Multiple Choice
In the Lewis dot structure of BCl_3, how many lone pairs are present on the central boron atom?
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11. Bonding & Molecular Structure
Lewis Dot Structures: Exceptions
Multiple Choice
Which of the following is the correct Lewis dot structure for XeF_4?
A
Xe atom in the center with four F atoms bonded to it, three lone pairs on Xe, and each F atom with one lone pair.
B
Xe atom in the center with four F atoms bonded to it, two lone pairs on Xe, and each F atom with three lone pairs.
C
Xe atom in the center with four F atoms bonded to it, no lone pairs on Xe, and each F atom with three lone pairs.
D
Xe atom in the center with four F atoms bonded to it, one lone pair on Xe, and each F atom with two lone pairs.
Verified step by step guidance1
Step 1: Determine the total number of valence electrons available for XeF_4. Xenon (Xe) is in group 18 and has 8 valence electrons, and each fluorine (F) atom is in group 17 with 7 valence electrons. Since there are 4 fluorine atoms, calculate the total valence electrons as: \(8 + 4 \times 7\).
Step 2: Draw the skeletal structure with the xenon atom in the center and four fluorine atoms bonded to it with single bonds. Each single bond represents 2 electrons.
Step 3: Subtract the electrons used in the bonds from the total valence electrons to find the remaining electrons to be placed as lone pairs.
Step 4: Distribute the remaining electrons as lone pairs first to the fluorine atoms to complete their octets (each fluorine needs 8 electrons total, including bonding electrons). Then place any leftover electrons as lone pairs on the xenon atom.
Step 5: Check the formal charges and the octet rule for each atom. Xenon can have an expanded octet because it is in period 5. The correct Lewis structure for XeF_4 has xenon with two lone pairs and four bonding pairs to fluorine atoms, and each fluorine atom has three lone pairs.
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