3.37 6th edition

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Lina Petrossian 1D
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Joined: Fri Oct 05, 2018 12:16 am

3.37 6th edition

Postby Lina Petrossian 1D » Sun Nov 11, 2018 11:12 pm

3.37 The following Lewis structure was drawn for a Period 3 element. Identify the element.

I got phosphorus, did anyone else get this as an answer?

Mindy Kim 4C
Posts: 65
Joined: Fri Sep 28, 2018 12:25 am

Re: 3.37 6th edition

Postby Mindy Kim 4C » Sun Nov 11, 2018 11:14 pm

Yes, I got phosphorus as well. The total number of valence electrons counted from the Lewis structure is 32 electrons. Subtracting 21 electrons for 3 chlorine atoms and 6 electrons for the oxygen, you obtain 5 valence electrons. The element in period 3 with 5 valence electrons is phosphorus.

Matthew Choi 2H
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Joined: Fri Sep 28, 2018 12:18 am

Re: 3.37 6th edition

Postby Matthew Choi 2H » Tue Nov 13, 2018 12:45 am

For these types of questions, first find out how many valence electrons the unknown element adds to the total amount of electrons in the molecule. Take the total amount of electrons in the molecule and subtract the number of valence electrons of the known elements. For this specific problem, you end up with the unknown element having 5 valence electrons, and phosphorous is the one that fits the description.

Vanadium Wang 4H
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Joined: Fri Sep 28, 2018 12:19 am

Re: 3.37 6th edition

Postby Vanadium Wang 4H » Wed Nov 14, 2018 12:59 am

The solution manual identifies the element as phosphorus (P).

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