JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 03

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Multiple Choice Questions

Question 1: Which of the following sets are Dobereiner triads? (a) P, As, Sb (b) Cu, Ag, Au (c) Fe, Co, Ni (d) S, Se, Te
(a) a and b
(b) b and c
(c) a and d
(d) All
Show Answer & Explanation

Answer: (c) a and d

Explanation:
1. In a triad the atomic mass of the middle element is nearly the mean of the other two.
2. P, As, Sb: (31 + 122)/2 = 76.5, near As = 75. S, Se, Te: (32 + 128)/2 = 80, near Se = 79.
3. Cu, Ag, Au: (63.5 + 197)/2 = 130, far from Ag = 108. Fe, Co, Ni have similar, not progressive, masses.

Teacher's Note:
1. Check the mean-mass rule numerically.
2. Ca, Sr, Ba and Cl, Br, I are the classic triads.

Question 2: The atomic weights of P and Sb are 31 and 120. According to Dobereiner's triad rule, the atomic weight of As is
(a) 151
(b) 75.5
(c) 89.5
(d) Unpredictable
Show Answer & Explanation

Answer: (b) 75.5

Explanation:
1. Mean = (31 + 120)/2 = 75.5.

Teacher's Note:
1. The true value is 74.9.

Question 3: The places left empty by Mendeleev were for
(a) Aluminium and silicon
(b) Gallium and germanium
(c) Arsenic and antimony
(d) Molybdenum and tungsten
Show Answer & Explanation

Answer: (b) Gallium and germanium

Explanation:
1. He left gaps for eka-aluminium (Ga) and eka-silicon (Ge), and predicted their properties.

Teacher's Note:
1. Their later discovery confirmed his periodic law.

Question 4: The law of triads is applicable to
(a) Os, Ir, Pt
(b) Ca, Sr, Ba
(c) Fe, Co, Ni
(d) Ru, Rh, Pd
Show Answer & Explanation

Answer: (b) Ca, Sr, Ba

Explanation:
1. Ca (40), Sr (88), Ba (137): mean of Ca and Ba = 88.5, which is close to Sr.
2. The other sets are not triads by the mass rule.

Teacher's Note:
1. Dobereiner's triads were the first attempt to classify elements.

Question 5: Elements that occupy peaks of the Lothar Meyer curve (atomic volume against atomic weight) are
(a) Alkali metals
(b) Highly electropositive elements
(c) Elements with large atomic volume
(d) All of the above
Show Answer & Explanation

Answer: (d) All of the above

Explanation:
1. Alkali metals are at the peaks and are highly electropositive with the largest atomic volumes.

Teacher's Note:
1. Each peak starts a new period.

Question 6: Which statement is wrong?
(a) The 2nd period contains 8 elements
(b) The 3rd period contains 18 elements
(c) The 1st period contains two non-metals
(d) The p-block has metals, non-metals and metalloids
Show Answer & Explanation

Answer: (b) The 3rd period contains 18 elements

Explanation:
1. The 3rd period has 8 elements (Na to Ar), not 18.
2. The other statements are correct (H and He are non-metals).

Teacher's Note:
1. Period lengths: 2, 8, 8, 18, 18, 32.

Question 7: Which of the following elements was absent in Mendeleev's periodic table?
(a) Tc
(b) Si
(c) B
(d) F
Show Answer & Explanation

Answer: (a) Tc

Explanation:
1. Tc (technetium) was discovered in 1937; Si, B and F were known.

Teacher's Note:
1. Technetium was the first artificially produced element.

Question 8: Which statement is wrong for the long form of the periodic table?
(a) There are 7 periods and 18 groups
(b) The number of valence shell electrons in a period is the same
(c) The group IIIB contains 32 elements
(d) Lanthanides and actinides are placed in the same group
Show Answer & Explanation

Answer: (b) The number of valence shell electrons in a period is the same

Explanation:
1. Valence electrons increase from left to right across a period, so they are not the same.
2. Group IIIB (group 3) includes the 14 lanthanides and 14 actinides along with Sc, Y, La, Ac: 32 elements in the extended view.

Teacher's Note:
1. Elements of a group (not a period) have the same valence configuration.

Question 9: The elements Ga and Sc are cited as examples proving the validity of the periodic law because
(a) They were predicted by Mendeleev (eka-aluminium and eka-boron)
(b) They are alkali metals
(c) They are noble gases
(d) They are lanthanides
Show Answer & Explanation

Answer: (a) They were predicted by Mendeleev (eka-aluminium and eka-boron)

Explanation:
1. Mendeleev predicted their existence and properties, and later discoveries matched.

Teacher's Note:
1. Predicted: eka-aluminium (Ga), eka-boron (Sc), eka-silicon (Ge).

Question 10: Which statement is false?
(a) Elements with ns2np6 configuration lie in the 1st to 6th period
(b) Typical elements lie in the 3rd period
(c) The seventh period can accommodate 32 elements
(d) Boron and silicon are diagonally related
Show Answer & Explanation

Answer: (a) Elements with ns2np6 configuration lie in the 1st to 6th period

Explanation:
1. Noble gases with ns2np6 lie in periods 2 to 6; helium is 1s2.
2. The other statements are correct.

Teacher's Note:
1. Diagonal relationship: Li-Mg, Be-Al, B-Si.

Question 11: Among the lanthanides the one obtained by the synthetic method is
(a) Lu
(b) Pm
(c) Pr
(d) Ce
Show Answer & Explanation

Answer: (b) Pm

Explanation:
1. Promethium (Z = 61) has no stable isotope and is made artificially.

Teacher's Note:
1. Tc is the lightest synthetic element.

