5. Correct match: A-4, B-1, C-2, D-3.

Teacher's Note:
1. Identify each element from its configuration first, then use family behaviour.
2. A positive value means energy must be supplied to add an electron, as in noble gases.

Answer: (a) Li < K < Na < Rb < Cs

Explanation:
1. On moving down a group, density generally increases, but the density of K is lower than that of Na because of the abnormal increase in the size of the K atom.
2. Densities: Li = 0.53, Na = 0.97, K = 0.86, Rb = 1.53 and Cs = 1.90 g/cm³.
3. So the order of density is Li < K < Na < Rb < Cs.

Teacher's Note:
1. K is the exception: it is less dense than Na.
2. Li, Na and K are all less dense than water and float on it.

Answer: (d) electronegativity

Explanation:
1. A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity.

Teacher's Note:
1. Electronegativity applies to a bonded atom; electron affinity applies to an isolated gaseous atom.
2. Electronegativity has no units, while electron affinity is measured in kJ/mol.

Answer: (a) II > III > I > IV

Explanation:
1. Electronegativity increases along a period and decreases down a group.
2. Pauling values: O = 3.5, N = 3.0, Be = 1.5, Mg = 1.2.
3. So the order is O > N > Be > Mg, i.e. II > III > I > IV.

Teacher's Note:
1. Non-metals on the right of a period are far more electronegative than the s-block metals.
2. Mg is below Be in group 2, so it is less electronegative.

Answer: (d) F > O > N

Explanation:
1. Electronegativity increases on moving from left to right in a period.
2. So the order of electronegativity of N, O and F is F > O > N (4.0 > 3.5 > 3.0).

Teacher's Note:
1. Fluorine is the most electronegative element (4.0 on the Pauling scale).
2. Unlike electron affinity, electronegativity shows no N/O anomaly here.

Answer: (d) 2.598

Explanation:
1. Pauling relation (energies in kcal mol⁻¹): \( \chi_C - \chi_H = 0.208\sqrt{\Delta} \)
2. where \( \Delta = E_{C-H} - \sqrt{E_{C-C} \times E_{H-H}} \)
3. \( \Delta = 98.8 - \sqrt{83.1 \times 104.2} = 98.8 - 93.05 = 5.75 \)
4. \( \chi_C - 2.1 = 0.208 \times \sqrt{5.75} = 0.208 \times 2.398 = 0.499 \)
5. χ_C = 2.1 + 0.499 ≈ 2.598

Teacher's Note:
1. 0.498 is only the difference χ_C − χ_H; remember to add χ_H.
2. The factor 0.208 is used when bond energies are in kcal mol⁻¹.

Answer: (c) Si, P, C, N

Explanation:
1. Electronegativity increases from left to right in a period and decreases down a group.
2. Pauling values: Si = 1.8, P = 2.1, C = 2.5, N = 3.0.
3. So the increasing order is Si < P < C < N.

Teacher's Note:
1. Period 2 elements C and N are more electronegative than period 3 elements Si and P.
2. Within each period, the element further right (P, N) is more electronegative.

Answer: (d) A-4, B-1, C-2, D-3

Explanation:
1. He (1s²) → highest ionisation energy, as it is a noble gas with a very small size (4).
2. F (1s² 2s² 2p⁵) → high electronegativity due to its small size (1).
3. Rb ([Kr] 5s¹) → most electropositive of the given atoms due to its large atomic size (2).
4. Li (1s² 2s¹) → strongest reducing agent in aqueous solution, due to its very high hydration enthalpy (most negative E° value) (3).
5. Correct match: A-4, B-1, C-2, D-3.

Teacher's Note:
1. Li is the strongest reducing agent in water even though Rb loses its electron more easily in the gas phase.
2. The high hydration enthalpy of the tiny Li⁺ ion explains this anomaly.

Answer: (c) A-2, B-3, C-5, D-1, E-4

Explanation:
1. F → maximum electronegativity (2).
2. Cl → maximum electron affinity (3).
3. Fe → variable valence (5).
4. He → maximum ionisation energy (1).
5. Ds (darmstadtium, Z = 110) → recently named by IUPAC (4).
6. Correct match: A-2, B-3, C-5, D-1, E-4.

Teacher's Note:
1. F is the most electronegative, but Cl has the highest electron affinity.
2. Variable valence is typical of transition metals such as Fe (+2, +3).