Question

draw (on paper) a lewis structure for nh₄⁺ and answer the following questions based on your drawing. do not draw double bonds unless they are needed for the central atom to obey the octet rule.

1. for the central nitrogen atom:

the number of lone pairs =
the number of single bonds =
the number of double bonds =

1. the central nitrogen atom

a. obeys the octet rule.
b. has an incomplete octet.
c. has an expanded octet.

Explanation:

Sub - question 1

Step 1: Determine valence electrons of atoms

Nitrogen (\(N\)) has 5 valence electrons, and each hydrogen (\(H\)) has 1 valence electron. The ion is \(NH_4^+\), so we subtract 1 electron for the positive charge. Total valence electrons \(= 5+4\times1 - 1=8\).

Step 2: Draw Lewis structure and analyze bonds/lone pairs

In the Lewis structure of \(NH_4^+\), the central \(N\) atom is bonded to 4 \(H\) atoms with single bonds. Each single bond uses 2 electrons. The total number of electrons used in bonding is \(4\times2 = 8\), which matches the total valence electrons. So, the number of lone pairs on \(N\) is 0 (since all 8 electrons are in bonding), the number of single bonds is 4, and the number of double bonds is 0.

The octet rule states that atoms tend to have 8 valence electrons. The central \(N\) in \(NH_4^+\) has 4 single bonds (8 electrons in bonding), so it has a complete octet and obeys the octet rule. Option A is correct as it obeys the octet rule, option B is wrong (it has a complete octet), and option C is wrong (it does not have an expanded octet as it has 8 electrons around it).

Answer:

• The number of lone pairs \(=\boldsymbol{0}\)
• The number of single bonds \(=\boldsymbol{4}\)
• The number of double bonds \(=\boldsymbol{0}\)

Sub - question 2