Question

1. choose the best answer.

which of these represents a correctly drawn lewis structure for nitrate no₃¹⁻?

Explanation:

Step1: Calculate the total number of valence electrons

Nitrogen has 5 valence electrons, each oxygen has 6 valence electrons, and there is an extra electron due to the -1 charge. So the total number of valence electrons is \(5 + 3\times6+ 1=24\).

Step2: Analyze the first structure

The first structure has too few electrons around the oxygen atoms (only 2 lone - pair electrons on one oxygen, and double - bonded oxygen has no lone pairs shown properly).

Step3: Analyze the second structure

The second structure has the correct distribution of electrons. Each single - bonded oxygen has 6 lone - pair electrons (3 lone pairs) and the double - bonded oxygen has 4 lone - pair electrons (2 lone pairs). The total number of electrons can be counted as follows:

• For the single - bonded oxygens: Each single - bonded \(O\) has 3 lone pairs (\(6\) electrons) and 1 bond (counted as \(2\) electrons for the bond).
• For the double - bonded \(O\): It has 2 lone pairs (\(4\) electrons) and 2 bonds (counted as \(4\) electrons for the bonds).
• For \(N\): It has 3 bonds (counted as \(6\) electrons).

The sum of electrons: \(2\times(6 + 2)+(4 + 4)+6=24\)

Step4: Analyze the third structure

The third structure has no lone pairs on any of the atoms, so it does not account for all 24 valence electrons.

Step5: Analyze the fourth structure

The fourth structure has an incorrect bonding pattern (only one double bond and not all electrons accounted for properly in terms of the total valence - electron count).

Answer:

The second structure (the one with lone pairs on all oxygen atoms and the correct total electron count) is the correct Lewis structure for \(NO_{3}^{-}\).