JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 21

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Multiple Choice Questions

Question 1: Which element can have a negative oxidation state?
(a) Fe
(b) Al
(c) Ca
(d) B
Show Answer & Explanation

Answer: (d) B

Explanation:
1. Boron is more electronegative than metals like Mg and shows -3 in borides such as Mg3B2.
2. Al, Ca and Fe are metals that form only positive states.

Teacher's Note:
1. Metals do not form negative states.
2. Boron is a metalloid.

Question 2: The lanthanoid contraction refers to the steady decrease in
(a) the density of the series
(b) the electronegativity of the series
(c) valence electrons of the series
(d) atomic and ionic radii of the series
Show Answer & Explanation

Answer: (d) atomic and ionic radii of the series

Explanation:
1. Poor shielding by 4f electrons makes Zeff rise along the series, so the size contracts.

Teacher's Note:
1. Contraction is in size.
2. Density and EN rise as a result.

Question 3: Which group of atoms has nearly the same atomic radius?
(a) Fe, Co, Mn
(b) Li, Be, B, C
(c) F, Cl, Br, I
(d) Na, K, Rb, Cs
Show Answer & Explanation

Answer: (a) Fe, Co, Mn

Explanation:
1. In the 3d series the increased nuclear charge is offset by screening from the added 3d electrons, so radii are nearly constant (127, 126, 125 pm).

Teacher's Note:
1. d-block series have nearly constant radii in the middle.
2. Other sets vary greatly.

Question 4: The second ionisation enthalpy of an element is always higher than its first because
(a) it is easier to remove an electron from a cation
(b) the cation always has a stable half-filled shell
(c) ionisation is an endothermic process
(d) the cation is smaller than the parent atom, so the electron is held more tightly
Show Answer & Explanation

Answer: (d) the cation is smaller than the parent atom, so the electron is held more tightly

Explanation:
1. As electrons are removed the proton to electron ratio rises, the radius falls and the remaining electrons are more strongly attracted.
2. So IE2 is greater than IE1.

Teacher's Note:
1. Successive IEs always rise.
2. The reason is smaller size and higher Zeff.

Question 5: Which of these first ionisation energy orders are correct? (a) Li < C, (b) O < N, (c) Be < N < Ne
(a) (b) and (c) only
(b) (a), (b) and (c)
(c) (a) and (b) only
(d) (a) and (c) only
Show Answer & Explanation

Answer: (b) (a), (b) and (c)

Explanation:
1. IE1 (kJ/mol): Li 520 < C 1086; O 1314 < N 1402; Be 899 < N 1402 < Ne 2081.
2. All three orders are correct.

Teacher's Note:
1. Check each against tabulated values.
2. N is above O.

Question 6: Which statement is correct?
(a) All of these: elements with large negative egH are strong oxidising agents, elements with low IE are strong reducing agents, and formation of S2-(g) from S(g) is endothermic
(b) Only the first statement
(c) Only the second statement
(d) None of these
Show Answer & Explanation

Answer: (a) All of these: elements with large negative egH are strong oxidising agents, elements with low IE are strong reducing agents, and formation of S2-(g) from S(g) is endothermic

Explanation:
1. Halogens have large negative egH and are strong oxidants; alkali metals have low IE and are strong reducers.
2. Adding the second electron to S- needs energy, so overall S to S2- is endothermic.

Teacher's Note:
1. Oxidising power relates to egH and reducing power to IE.
2. Second egH is positive.

Question 7: For the electron gain enthalpy of chalcogens and halogens, which of the following is correct?
(a) O < Cl
(b) S < Se
(c) Br > F
(d) S > F
Show Answer & Explanation

Answer: (a) O < Cl

Explanation:
1. Magnitudes (kJ/mol): O 141, Cl 349, Br 325, F 328, S 200, Se 195.
2. Only O < Cl is true; Br < F, S < F and S > Se.

Teacher's Note:
1. Halogens have higher affinity than chalcogens.
2. S is the highest chalcogen.

Question 8: O(g) + e- gives O-(g) with egH = -141 kJ/mol, while O-(g) + e- gives O2-(g) with egH = +780 kJ/mol. The formation of O2- in the gas phase is unfavourable although O2- is isoelectronic with neon, because
(a) oxygen has a high electron gain enthalpy
(b) oxygen is more electronegative
(c) electron repulsion outweighs the stability gained by reaching the noble gas configuration
(d) the O- ion is smaller than the O atom
Show Answer & Explanation

Answer: (c) electron repulsion outweighs the stability gained by reaching the noble gas configuration

Explanation:
1. The added electron is repelled by the negative charge of O-, and the repulsion exceeds the gain from a neon-like shell.

