2A.17 7th edition

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Vikramjeet Gill 1C
Posts: 30
Joined: Fri Sep 28, 2018 12:15 am

2A.17 7th edition

Postby Vikramjeet Gill 1C » Wed Oct 31, 2018 12:16 pm

Predict the number of valence electrons present for each of the following ions a)Mn^4+, c)Co^3+, d)P^3+

can someone explain the steps to solve this problem.

Andre_Galenchik_2L
Posts: 69
Joined: Fri Sep 28, 2018 12:23 am

Re: 2A.17 7th edition

Postby Andre_Galenchik_2L » Wed Oct 31, 2018 1:28 pm

I am not exactly sure about a and c, but for d, a +3 charge means the element is losing 3 electrons. Knowing this, phosphorus if it lost 3 electrons would have the configuration of magnesium which has 2 valence electrons.

aisteles1G
Posts: 117
Joined: Fri Sep 28, 2018 12:15 am

Re: 2A.17 7th edition

Postby aisteles1G » Wed Oct 31, 2018 2:19 pm

First look at the electron configuration, for Mn+4 you see Mn has in its outer most shell 3d^5 4s^2 meaning at this moment it has 2 in its outer most shell (4s). then because of the +4 charge, subtract 4 electrons starting from the outer most shell, in this case the 4s one. You will end up with 3d^3 and now 3d is your valence/outer most shell and contains 3 electrons. Same with part C, for Co you have 3d^7 4s^2 and need to take 3 away. Start in the outer most shell (4s) and subtract 3, you should end up with 3d^6 so now 6 valence electrons. This is more difficult if you dont have the periodic table with electron configurations written out bc then you have to write them yourself, but just start out in the outermost shell and subtract the charge. P^3+ : P is 3s^2 3p^3 so start at 3p^3 and take away 3 you end up with 3s^2 meaning 2 valence electrons. Hope this helps!


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