2B. 3 part d

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Aashka Popat 1A
Posts: 50
Joined: Sat Sep 07, 2019 12:16 am

2B. 3 part d

Postby Aashka Popat 1A » Sat Nov 02, 2019 6:38 pm

I'm having trouble with part d of this question, which asks us to draw the lewis structure of BrF3. I don't understand how to arrange the electrons to give each atom an octet?

Samuel Tzeng 1B
Posts: 103
Joined: Sat Aug 24, 2019 12:15 am

Re: 2B. 3 part d

Postby Samuel Tzeng 1B » Sat Nov 02, 2019 6:48 pm

Br has three bonds (one with each F) and two pairs of lone electrons, while the F's have 3 pairs of lone electrons

Osvaldo SanchezF -1H
Posts: 122
Joined: Wed Sep 18, 2019 12:21 am

Re: 2B. 3 part d

Postby Osvaldo SanchezF -1H » Sat Nov 02, 2019 7:55 pm

The best way to think about the problems such as this is to first how many electrons there are in the equations(valence electrons). Then put the element with the least electronegativity in the center and everything else in the outside. give everything an octet just by standard rule. After make sure that the number of electrons you added matches up with how much there is supposed to be. After check the charge on each element to make sure that there is the least amount of charge possible and if not, manipulate it so that you fixed it. So when doing BF3, if you follow these steps it should be easier to visualize. Hope this helps!

chari_maya 3B
Posts: 108
Joined: Sat Sep 07, 2019 12:18 am

Re: 2B. 3 part d

Postby chari_maya 3B » Sat Nov 02, 2019 10:03 pm

Why can Br have 10 electrons, instead of 8?

Ixcel Guzman 4A
Posts: 16
Joined: Wed Sep 25, 2019 12:17 am

Re: 2B. 3 part d

Postby Ixcel Guzman 4A » Sat Nov 02, 2019 10:29 pm

to give it an octect, the Br is surrounded by 3 F with single bonds. F will have 6 electrons around them and Br will have 2 paired electrons. thats how they all form an octect.

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