Teacher's Note:
1. Second egH is always positive.
2. Oxide ions are stabilised only in a lattice.

Question 9: Which property is not common to both group 1 and group 17 elements?
(a) Electropositive character increases down the group
(b) Atomic radius increases with atomic number
(c) Reactivity decreases from top to bottom in the group
(d) Electronegativity decreases down the group
Show Answer & Explanation

Answer: (c) Reactivity decreases from top to bottom in the group

Explanation:
1. Reactivity of alkali metals increases down the group, but that of halogens decreases.
2. The other three properties follow the same trend in both groups.

Teacher's Note:
1. Metals react by losing electrons, non-metals by gaining.
2. Reactivity trends are opposite.

Question 10: The ionisation enthalpy is highest for the electron removed from the orbital with azimuthal quantum number l equal to (other factors equal)
(a) 1
(b) 3
(c) 2
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. For the same shell, penetration is s > p > d > f, so s electrons are held most tightly.

Teacher's Note:
1. Higher penetration means higher IE.
2. l = 0 is an s electron.

Question 11: Among elements with atomic number 23, 24, 25 and 26, the one with the highest second ionisation enthalpy has atomic number
(a) 24
(b) 26
(c) 25
(d) 23
Show Answer & Explanation

Answer: (a) 24

Explanation:
1. Removing one electron from Cr gives Cr+ with a stable 3d5 configuration, so the second electron is hard to remove (1592 kJ/mol).

Teacher's Note:
1. Half-filled d5 is stable.
2. Cr has configuration 3d54s1.

Question 12: An element with the electronic configuration [Ar]3d104s1 belongs to group
(a) 1
(b) 10
(c) 12
(d) 11
Show Answer & Explanation

Answer: (d) 11

Explanation:
1. The group of a d-block element = number of (n-1)d electrons + ns electrons = 10 + 1 = 11.

Teacher's Note:
1. Cu is in group 11.
2. Count d + s electrons.

Question 13: Which of the following ions has the highest ionic radius?
(a) F-
(b) Li+
(c) B3+
(d) O2-
Show Answer & Explanation

Answer: (d) O2-

Explanation:
1. O2- and F- are isoelectronic and O2- has the lower nuclear charge, so it is larger; Li+ and B3+ are small cations.

Teacher's Note:
1. Isoelectronic anions: radius is larger for smaller Z.
2. Cations are small.

Question 14: In which arrangement is the order NOT according to the property indicated?
(a) Al3+ < Mg2+ < Na+ < F-, increasing ionic size
(b) B < C < N < O, increasing first ionisation enthalpy
(c) Li < Na < K < Rb, increasing metallic radius
(d) I < Br < F < Cl, increasing electron gain enthalpy (with negative sign)
Show Answer & Explanation

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

Explanation:
1. IE1 (kJ/mol): B 801, C 1086, N 1402, O 1314.
2. N is higher than O, so the stated order is wrong.

Teacher's Note:
1. N has a half-filled 2p.
2. The other three orders are correct.

Question 15: The lanthanoid contraction is responsible for the fact that
(a) Zr and Nb have similar oxidation states
(b) Zr and Zn have the same oxidation state
(c) Zr and Y have about the same radius
(d) Zr and Hf have about the same radius
Show Answer & Explanation

Answer: (d) Zr and Hf have about the same radius

Explanation:
1. The 4f electrons shield poorly, so the 5d elements (Hf) are almost the same size as their 4d congeners (Zr).

Teacher's Note:
1. Contraction makes second and third transition series members alike in size.
2. This makes Zr and Hf difficult to separate.

Question 16: Lanthanoid contraction is caused by
(a) the appreciable shielding of outer electrons by 4f electrons
(b) the same effective nuclear charge from Ce to Lu
(c) the appreciable shielding by 5f electrons
(d) the imperfect shielding of outer electrons by 4f electrons from the nuclear charge
Show Answer & Explanation

Answer: (d) the imperfect shielding of outer electrons by 4f electrons from the nuclear charge

Explanation:
1. The diffuse f orbitals shield poorly, so Zeff grows as 4f electrons are added and the size shrinks.

Teacher's Note:
1. Poor shielding by f electrons.
2. Zeff increases from Ce to Lu.

Question 17: The correct order of ionic radius is
(a) Li+ > Be2+ > Na+ > Mg2+
(b) Li+ > Na+ > Mg2+ > Be2+
(c) Mg2+ > Be2+ > Li+ > Na+
(d) Na+ > Li+ > Mg2+ > Be2+
Show Answer & Explanation

Answer: (d) Na+ > Li+ > Mg2+ > Be2+

Explanation:
1. Radii (pm): Na+ 102, Li+ 76, Mg2+ 72, Be2+ 45.

