Total number of non bonded electrons present in NO2^- ion based on Lewis theory is ________.

Numerical Answer Type:
Enter a numerical value Answer: 12 to 12 +4 marks

Solution & Explanation

### Core Logic Let's compute the total valence electrons for the nitrite ion (NO_2^-): textValence electrons = 5text (from N) + 2 times 6text (from O) + 1text (negative charge) = 18text electrons In the valid Lewis structural representation: * The central nitrogen atom forms one single bond and one double bond with the terminal oxygens, consuming 2 + 4 = 6 bonding electrons. * Remaining non-bonded valence electrons = 18 - 6 = 12 electrons. ### Step 1: Account for Lone Pairs Distribution of non-bonded electrons across the individual atoms: * Central Nitrogen atom has 1 lone pair (2 electrons). * Single-bonded Oxygen atom has 3 lone pairs (6 electrons). * Double-bonded Oxygen atom has 2 lone pairs (4 electrons). textTotal non-bonded electrons = 2 + 6 + 4 = 12 ### Pattern Recognition Non-bonded electrons can always be obtained directly by subtracting total bonding electrons from total valence electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 7

Q85 jee_main_2024_31_jan_morning Hybridization
The number of species from the following in which the central atom uses sp^3 hybrid orbitals in its bonding is NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3
Numerical Answer. Answer: 4 to 4

Solution

### Core Logic Analyzing the hybridization of the central atom in each species: - NH_3: 3 bp + 1 lp = 4 electron domains rightarrow sp^3 - SO_2: 2 bp + 1 lp = 3 electron domains rightarrow sp^2 - SiO_2: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally rightarrow sp^3 - BeCl_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - CO_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - H_2O: 2 bp + 2 lp = 4 electron domains rightarrow sp^3 - CH_4: 4 bp + 0 lp = 4 electron domains rightarrow sp^3 - BF_3: 3 bp + 0 lp = 3 electron domains rightarrow sp^2 Total species with sp^3 hybridization: NH_3, SiO_2, H_2O, CH_4. Total count = 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

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