2B.19 (7th ed)

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Duby3L
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Joined: Fri Sep 28, 2018 12:26 am

2B.19 (7th ed)

Postby Duby3L » Fri Nov 02, 2018 8:41 pm

Draw the Lewis structure and determine the formal charge on each atom.

So far CN- how come we couldn't have done 5 bonds with 1 lone pair on nitrogen. The solution shows 3 bonds and 1 lone pair on C and 1 lone pair on nitrogen but I don't understand why they have that structure?

Samantha Hoegl Roy 2C
Posts: 81
Joined: Fri Sep 28, 2018 12:15 am

Re: 2B.19 (7th ed)

Postby Samantha Hoegl Roy 2C » Fri Nov 02, 2018 9:20 pm

For CN- you have 4e- + 5e- + 1e- = 10 e-

You can not violate the octate rule with these specific elements. If you give them five bonds, this goes against the octate rule. Thus, the triple bond with the two lone pairs in the only solution that works

Abby De La Merced 3F
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Re: 2B.19 (7th ed)

Postby Abby De La Merced 3F » Fri Nov 02, 2018 9:34 pm

I am kind of confused about this too. Does having two lone pairs instead of one make the structure more stable?


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