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MCQEasyJEE 2025Covalent Bonding & Lewis Structures

JEE Chemistry 2025 Question with Solution

Total number of non-bonded electrons present in NO2−\text{NO}_2^- ion based on Lewis theory is:

  • A

    1010

  • B

    1212

  • C

    88

  • D

    66

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The ion is NO2−\text{NO}_2^-.

Find: The total number of non-bonded electrons present in its Lewis structure.

First count the total valence electrons:

5+2×6+1=185 + 2 \times 6 + 1 = 18

Here, nitrogen contributes 55 valence electrons, two oxygen atoms contribute 1212 valence electrons, and the negative charge contributes 11 extra electron.

In the Lewis structure of NO2−\text{NO}_2^-, resonance is represented as:

O=N−O−↔O−−N=O\text{O} = \text{N} - \text{O}^- \leftrightarrow \text{O}^- - \text{N} = \text{O}

Now count the non-bonded electrons:

  • Double-bonded oxygen has 22 lone pairs =4= 4 electrons.
  • Single-bonded oxygen has 33 lone pairs =6= 6 electrons.
  • Nitrogen has 11 lone pair =2= 2 electrons.

So,

4+6+2=124 + 6 + 2 = 12

Therefore, the total number of non-bonded electrons is 1212. Hence, the correct option is B.

The solution states the correct species as NO2−\text{NO}_2^-, which resolves the formatting ambiguity in the question text.

Electron Counting by Lewis Structure

Given: NO2−\text{NO}_2^- ion.

Find: Number of non-bonded electrons.

Step 1: Determine total valence electrons.

Total valence electrons=5+(2×6)+1=18\begin{aligned} \text{Total valence electrons} &= 5 + (2 \times 6) + 1 \\ &= 18 \end{aligned}

Step 2: Construct the Lewis structure with nitrogen as the central atom. One oxygen is double bonded and the other is single bonded, with resonance between the two forms.

Step 3: Count lone-pair electrons.

Double-bonded O=4Single-bonded O=6N atom=2\begin{aligned} \text{Double-bonded O} &= 4 \\ \text{Single-bonded O} &= 6 \\ \text{N atom} &= 2 \end{aligned}

Step 4: Add them.

12=4+6+212 = 4 + 6 + 2

Therefore, the total number of non-bonded electrons present in NO2−\text{NO}_2^- is 1212.

Common mistakes

  • Counting total valence electrons without adding the extra electron for the negative charge. This is wrong because NO2−\text{NO}_2^- has one more electron than neutral NO2\text{NO}_2. Always include the charge while calculating total valence electrons.

  • Ignoring resonance and drawing an incorrect Lewis structure. This leads to wrong lone-pair counting. Use the correct nitrite ion structure with one single bond, one double bond, and resonance between equivalent forms.

  • Counting bonds instead of non-bonded electrons. Bonding electrons are shared in covalent bonds and must not be included here. Only lone-pair electrons on nitrogen and oxygen should be counted.

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