JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 04

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Chapter-wise Objective Questions: Classification Of Elements and Periodicity In Properties

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Multiple Choice Questions

Question 1: The formula of effective nuclear charge is (sigma = screening constant)
(a) Z - sigma
(b) Z + sigma
(c) Z / sigma
(d) Z x sigma
Show Answer & Explanation

Answer: (a) Z - sigma

Explanation:
1. Effective nuclear charge Zeff = Z - sigma.

Teacher's Note:
1. Screening reduces the nuclear attraction felt by a valence electron.

Question 2: In the sodium atom, the screening of the valence electron is due to
(a) 3s2, 3p6
(b) 2s1
(c) 1s2, 2s2, 2p6
(d) 1s2, 2s2
Show Answer & Explanation

Answer: (c) 1s2, 2s2, 2p6

Explanation:
1. The valence electron is 3s1 and the inner 10 electrons (1s2 2s2 2p6) screen it.

Teacher's Note:
1. Na: Z = 11, sigma about 8.8 by Slater, Zeff about 2.2.

Question 3: Which of the following orders of radii is correct?
(a) Li < Be < Mg
(b) H+ < Li+ < H-
(c) O < F < Ne
(d) Na+ > F- > O2-
Show Answer & Explanation

Answer: (b) H+ < Li+ < H-

Explanation:
1. H+ (no electrons) is tiny, Li+ is 76 pm and H- is about 208 pm.
2. Be is smaller than Li; Mg is larger than Be; in isoelectronic Na+ < F- < O2-.

Teacher's Note:
1. Anions are larger than their parent atoms; cations smaller.

Question 4: K+, Ar, Ca2+ and S2- contain
(a) Same electronic configuration and atomic volume
(b) Different electronic configuration but same IP
(c) Same electronic configuration but different atomic volume
(d) None
Show Answer & Explanation

Answer: (c) Same electronic configuration but different atomic volume

Explanation:
1. All have 18 electrons (3s2 3p6) but different nuclear charge, so radii differ.

Teacher's Note:
1. More protons mean smaller radius for the same number of electrons.

Question 5: Which group of atoms have nearly the same atomic radius?
(a) Na, K, Rb, Cs
(b) Li, Be, B, C
(c) Fe, Co, Ni
(d) F, Cl, Br, I
Show Answer & Explanation

Answer: (c) Fe, Co, Ni

Explanation:
1. In the 3d series the radii of Fe, Co and Ni (about 126, 125, 124 pm) are almost equal due to increased screening by d-electrons.

Teacher's Note:
1. Constant radii occur in the middle of the d-series.

Question 6: The atomic radii of fluorine and neon in angstrom units are given by
(a) 0.72, 1.60
(b) 1.60, 1.60
(c) 0.72, 0.72
(d) None
Show Answer & Explanation

Answer: (a) 0.72, 1.60

Explanation:
1. The radius for F is the covalent radius (0.72 A) and for Ne it is the van der Waals radius (1.60 A).

Teacher's Note:
1. Noble gases have only van der Waals radii.

Question 7: Which order of atomic/ionic radius is NOT correct?
(a) I- > I > I+
(b) Mg2+ > Na+ > F-
(c) P5+ < P3+
(d) Li > Be > B
Show Answer & Explanation

Answer: (b) Mg2+ > Na+ > F-

Explanation:
1. Among isoelectronic ions radius decreases with nuclear charge: F- > Na+ > Mg2+.
2. The other orders are correct.

Teacher's Note:
1. Isoelectronic: more protons, smaller ion.

Question 8: In the lithium atom, the screening effect on the valence electron is caused by
(a) Electrons of K and L shell
(b) Electrons of K shell only
(c) Two electrons of 1st and one of the 2nd shell
(d) None
Show Answer & Explanation

Answer: (b) Electrons of K shell only

Explanation:
1. The valence electron is 2s1; only the two 1s electrons (K shell) screen it effectively.

