## HW 3.37

Maya Khoury
Posts: 29
Joined: Fri Apr 06, 2018 11:03 am

### HW 3.37

The answer in the solutions manual says "Therefore, E must be a member of the nitrogen family and since it is a third period element, E must be phosphorous". Can someone explain this to me?

Odalys Cuevas 1C
Posts: 31
Joined: Fri Apr 06, 2018 11:02 am

### Re: HW 3.37

The way that I did this was that I counted the total number of electrons on the whole structure (32 electrons) then I saw which elements were given and then counted the amount of valence electron each element had and multiplied it by the number there was on the structure (3 * 7 for Cl and 1 * 6 for O). I then subtracted the total number of electrons by the electrons used by those two elements which gave me a total of 5 valence electrons. The problem said that the element was in the 3rd period, and from that I knew that the element was Phosphorus due to the 5 valence electrons the element must have in this structure.

Hope that makes sense

Joshua Yang 1H
Posts: 22
Joined: Mon Jan 08, 2018 8:19 am

### Re: HW 3.37

total number of e- = # of dots + 2 x number of lines = 22 + 10 = 32 e-

O = 6 valence e-
Cl3 = 7 valence e- x 3 = 21e-

therefore we have 27e- accounted for.

32 - 27 = 5e-

which element in period 3 has 5 valence electrons? = phosphorus