Chapter 7
Q. 7.1
Draw Lewis structures of the hypochlorite ion, OCl^{-} (Figure 7.4), and the ethane, C_{2} H_{6}, molecule.
STRATEGY
1. Follow the steps outlined in Figure 7.3.
2. For ethane, hydrogen must be a terminal atom since it cannot form double bonds. Carbon ordinarily forms four bonds.


Step-by-Step
Verified Solution
OCl^{-} skeleton [ OCl]^{-}
VE 6 (for O) + 7 (for Cl) + 1(-1 charge) = 14
AE AE = VE – 2(bonds) = 14 – 2(1 bond) = 12
NE 6 (for O to have an octet) + 6 (for Cl to have an octet) = 12
AE = NE ? Yes; distribute electrons.
Lewis structure \left[:\overset{..}{\underset{..}{O} } -\underset{..}{\overset{..}{O}: } \right] ^{-}
C_{2} H_{6} skeleton \begin{matrix}\underset{|}{H} \ \ \ \ \underset{|}{H}\\ H-C-C-H\\ \overset{|}{H} \ \ \ \ \overset{|}{H} \end{matrix}
VE 2 × 4 (for C) + 6 × 1 (for H) = 14
AE AE = VE – 2(bonds) = 14 – 2(7 bonds) = 0
NE 0 : All the H atoms have duets and both C atoms have octets.
AE = NE ? Yes; distribute electrons.
Lewis structur \begin{matrix}\underset{|}{H} \ \ \ \ \underset{|}{H}\\ H-C-C-H\\ \overset{|}{H} \ \ \ \ \overset{|}{H} \end{matrix}
END POINT
After you have written the Lewis structure, it is a good idea to add the number of unshared electron pairs and bonding electrons. This sum must equal the number of valence electrons (VE).