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Multiple Choice
Which of the following best describes the Lewis dot structure of a neutral H2O molecule, including nonbonding electron pairs on the oxygen atom?
A
Oxygen is bonded to two hydrogen atoms with single bonds and has one pair of nonbonding electrons.
B
Oxygen is bonded to two hydrogen atoms with double bonds and has no nonbonding electrons.
C
Oxygen is bonded to two hydrogen atoms with single bonds and has two pairs of nonbonding electrons.
D
Oxygen is bonded to one hydrogen atom with a single bond and has three pairs of nonbonding electrons.
Verified step by step guidance
1
Recall that a Lewis dot structure represents the valence electrons of atoms in a molecule, showing bonding pairs (shared electrons) and nonbonding pairs (lone pairs).
Determine the total number of valence electrons in the H2O molecule: Oxygen has 6 valence electrons, and each Hydrogen has 1 valence electron, so total electrons = 6 + 2(1) = 8 electrons.
Draw the skeleton structure with Oxygen as the central atom bonded to two Hydrogen atoms with single bonds. Each single bond represents 2 shared electrons, so 2 bonds use 4 electrons.
Distribute the remaining electrons as lone pairs on the Oxygen atom. Since 4 electrons are used in bonds, 8 - 4 = 4 electrons remain, which form two lone pairs (each pair has 2 electrons).
Verify that each atom satisfies the octet (or duet for Hydrogen) rule: Oxygen has 2 bonding pairs + 2 lone pairs = 8 electrons, and each Hydrogen has 2 electrons from the single bond.