Teacher's Note:
1. Radius increases down a group and falls with higher charge.
2. Diagonal pairs are similar in size.

Question 18: The outer electronic configuration of Gd (Z = 64) is
(a) 4f35d56s2
(b) 4f85d06s2
(c) 4f45d46s2
(d) 4f75d16s2
Show Answer & Explanation

Answer: (d) 4f75d16s2

Explanation:
1. A half-filled 4f7 subshell is extra stable, so Gd takes one electron into 5d rather than 4f8.

Teacher's Note:
1. Half-filled subshells give extra stability.
2. Gd is an exception to the aufbau order.

Question 19: The correct order of electron gain enthalpy (with negative sign) of F, Cl, Br and I is
(a) Br > Cl > I > F
(b) I > Br > Cl > F
(c) Cl > F > Br > I
(d) F > Cl > Br > I
Show Answer & Explanation

Answer: (c) Cl > F > Br > I

Explanation:
1. Cl (349) > F (328) > Br (325) > I (295 kJ/mol).
2. F is lowered by repulsion in its small 2p subshell.

Teacher's Note:
1. Cl has the highest affinity of all elements.
2. F is an exception.

Question 20: The correct order of increasing first ionisation enthalpy for Ca, Ba, S, Se and Ar is
(a) Ba < Ca < Se < S < Ar
(b) S < Se < Ca < Ba < Ar
(c) Ca < Ba < S < Se < Ar
(d) Ca < S < Ba < Se < Ar
Show Answer & Explanation

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

Explanation:
1. IE1 (kJ/mol): Ba 503, Ca 590, Se 941, S 1000, Ar 1521.
2. IE falls down a group, so Ba < Ca and Se < S.

Teacher's Note:
1. Metals have low IE.
2. Noble gas has the highest.

Question 21: The first ionisation potential of Na is 5.1 eV. The electron gain enthalpy of Na+ is
(a) -5.1 eV
(b) -10.2 eV
(c) +2.55 eV
(d) -2.55 eV
Show Answer & Explanation

Answer: (a) -5.1 eV

Explanation:
1. Na+ + e- gives Na is the reverse of Na gives Na+ + e-.
2. So its enthalpy change is -5.1 eV.

Teacher's Note:
1. Electron gain by a cation = minus the ionisation energy of the atom.
2. Same magnitude, opposite sign.

Question 22: Which atom has the highest first ionisation energy: Na, K, Sc or Rb?
(a) Na
(b) K
(c) Rb
(d) Sc
Show Answer & Explanation

Answer: (d) Sc

Explanation:
1. IE1 (kJ/mol): Sc 633, Na 496, K 419, Rb 403.
2. Sc is a d-block element with a smaller size and higher Zeff.

Teacher's Note:
1. Alkali metals have the lowest IE in their periods.
2. Sc has a higher nuclear charge.

Question 23: Which group contains isoelectronic species?
(a) O-, F-, Na, Mg+
(b) O2-, F-, Na+, Mg2+
(c) O-, F-, Na+, Mg2+
(d) O2-, F-, Na, Mg2+
Show Answer & Explanation

Answer: (b) O2-, F-, Na+, Mg2+

Explanation:
1. O2-, F-, Na+ and Mg2+ each have 10 electrons.
2. In the other sets, at least one species has a different electron count.

Teacher's Note:
1. Count electrons: Z minus charge.
2. All 10 here.

Question 24: The oxidation state of chromium in K2Cr2O7 is
(a) +4
(b) -4
(c) +6
(d) +3
Show Answer & Explanation

Answer: (c) +6

Explanation:
1. 2(+1) + 2x + 7(-2) = 0, so x = +6.

Teacher's Note:
1. Sum of oxidation numbers is zero.
2. Dichromate has Cr(VI).

Question 25: The smallest species among N3-, F-, O2- and Na+ is
(a) O2-
(b) N3-
(c) F-
(d) Na+
Show Answer & Explanation

Answer: (d) Na+

Explanation:
1. All have 10 electrons and Na+ has the highest nuclear charge (+11), so it is the smallest.

Teacher's Note:
1. Higher Z gives smaller isoelectronic ion.
2. N3- is largest.

Question 26: The formation of an anion from a neutral atom X is favoured by
(a) large size of X
(b) high charge on the anion
(c) low ionisation potential
(d) high electron affinity
Show Answer & Explanation

Answer: (d) high electron affinity

Explanation:
1. A large negative electron gain enthalpy (high electron affinity) means more energy is released when X gains an electron.

Teacher's Note:
1. Anion formation is related to electron gain enthalpy.
2. Large charge on the anion is unfavourable.

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