Chemistry · Classification of elements and periodicity · chapter 3 of 9

Class 11 Chemistry Chapter 3: Classification of Elements and Periodicity in Properties MCQs with answers

10 multiple-choice questions on Classification of Elements and Periodicity in Properties, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
Sulphur (atomic weight 32), selenium and tellurium (atomic weight 127.6) behave as a Dobereiner triad. The atomic weight the Law of Triads predicts for selenium is
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Answer: (b) 79.8

In a triad the middle element's atomic weight is about half way between the other two: (32 + 127.6)/2 = 79.8. 95.6 is the difference, 47.8 half of it, 159.6 the sum.

Question 2
Newlands' Law of Octaves was found to hold only for the lighter elements. The element beyond which it failed is
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Answer: (c) calcium

Newlands' arrangement, in which every eighth element resembles the first, seemed to be true only for elements up to calcium.

Question 3
In Mendeleev's table tellurium (atomic weight 127.6) stands before iodine (atomic weight 126.9), against the order of atomic weights. In the modern periodic table this order is
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Answer: (a) kept, because tellurium (Z = 52) has a lower atomic number than iodine (Z = 53)

The modern table is arranged by atomic number. Te has Z = 52 and I has Z = 53, so Te comes first, and the inversion Mendeleev made on the basis of properties is no longer an exception. Te stays in group 16 and I in group 17.

Question 4
Mendeleev predicted that eka-aluminium would form an oxide E₂O₃ long before gallium was discovered. The basis of this prediction was that
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Answer: (d) the gap lay in aluminium's group, and elements of a group form compounds of similar empirical formula, such as Al₂O₃

Mendeleev relied on the similarities in the empirical formulas of the compounds of elements in a group. The gap under aluminium therefore had to form E₂O₃ like Al₂O₃. Gallium itself was unknown, so nothing about it could have been measured.

Question 5
Moseley plotted the square root of the frequency of the characteristic X-rays of elements against another property and obtained a straight line. That property is
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Answer: (c) atomic number

The plot of √ν against atomic number (Z) is a straight line; against atomic mass it is not. This showed atomic number to be the more fundamental property.

Question 6
The total number of elements in the first four periods of the long form of the periodic table is
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Answer: (a) 36

The first four periods hold 2, 8, 8 and 18 elements: 2 + 8 + 8 + 18 = 36. 54 includes the fifth period too.

Question 7
An element X has atomic number 38. The element whose chemical properties would be most like those of X has the atomic number
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Answer: (b) 20

Z = 38 is [Kr]5s²: period 5, group 2. Z = 20 is [Ar]4s², also group 2, with the same ns² valence configuration and so similar properties. 37 and 39 are its neighbours in the period (groups 1 and 3), and 36 is a noble gas.

Question 8
Helium has the configuration 1s², yet it is placed in group 18 and not with the s-block elements because
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Answer: (d) its valence shell is completely filled, so it behaves like the other noble gases

Strictly helium belongs to the s-block, but its completely filled valence shell (1s²) gives it the properties of the other noble gases, which justifies its place in the p-block with group 18.

Question 9
Which of the following elements is a metalloid?
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Answer: (a) germanium

Germanium borders the zig-zag line and shows properties of both metals and non-metals. Gallium and tin are metals; selenium is a non-metal.

Question 10
Which of the following pairs is isoelectronic?
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Answer: (c) N³⁻ and Ne

N³⁻ has 7 + 3 = 10 electrons and Ne has 10. Na⁺ (10) and Mg (12), O²⁻ (10) and Cl⁻ (18), K⁺ (18) and Ne (10) differ.

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