2A 1

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Ziyan Wang 3J
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Joined: Wed Sep 18, 2019 12:22 am

2A 1

Postby Ziyan Wang 3J » Thu Oct 31, 2019 12:56 pm

Give the number of valence electrons (including d electrons) for each of the following elements. (a) Sb (b) Si (c) Mn (d) B.

I wonder why the answer for (a) is 5 but that for (c) is 7.

Vanessa Chuang 4F
Posts: 51
Joined: Sat Aug 24, 2019 12:18 am

Re: 2A 1

Postby Vanessa Chuang 4F » Thu Oct 31, 2019 2:30 pm

For (a) Sb, the reason why it has 5 valence electrons is because it is in the Nitrogen group. It is 3 electrons from a noble gas (full shell), thus it already has 5 electrons.

For (c) Mn, I counted the number of columns from the left. Since Mn is in the 7th column from the left, it has 7 valence electrons.

LReedy_3C
Posts: 61
Joined: Fri Aug 30, 2019 12:17 am

Re: 2A 1

Postby LReedy_3C » Thu Oct 31, 2019 2:34 pm

It's because the d block is valence in c), but in a) it is full, so the energy is lower and it is not a valence shell. In a), only the s and p orbitals contain the valence electrons.

A Raab 1K
Posts: 56
Joined: Sat Aug 24, 2019 12:16 am

Re: 2A 1

Postby A Raab 1K » Thu Oct 31, 2019 3:42 pm

The electron configuration for Sb is [Kr] 4D10 5S2 5P3. If you count the electrons in the highest level (5) then you can see that there's 5 valence electrons.

Anthony Hatashita 4H
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Joined: Wed Sep 18, 2019 12:21 am

Re: 2A 1

Postby Anthony Hatashita 4H » Thu Oct 31, 2019 8:01 pm

Mn has a ground state of [Ar]3d^7, hence the 7 valence electrons while Sb is [Kr]4d^10 5s^2 5p^3 with 5 valence electrons in the higher energy state.

Victoria Otuya 4F
Posts: 50
Joined: Wed Sep 18, 2019 12:21 am

Re: 2A 1

Postby Victoria Otuya 4F » Thu Oct 31, 2019 9:52 pm

I learned that the row from Hydrogen to Francium has 1 valence electron. Then the second row from Beryllium to Radium has two valence electrons. Then skipping over to the row from Boron to Nihonium has 3 valence electrons and it goes on to the last row from Helium to Oganesson with a valence electron of 8.


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