Teacher's Note:
1. Electrons in the same shell screen only weakly.

Question 9: The radius of the potassium atom is 0.203 nm. The radius of the potassium ion in nanometre will be
(a) 0.133
(b) 0.231
(c) 0.234
(d) 0.251
Show Answer & Explanation

Answer: (a) 0.133

Explanation:
1. Cations are smaller than atoms; the K+ radius is about 0.133 nm.

Teacher's Note:
1. Loss of the outer shell shrinks the ion a lot.

Question 10: S2- is NOT isoelectronic with
(a) Ar
(b) Cl-
(c) HS-
(d) Ti3+
Show Answer & Explanation

Answer: (d) Ti3+

Explanation:
1. S2-, Ar, Cl- and HS- each have 18 electrons; Ti3+ has 22 - 3 = 19.

Teacher's Note:
1. Count the electrons.

Question 11: In an anion, the radius compared with the neutral atom is
(a) Smaller because protons decrease
(b) Smaller because protons are more
(c) Larger because Zeff is more
(d) Larger because of greater electron repulsion
Show Answer & Explanation

Answer: (d) Larger because of greater electron repulsion

Explanation:
1. An anion has more electrons than protons, so repulsion rises and Zeff per electron falls: the radius is larger.

Teacher's Note:
1. Anion radius is greater than atomic radius.

Question 12: The maximum size of the first member of a period is due to
(a) Maximum number of shells
(b) Maximum screening effect
(c) Minimum Zeff
(d) All of the above
Show Answer & Explanation

Answer: (c) Minimum Zeff

Explanation:
1. Alkali metals have the smallest Zeff in the period, which makes them the largest.

Teacher's Note:
1. The number of shells is the same across a period.

Question 13: In which pair is the radius of the second species smaller than that of the first species?
(a) Li, Na
(b) Na+, F-
(c) N3-, Al3+
(d) Mn7+, Mn4+
Show Answer & Explanation

Answer: (c) N3-, Al3+

Explanation:
1. N3- and Al3+ are isoelectronic; Al3+ has more protons, hence smaller.
2. In (a) Na is larger; in (b) F- is larger; in (d) Mn4+ is larger.

Teacher's Note:
1. Use charge/size rules.

Question 14: Which is the incorrect order of atomic radii?
(a) Cu > Zn
(b) Cl > F
(c) P > S
(d) Sc > Ti
Show Answer & Explanation

Answer: (a) Cu > Zn

Explanation:
1. Cu (128 pm) is smaller than Zn (134 pm), so the order Cu > Zn is incorrect.
2. The others (Cl > F, P > S, Sc > Ti) are correct.

Teacher's Note:
1. In the d-series radii first decrease then increase slightly at the end.

Question 15: The correct decreasing order of size of ions is
(a) Br- > S2- > Cl- > N3-
(b) N3- > S2- > Cl- > Br-
(c) Br- > Cl- > S2- > N3-
(d) N3- > Cl- > S2- > Br-
Show Answer & Explanation

Answer: (a) Br- > S2- > Cl- > N3-

Explanation:
1. Ionic radii (pm): Br- 196, S2- 184, Cl- 181, N3- 146.

Teacher's Note:
1. Larger period and lower charge gives different sizes: compare radii.

Question 16: Which is an incorrect order of ionic radius?
(a) La3+ > Gd3+ > Eu3+ > Lu3+
(b) V2+ > V3+ > V4+ > V5+
(c) In+ > Sn2+ > Sb3+
(d) K+ > Sc3+ > V5+ > Mn7+
Show Answer & Explanation

Answer: (a) La3+ > Gd3+ > Eu3+ > Lu3+

Explanation:
1. Along the lanthanoids the ionic radius decreases with atomic number: La3+ > Eu3+ > Gd3+ > Lu3+.
2. So placing Gd3+ before Eu3+ is wrong.

Teacher's Note:
1. Lanthanoid contraction.

