JEE Chemistry Classification Of Elements and Periodicity In Properties MCQs Set 17

Answer: (a) Period = 3, block = p, group = 16

Explanation:
1. The highest principal quantum number is 3, so the period is 3.
2. The last electron enters the 3p subshell, so it is a p-block element.
3. Group number = 10 + number of valence electrons (2 + 4 = 6), giving group 16 (sulphur).

Teacher's Note:
1. For p-block elements, group = 10 + (ns and np electrons).
2. The period is always the outermost shell number.

Answer: (a) [Rn] 5f¹⁴ 6d⁴ 7s²

Explanation:
1. Radon has Z = 86, so the extra 20 electrons go: 7s² (2), 5f¹⁴ (14), 6d⁴ (4) = 20.
2. Element 106 lies in group 6 of the 7th period, a d-block element, so its configuration is [Rn] 5f¹⁴ 6d⁴ 7s².

Teacher's Note:
1. Count the electrons beyond the noble-gas core and fill 7s, then 5f, then 6d.
2. Element 106 is in the same group as Cr, Mo and W.

Answer: (c) Similar size, similar electronegativity and similar high polarising power

Explanation:
1. Li and Mg lie diagonally in the periodic table.
2. The smaller size of Li⁺ is offset by Mg²⁺ having a higher charge, so their charge/size ratios (polarising power) are similar.
3. Their electronegativities are also close, so they resemble each other (for example both form nitrides and their carbonates decompose on heating).

Teacher's Note:
1. Diagonal relationships arise from similar charge-to-size ratio.
2. Other diagonal pairs are Be-Al and B-Si.

Answer: (b) 3, 11

Explanation:
1. Elements with the same outer configuration share chemical properties.
2. Z = 3 (Li, 2s¹) and Z = 11 (Na, 3s¹) are both alkali metals of group 1.
3. The other pairs are Al/Ti, Be/Cr and He/Be, which have different valence configurations.

Teacher's Note:
1. Same group means the same valence-shell configuration.
2. Write the configurations quickly: 2,1 and 2,8,1.

Answer: (c) Z

Explanation:
1. X is Si (a metalloid), Y is Ar (a noble gas), Z is K (an alkali metal) and T is Cl (a non-metal).
2. Metallic character is greatest for elements with few valence electrons and a large size, so potassium is the most metallic.

Teacher's Note:
1. Identify each element from its atomic number: 14, 18, 19 and 17.
2. Alkali metals are the most metallic elements of their periods.

Answer: (d) Be

Explanation:
1. Na is 3s¹, Cu is 3d¹⁰4s¹ and Au is 5d¹⁰6s¹.
2. Be is 1s²2s², so it has two valence electrons.

Teacher's Note:
1. Cu and Au (coinage metals) have one s electron because of the stable d¹⁰ arrangement.
2. Group 2 elements always have ns².

Answer: (a) 1 and 17

Explanation:
1. Ionic bonding is strongest when one element has very low ionisation energy and the other very high electron affinity.
2. Group 1 (alkali metals) and group 17 (halogens) fit best, as in CsF.

Teacher's Note:
1. A large electronegativity difference gives an ionic bond.
2. Noble gases (group 18) rarely form compounds.

Answer: (d) N, P, V

Explanation:
1. Set (a): Cl^- has 0, Fe³⁺ (3d⁵) has 5, Cr³⁺ (3d³) has 3. Not equal.
2. Set (b): Na⁺ 0, Mg²⁺ 0, Al (3p¹) 1. Not equal.
3. Set (c): Na 1, P (3p³) 3, Cl 1. Not equal.
4. Set (d): N (2p³) 3, P (3p³) 3, V (3d³4s²) 3. All equal.

Teacher's Note:
1. Write the valence configuration of each species and count half-filled orbitals.
2. Half-filled p³ and d³ both give three unpaired electrons.

Answer: (a) H₂O

Explanation:
1. The bond angles are H₂O about 104.5 degrees, NH₃ about 107 degrees, CH₄ 109.5 degrees and CO₂ 180 degrees.
2. Water has two lone pairs on oxygen, which compress the bond angle the most.

Teacher's Note:
1. More lone pairs on the central atom reduce the bond angle.
2. Linear molecules have 180 degrees.

Answer: (b) Na₂CO₃

Explanation:
1. MgCO₃ gives MgO + CO₂, Li₂CO₃ gives Li₂O + CO₂, and Ca(HCO₃)₂ gives CaCO₃ + H₂O + CO₂.
2. Sodium carbonate is thermally stable because the large, weakly polarising Na⁺ cannot distort the carbonate ion.

Teacher's Note:
1. Thermal stability of carbonates increases with cation size down a group.
2. Lithium is anomalous and behaves like magnesium.

Answer: (b) Tungsten

Explanation:
1. Mercury is liquid at room temperature and silver melts near 962 degrees C.
2. Chromium melts at about 1907 degrees C, whereas tungsten melts at about 3422 degrees C, the highest of all metals.

Teacher's Note:
1. Strong metallic bonding by many unpaired d electrons gives very high melting points.
2. Diamond is not a metal.

Answer: (a) 1/2

Explanation:
1. For hydrogen-like species r = 0.529 n²/Z angstrom.
2. For H, Z = 1 and n = 1; for He⁺, Z = 2 and n = 1.
3. Ratio = (1/2)/(1/1) = 1/2.

Teacher's Note:
1. Radius of a one-electron species is inversely proportional to Z at fixed n.
2. Both species have n = 1 in the ground state.

