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## 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.

## 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).