Total number of non-bonded electrons present | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The total number of valence electrons in $NO_2^{-}$ is calculated as follows:Valence electrons of $N = 5$Valence electrons of $O = 6 	imes 2 = 12

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The total number of valence electrons in $NO_2^{-}$ is calculated as follows:Valence electrons of $N = 5$Valence electrons of $O = 6 	imes 2 = 12$Negative charge = $1$Total valence electrons = $5 + 12 + 1 = 18$.In the $NO_2^{-}$ structure,the nitrogen atom is bonded to two oxygen atoms with one single bond and one double bond $(O=N-O^{-})$.Each double bond involves $4$ electrons (bonding) and each single bond involves $2$ electrons (bonding).Total bonding electrons = $4 + 2 = 6$.Total non-bonded (lone pair) electrons = Total valence electrons - Total bonding electrons = $18 - 6 = 18 - 6 = 12$.Thus,the total number of non-bonded electrons is $12$.

Total number of non-bonded electrons present in $NO_2^{-}$ ion based on Lewis theory is $........$

Similar Questions

What is true about $S_8$ molecule?
$I.$ There are no $p_{\pi}-p_{\pi}$ bonds present in the molecule.
$II.$ There are eight lone pairs of electrons.
$III.$ Each $S$ atom is $sp^3$ hybridized.

Which of the following is the correct order of increasing $O-O$ bond length in the given compounds?

Which of the following contains both ionic and covalent bonds?

The Lewis structure for $O_3$ molecule is given below. The correct formal charges on oxygen atoms labelled $1$,$2$,$3$ are respectively.

Assertion : Bond angle of $H_2S$ is smaller than $H_2O$.
Reason : Electronegativity of the central atom increases,bond angle decreases.

Vedclass Test Series

Exam Paper Generator

Online Exam Module