## 2C.9

Moderators: Chem_Mod, Chem_Admin

ayushibanerjee06
Posts: 177
Joined: Thu Jul 11, 2019 12:16 am
Been upvoted: 1 time

### 2C.9

For 2C.9, when I did the correct Lewis Structure, I did not get a formal charge of 0 for Xe. What am I doing wrong?

ayushibanerjee06
Posts: 177
Joined: Thu Jul 11, 2019 12:16 am
Been upvoted: 1 time

### Re: 2C.9

Sorry, I meant 2C.11 part c

Brooke Yasuda 2J
Posts: 102
Joined: Sat Jul 20, 2019 12:17 am

### Re: 2C.9

Remember that formal charge equation is #valance electrons - ((bonding electrons/2) + electrons in lone pairs)

Baoying Li 1B
Posts: 113
Joined: Sat Aug 17, 2019 12:18 am

### Re: 2C.9

You can count the valences of electron of the Lewis structure and check that if it's equal to the number of valence of electrons you have calculated. Since there are 42 valences of electron in XeOF2, Xe has one lone pair.

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: 2C.9

If your lewis structure is correct, there will be a double bond between Xe and O, single bonds between each F and the Xe, and a lone pair on the central atom, Xe.

Now you take the equation for formal charge = V - (L + 1/2*B)
Since Xe is in the very last group it has a value of V = 8.
There are 2 lone pairs (L=2) and 12 electrons in bonded pairs (1/2*12=6)

8 - (2+6) = 0.

Return to “Octet Exceptions”

### Who is online

Users browsing this forum: No registered users and 0 guests