Multiple Choice
How many lone pairs of electrons are present on the iodine atom in the Lewis dot structure of IF_4^+?
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11. Bonding & Molecular Structure
Lewis Dot Structures: Ions
Multiple Choice
How many lone electron pairs are present in the Lewis dot structure of the NH_4^+ ion?
A
4
B
0
C
1
D
2
Verified step by step guidance1
Step 1: Identify the total number of valence electrons available for the NH_4^+ ion. Nitrogen (N) has 5 valence electrons, each hydrogen (H) has 1 valence electron, and since the ion has a +1 charge, subtract one electron from the total count.
Step 2: Calculate the total valence electrons: Total electrons = 5 (from N) + 4 × 1 (from H) - 1 (due to positive charge).
Step 3: Draw the skeletal structure with nitrogen as the central atom bonded to four hydrogen atoms. Each N-H bond represents a pair of shared electrons.
Step 4: Distribute the remaining electrons (if any) as lone pairs on the nitrogen atom to satisfy the octet rule. Remember that hydrogen atoms only need 2 electrons to be stable.
Step 5: Count the number of lone electron pairs on the nitrogen atom after bonding. Since all valence electrons are used in bonding, there should be zero lone pairs on nitrogen in NH_4^+.
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