Answer: (a) O^2- > F^- > O > F

Explanation:
1. An anion is larger than its parent atom, and between isoelectronic anions the one with lower nuclear charge is larger, so O^2- > F^-.
2. Among the neutral atoms, O is larger than F (higher nuclear charge in F).
3. Order: O^2- (1.40 A) > F^- (1.36 A) > O (0.74 A) > F (0.72 A).

Teacher's Note:
1. Compare anions with each other first, then atoms.
2. In a period, radius falls from left to right.

Answer: (c) F^-

Explanation:
1. All four ions are isoelectronic (10 electrons).
2. Size decreases as the nuclear charge increases: C^4- (Z = 6) > N^3- (7) > O^2- (8) > F^- (9).
3. F^- has the highest nuclear charge and is the smallest.

Teacher's Note:
1. For isoelectronic species, size is inversely related to Z.
2. Write the atomic numbers in order.

Answer: (a) Al

Explanation:
1. Removing electrons increases the effective nuclear charge per electron and shrinks the species.
2. The neutral atom Al is largest; the size decreases as Al > Al⁺ > Al²⁺ > Al³⁺.

Teacher's Note:
1. A cation is always smaller than its parent atom.
2. More positive charge means smaller size.

Answer: (d) I^- > I > I⁺

Explanation:
1. The anion I^- has an extra electron, so electron repulsion enlarges it beyond the atom.
2. The cation I⁺ has one electron less and a higher effective nuclear charge, so it is smaller than I.
3. Order: I^- > I > I⁺.

Teacher's Note:
1. Anion > atom > cation for the same element.
2. Charge-to-electron ratio explains the trend.

Answer: (a) Mg²⁺ < Na⁺ < F^- < Al

Explanation:
1. Approximate radii: Mg²⁺ 0.72 A, Na⁺ 0.95 A, F^- 1.36 A, Al atom 1.43 A (metallic radius).
2. The increasing order is therefore Mg²⁺ < Na⁺ < F^- < Al.

Teacher's Note:
1. Isoelectronic ions: higher charge means smaller size.
2. A neutral atom of the third period is larger than a second-period anion.

Answer: (a) N^3- > O^2- > F^- > Na⁺

Explanation:
1. All four are isoelectronic with 10 electrons.
2. Radius decreases as the nuclear charge increases: N (7), O (8), F (9), Na (11).
3. So N^3- > O^2- > F^- > Na⁺.

Teacher's Note:
1. Use Z/e as the guide for isoelectronic species.
2. Anions with low Z are the largest.

Answer: (c) 1.71, 1.40, 1.36

Explanation:
1. The three ions are isoelectronic, so radius decreases N^3- > O^2- > F^-.
2. The experimental values are 1.71, 1.40 and 1.36 angstrom.

Teacher's Note:
1. The ion with the smallest nuclear charge is the largest.
2. Order of Z is 7 < 8 < 9.

Answer: (a) K⁺ > Na⁺ > Li⁺

Explanation:
1. Going down group 1, each ion has one more shell.
2. Hence Li⁺ (0.76 A) < Na⁺ (1.02 A) < K⁺ (1.38 A).

Teacher's Note:
1. Ions of the same charge increase in size down a group.

Answer: (c) NaCl < LiCl < BeCl₂

Explanation:
1. By Fajans' rules a small, highly charged cation polarises the anion more and increases covalent character.
2. Na⁺ is largest and least polarising, Li⁺ is smaller, and Be²⁺ is smallest with a 2+ charge.
3. Covalent character: NaCl < LiCl < BeCl₂.

Teacher's Note:
1. Polarising power is proportional to charge/size.
2. BeCl₂ is a covalent, volatile compound.

Answer: (b) The second ionisation energy of Mg is greater than that of Na

Explanation:
1. (a) Al (577) < Mg (738): correct, because Al removes a 3p electron.
2. (b) IE₂ of Mg is 1451 kJ/mol while IE₂ of Na is 4562 kJ/mol (removal from the neon core), so Mg < Na and the statement is wrong.
3. (c) Na (496) < Mg (738): correct.
4. (d) IE₃ of Mg (7733) > IE₃ of Al (2745): correct.

Teacher's Note:
1. Removing a second electron from an alkali metal breaks a noble-gas core.
2. Check each statement against the configuration.

Answer: (c) [Ne] 3s² 3p³

Explanation:
1. The four species are Al, Si, P and Ge.
2. P has a half-filled 3p subshell, which is extra stable, so its IE (1012 kJ/mol) is the highest; Ge is larger and has lower IE than Si.

Teacher's Note:
1. IE rises across a period and falls down a group, apart from half-filled subshell effects.
2. Identify the elements first.

Answer: (c) Nitrogen

Explanation:
1. N and P are in group 15 with half-filled p subshells, and O and S are in group 16.
2. Within a group, IE falls on descending, so N > P and O > S.
3. N has the half-filled 2p³ and a smaller size, so its IE is the highest (1402 kJ/mol).

Teacher's Note:
1. Compare N with O: nitrogen is higher because of half-filled stability.
2. Smaller atoms hold electrons more tightly.

Answer: (c) As < S < P

Explanation:
1. Values: As 947, S 1000 and P 1012 kJ/mol.
2. P is higher than S because of its half-filled 3p subshell, and S is higher than As because As is larger.
3. Order: As < S < P.

Teacher's Note:
1. P > S is the usual half-filled exception.
2. Larger atoms have lower IE.