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Class XI · Chemistry

Classification of Elements and Periodicity in Properties: NEET MCQ practice

477 practice questions on Classification of Elements and Periodicity in Properties, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

477

Practice questions

316

Easy

159

Medium

2

Hard

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What Classification of Elements and Periodicity in Properties covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Modern periodic law and present form of periodic table
  • s, p, d and f block elements
  • Periodic trends- atomic/ionic radii, ionization enthalpy, electron gain enthalpy, valence, oxidation states, chemical reactivity

Sample questions on Classification of Elements and Periodicity in Properties

Try each one, then open the answer and explanation.

Question 1 · Practice question

Which of the following represents the decreasing order of van der Waals’ forces in halogens?

  1. A.F₂ > Cl₂ > Br₂ > I₂
  2. B.I₂ > Br₂ > Cl₂ > F₂
  3. C.Br₂ > Cl₂ > F₂ > I₂
  4. D.Cl₂ > F₂ > I₂ > Br₂
Show answer

Answer: B. I₂ > Br₂ > Cl₂ > F₂

Van der Waals forces increase with molecular size down the halogen group because larger molecules have more polarizable electron clouds, leading to stronger dispersion forces. I₂ is largest and F₂ is smallest, so the correct decreasing order is I₂ > Br₂ > Cl₂ > F₂. The tempting wrong option (a) reverses the trend, mistakenly suggesting fluorine has stronger forces than iodine.

Question 2 · Practice question

The correct order of in creasing order of radiiions Br -, F -, O 2- and S 2- is as follow

  1. A.Br -, -, 2- 2-
  2. B.S 2- 2- - -
  3. C.F - 2- 2- -
  4. D.F - - 2- 2-
Show answer

Answer: C. F - 2- 2- -

The order of increasing ionic radii is F⁻ < O²⁻ < S²⁻ < Br⁻. This depends on two periodic trends: For isoelectronic anions (F⁻ and O²⁻ both have 10 electrons), the ion with lower nuclear charge (O²⁻) is larger. Then, moving down a group (O²⁻ to S²⁻) and to a higher period (S²⁻ to Br⁻), size increases due to extra shells. Option C matches this, while others either misplace Br⁻ or reverse the isoelectronic pair order.

Question 3 · Practice question

Direction (Q. Nos. 16 and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d). Passage Group 17 elements are known as halogens ( sea salt forming). Their general configuration is ns 2 np 5 where, n = 2 to 7. Fluorine is the first member of the group differs in several ways from the rest of the group. Halogens are highly reactive elements having strong affinity for hydrogen. All the halogens form both ionic and covalent compounds .Fluorine is always monovalent and shows -1 oxidation state in its compounds .The other halogens exhibit 1, 3, 5, 7 valencies in their compounds and show - 1 to + 7 oxidation states. Q. Valence shell configuration that belongs to most reactive non-metal is

  1. A.(n-1)s 2 p 6 ns 1
  2. B.ns 2 np 5
  3. C.ns 2 np 6
  4. D.(n-1)s 2 p 6 ns 2
Show answer

Answer: B. ns 2 np 5

The passage states that halogens (Group 17 elements) have the valence shell configuration ns²np⁵ and are highly reactive non-metals. Option B matches this configuration exactly. The other options represent different groups: A is alkali metals (reactive but metals), C is noble gases (non-reactive), and D is alkaline earth metals (metals).

Question 4 · Practice question

Nitrogen is relatively inactive element because:

  1. A.Its atoms has a stable electronic configuration
  2. B.It has low atomic radius
  3. C.Its electronegativity is fairly high
  4. D.Dissociation energy of its molecule is fairly high
Show answer

Answer: D. Dissociation energy of its molecule is fairly high

Nitrogen is inactive because the N≡N triple bond is very strong, requiring high energy to break, making N2 molecule stable. Options (a) and (c) are incorrect because nitrogen's electronic configuration is not fully stable, and high electronegativity usually increases reactivity, not inactivity. Option (b) is not the primary reason for nitrogen's inertness.

Question 5 · Practice question

The total number of positive oxidation states shown by fluorine is

  1. A.5
  2. B.2
  3. C.4
  4. D.1
Show answer

Answer: D. 1

Fluorine is the most electronegative element and in all compounds it shows only –1 oxidation state, and 0 in the elemental form. Unlike other halogens, fluorine exhibits no positive oxidation states. While '2' refers to its total oxidation states (0 and –1), the question likely intends this given the distractors (such as 4 or 5 for other halogens) test awareness of fluorine's unique behavior.

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