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Valence, oxidation states and chemical reactivity trends: NEET MCQs with answers

50 questions on this topic of Classification of Elements and Periodicity in Properties (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following elements shows an anomalous valency and cannot expand its octet beyond 4 due to the total absence of d-orbitals in its valence shell?

  1. Silicon
  2. Sulfur
  3. Nitrogen
  4. Phosphorus
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Answer: (C) Nitrogen

Nitrogen belongs to the second period and has only 2s and 2p orbitals available in its valence shell (total 4 orbitals). Therefore, it cannot expand its valence shell beyond four electron pairs, unlike third-period elements like P, S, and Si which have vacant 3d-orbitals.

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2. Consider the reactivity of halogens with water. Which halogen spontaneously oxidizes water to liberate dioxygen gas?

  1. Iodine
  2. Chlorine
  3. Bromine
  4. Fluorine
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Answer: (D) Fluorine

Fluorine is extremely electronegative with a very high reduction potential (E° = +2.87 V), enabling it to spontaneously oxidize water to O₂ (2F₂ + 2H₂O → 4HF + O₂). Chlorine and bromine disproportionate in water, while iodine cannot oxidize water.

hard

3. Given below are two statements: Statement I: In group 14, PbO₂ is a powerful oxidizing agent, whereas SnO₂ is not. Statement II: Due to the inert pair effect, the +2 oxidation state is more stable than the +4 oxidation state for lead, whereas for tin, the +4 state is more stable than the +2 state. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Because of the strong inert pair effect in lead, Pb(IV) is unstable and readily accepts two electrons to form the much more stable Pb(II), making PbO₂ a strong oxidizing agent. For tin, the inert pair effect is weaker, and Sn(IV) is more stable than Sn(II), so SnO₂ does not act as an oxidizing agent. Both statements are correct.

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4. Consider the elements with atomic numbers Z = 11 to 17 (Period 3). What is the formula and nature of the compound formed between the element with Z = 12 and the element with Z = 17?

  1. MgCl₂, ionic
  2. MgCl, covalent
  3. Mg₂Cl, ionic
  4. MgCl₂, covalent
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Answer: (A) MgCl₂, ionic

Z = 12 corresponds to magnesium (Group 2, valency = 2) and Z = 17 corresponds to chlorine (Group 17, valency = 1). The compound formed is MgCl₂. Because Mg has low ionization enthalpy and Cl has high electron gain enthalpy, the bond formed between them is predominantly ionic.

easy

5. What is the maximum valence of an element with atomic number 15 towards oxygen, and what is its valence towards hydrogen?

  1. 3 and 5
  2. 5 and 3
  3. 5 and 5
  4. 3 and 3
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Answer: (B) 5 and 3

The element with atomic number 15 is phosphorus (configuration: [Ne] 3s² 3p³). The valence towards oxygen corresponds to the total number of valence electrons, which is 5 (as in P₂O₅), whereas towards hydrogen it is given by (8 - valence electrons) = 8 - 5 = 3 (as in PH₃).

hard

6. Match List-I (Species) with List-II (Oxidation state of the central atom) and select the correct option: List-I: (A) CrO₅ (B) H₂SO₃ (C) ClO₃⁻ (D) XeF₄ List-II: (I) +4 (II) +5 (III) +6 (IV) +4

  1. (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  2. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  3. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  4. (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
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Answer: (A) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)

In CrO₅ (butterfly structure), there are two peroxo groups and one oxo group: Cr + 4(-1) + 1(-2) = 0 ⇒ Cr = +6. In H₂SO₃, 2(+1) + S + 3(-2) = 0 ⇒ S = +4. In ClO₃⁻, Cl + 3(-2) = -1 ⇒ Cl = +5. In XeF₄, Xe + 4(-1) = 0 ⇒ Xe = +4. Thus, (A)-(III), (B)-(I), (C)-(II), (D)-(IV).

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7. Given below are two statements: Statement I: The covalency of nitrogen in N₂O₅ is restricted to four. Statement II: Nitrogen cannot expand its covalency beyond four due to the non-availability of d-orbitals in its valence shell. In light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
  2. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct
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Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

In N₂O₅, each nitrogen atom forms four covalent bonds (covalency = 4) even though its oxidation state is +5. Second period elements like nitrogen possess only four valence orbitals (one 2s and three 2p) and lack valence d-orbitals, restricting their maximum covalency to four.

hard

8. Two oxides of nitrogen, NO₂ and N₂O, contain nitrogen in different formal oxidation states. What is the difference in the oxidation states of nitrogen between these two oxides?

  1. +4
  2. +2
  3. +1
  4. +3
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Answer: (D) +3

In NO₂, let the oxidation state of N be x: x + 2(-2) = 0 ⇒ x = +4. In N₂O (nitrous oxide), let the oxidation state of N be y: 2y + (-2) = 0 ⇒ y = +1. The difference between the oxidation states is (+4) - (+1) = +3.

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9. In which of the following pairs of compounds does the transition metal display the same oxidation state?

  1. [Cr(H₂O)₆]Cl₃ and K₂CrO₄
  2. KMnO₄ and K₂MnO₄
  3. K₂Cr₂O₇ and CrO₂Cl₂
  4. FeSO₄ and Fe₂(SO₄)₃
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Answer: (C) K₂Cr₂O₇ and CrO₂Cl₂

In K₂Cr₂O₇, let Cr be x: 2(+1) + 2x + 7(-2) = 0 ⇒ 2x = +12 ⇒ x = +6. In CrO₂Cl₂, let Cr be y: y + 2(-2) + 2(-1) = 0 ⇒ y = +6. Both compounds contain chromium in the +6 oxidation state.

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10. Within a period, chemical reactivity is generally highest for elements:

  1. In group 18
  2. At the two extremes (alkali metals on the left, halogens on the right)
  3. In the middle of the period
  4. In group 14 only
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Answer: (B) At the two extremes (alkali metals on the left, halogens on the right)

Alkali metals readily lose an electron and halogens readily gain one, so reactivity is highest at the extremes and lowest in the middle (noble gases are unreactive).

NCERT Class 11

easy

11. Which element of the third period forms a basic oxide, an amphoteric oxide, and an acidic oxide respectively when grouped with two other period 3 elements?

  1. Na (basic), Si (amphoteric), S (acidic)
  2. Mg (basic), Al (amphoteric), P (acidic)
  3. Al (basic), Mg (amphoteric), Cl (acidic)
  4. Mg (basic), P (amphoteric), S (acidic)
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Answer: (B) Mg (basic), Al (amphoteric), P (acidic)

In period 3, MgO is basic, Al₂O₃ is amphoteric, and oxides of phosphorus like P₄O₁₀ are acidic. Silica (SiO₂) is weakly acidic, not amphoteric.

easy

12. The maximum coordination number and valence of boron are restricted to 4, whereas aluminum in the same group can expand its coordination number up to 6 (for example, in [AlF₆]³⁻). What is the primary reason for this difference?

  1. Diagonal relationship between boron and silicon
  2. Higher first ionization enthalpy of boron compared to aluminum
  3. Larger electronegativity of aluminum than boron
  4. Availability of vacant 3d-orbitals in aluminum but not in boron
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Answer: (D) Availability of vacant 3d-orbitals in aluminum but not in boron

Boron belongs to the second period and has only 2s and 2p valence orbitals (a total of four orbitals), restricting its maximum covalency to 4. Aluminum belongs to the third period and has vacant 3d-orbitals available in its valence shell, allowing it to expand its coordination number up to 6.

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