Question 12: The expected electronic configuration of the element with atomic number 108 is
(a) [Rn] 5f14 6d6 7s2
(b) 6f14 7d8 7s2
(c) [Rn] 5f14 6d8 7s0
(d) [Xe] 4f14 5d8 6s2
Show Answer & Explanation

Answer: (a) [Rn] 5f14 6d6 7s2

Explanation:
1. Z = 108 (Hassium): 86 (Rn) + 14 (5f) + 6 (6d) + 2 (7s) = 108.

Teacher's Note:
1. Fill 7s, then 5f, then 6d in this region.

Question 13: In the 6th period, the electronic energy levels are filled in the order
(a) 6s, 4f, 5d, 6p
(b) 6s, 6p, 4f, 5d
(c) 4f, 5d, 6s, 6p
(d) None
Show Answer & Explanation

Answer: (a) 6s, 4f, 5d, 6p

Explanation:
1. The 6th period has 32 elements built from 6s, 4f, 5d and 6p.

Teacher's Note:
1. The order follows the (n + l) rule.

Question 14: The IUPAC name of the element after Db (Z = 105) is
(a) Unnilpentium
(b) Unununium
(c) Unnilhexium
(d) Unnilquadium
Show Answer & Explanation

Answer: (c) Unnilhexium

Explanation:
1. Z = 106: un-nil-hex-ium = Unnilhexium (seaborgium).

Teacher's Note:
1. Numerical roots: nil 0, un 1, bi 2, tri 3, quad 4, pent 5, hex 6, sept 7.

Question 15: The element with electronic configuration ns2 (n-1)s2p6d0 (n-2)s2p6d10 f7 belongs to the
(a) s-block
(b) p-block
(c) d-block
(d) f-block
Show Answer & Explanation

Answer: (d) f-block

Explanation:
1. The last electrons enter the (n-2)f orbitals, so it is an f-block element.

Teacher's Note:
1. The block is determined by the subshell receiving the last electron.

Question 16: An atom with valence shell electronic configuration 4s2 4p2 is in
(a) Group IIA, period 3
(b) Group IIB, period 4
(c) Group IVA, period 4
(d) Group IVA, period 3
Show Answer & Explanation

Answer: (c) Group IVA, period 4

Explanation:
1. The valence shell is n = 4, so period 4; 4 valence electrons (s2p2) place it in group 14 (IVA).

Teacher's Note:
1. This is germanium.

Question 17: The electronic configuration of d-block elements is generally
(a) ns(1-2) (n-1)d(1-10)
(b) ns2 (n-1)d10
(c) (n-1)d10 ns2
(d) ns2 np4
Show Answer & Explanation

Answer: (a) ns(1-2) (n-1)d(1-10)

Explanation:
1. The last electron enters the (n-1)d subshell and the outer shell has 1 or 2 s-electrons.

Teacher's Note:
1. Pd (4d10 5s0) is a rare exception.

Question 18: An element has the configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p3. It belongs to which group?
(a) 3rd
(b) 5th
(c) 15th
(d) 17th
Show Answer & Explanation

Answer: (c) 15th

Explanation:
1. The valence shell is 5s2 5p3: 5 valence electrons, p-block, group 15.

Teacher's Note:
1. It is antimony (Z = 51).

Question 19: An element has configuration 4d3 5s2. It belongs to group
(a) IIA
(b) IIB
(c) VB
(d) IIIB
Show Answer & Explanation

Answer: (c) VB

Explanation:
1. Z = 41 (Nb): 5 valence electrons (d3 s2) in a d-block element gives group 5 (VB).

Teacher's Note:
1. For d-block, group number = number of (n-1)d + ns electrons.

Question 20: Which of the following configurations belongs to inert gas elements?
(a) ns2 (n-1)d10
(b) ns2 (n-1)s2 p6
(c) ns2 np6
(d) None
Show Answer & Explanation

Answer: (c) ns2 np6

Explanation:
1. Noble gases have complete valence shells ns2np6 (He is 1s2).

Teacher's Note:
1. Ne, Ar, Kr, Xe, Rn all have ns2np6.

Question 21: The atom X has configuration 1s2 2s2 2p6 3s2 3p5 and Z has 1s2 2s2 2p5. What is the position of X relative to Z?
(a) Just below Z
(b) Just above Z
(c) Left to Z
(d) Right to Z
Show Answer & Explanation

Answer: (a) Just below Z

Explanation:
1. X is Cl and Z is F; both are in group 17 and Cl is in the period below F.

Teacher's Note:
1. Same group, next period down.

Question 22: The total number of transuranic elements in the periodic table, with uranium (Z = 92) the last natural element and Z = 112 the last known, is
(a) 21
(b) 20
(c) 11
(d) 12
Show Answer & Explanation

Answer: (b) 20

Explanation:
1. Elements from Z = 93 to 112: 112 - 92 = 20.

Teacher's Note:
1. Transuranic elements have Z greater than 92.

Question 23: The atomic number of the element with the maximum number of unpaired electrons in the 4p subshell is
(a) 33
(b) 17
(c) 53
(d) 15
Show Answer & Explanation

Answer: (a) 33

Explanation:
1. 4p3 has three unpaired electrons: Z = 33 (As).

Teacher's Note:
1. 4p can hold at most 3 unpaired electrons.

Question 24: The general configuration of the outermost and penultimate shell is (n-1)s2 (n-1)p6 (n-1)dx ns2. If n = 4 and x = 5, the number of protons in the nucleus is
(a) more than 25
(b) less than 24
(c) 25
(d) 30
Show Answer & Explanation

Answer: (c) 25

Explanation:
1. Configuration: 1s2 2s2 2p6 3s2 3p6 3d5 4s2 = 25 electrons.
2. So Z = 25 (Mn).

Teacher's Note:
1. Neutral atom: protons = electrons.

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