Question 17: The correct order of effective nuclear charge on the last electron according to Slater's rule for Li, Na and K is
(a) Li > Na > K
(b) K > Na > Li
(c) Na > Li > K
(d) Li < Na = K
Show Answer & Explanation

Answer: (d) Li < Na = K

Explanation:
1. Slater: Li 1.30; Na 2.20; K 2.20.

Teacher's Note:
1. Zeff increases from Li to Na then stays the same for K.

Question 18: The ionisation potential of isotopes of an element will be
(a) Same
(b) Different
(c) Depends on atomic masses
(d) Depends on number of neutrons
Show Answer & Explanation

Answer: (a) Same

Explanation:
1. Ionisation energy depends on nuclear charge and electron configuration, which are the same for isotopes.

Teacher's Note:
1. Isotopes differ only in neutron number.

Question 19: A sudden large jump between the second and third ionisation energies of an element is associated with the configuration
(a) 1s2 2s2 2p6 3s1
(b) 1s2 2s2 2p6 3s2 3p5
(c) 1s2 2s2 2p6 3s2 3p2
(d) 1s2 2s2 2p6 3s2
Show Answer & Explanation

Answer: (d) 1s2 2s2 2p6 3s2

Explanation:
1. 3s2 has two valence electrons: the third electron comes from the stable neon core, which needs much more energy.

Teacher's Note:
1. A large jump after IEn means the element has n valence electrons.

Question 20: A sudden jump between the first and second IE appears for which configuration?
(a) 1s2 2s2 2p6 3s1
(b) 1s2 2s2 2p6 3s2
(c) 1s2 2s2 2p6 3s2 3p1
(d) 1s2 2s2 2p6 3s2 3p6
Show Answer & Explanation

Answer: (a) 1s2 2s2 2p6 3s1

Explanation:
1. One valence electron (3s1) is lost easily and the second comes from the stable neon core.

Teacher's Note:
1. Alkali metals have a huge IE2.

Question 21: Which of the following has the least ionisation potential?
(a) Be3+
(b) H
(c) Li2+
(d) He+
Show Answer & Explanation

Answer: (b) H

Explanation:
1. These are hydrogen-like species with IE = 13.6 Z2 eV: Be3+ 217.6, H 13.6, Li2+ 122.4, He+ 54.4.

Teacher's Note:
1. IE of one-electron species goes as Z squared.

Question 22: Mg forms Mg2+ (and not Mg3+) because
(a) The oxidation state of Mg is +2
(b) The difference between IP1 and IP2 is greater than 16 eV
(c) There is only one electron in the outermost energy level
(d) The difference between IP1 and IP2 is less than 11 eV
Show Answer & Explanation

Answer: (d) The difference between IP1 and IP2 is less than 11 eV

Explanation:
1. IE1 and IE2 of Mg are 7.6 and 15.0 eV, a difference of about 7.4 eV, which is small, while IE3 is very large (80 eV).
2. So the second electron is lost easily and the third is not.

Teacher's Note:
1. The two 3s electrons are lost; the third is from the neon core.

Question 23: Which of the following has the minimum first ionisation energy?
(a) 1s2 2s2 2p5
(b) 1s2 2s2 2p6 3s2 3p2
(c) 1s2 2s2 2p6 3s1
(d) 1s2 2s2 2p6
Show Answer & Explanation

Answer: (c) 1s2 2s2 2p6 3s1

Explanation:
1. These are F, Si, Na and Ne; the alkali metal Na (3s1) has the lowest IE (496 kJ/mol).

Teacher's Note:
1. Alkali metals have the lowest IE in a period.

Question 24: The energy needed to remove one electron from a uni-positive ion is called
(a) 1st I.P.
(b) 3rd I.P.
(c) 2nd I.P.
(d) 1st E.A.
Show Answer & Explanation

Answer: (c) 2nd I.P.

Explanation:
1. Removing an electron from M+ gives M2+ and is the second ionisation energy.

Teacher's Note:
1. IE1 refers to the neutral atom.

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