Question

Draw electronic Configuration (a) Arsenic (by chlorina an state e) den e pected hybrid zat hole uran geometry and VE for the Central storm show below (by

Answer 1

To draw the configuration you need to know the electronic configuration first. It comes from the arrengement of the electrons according to the levels s, p, d, f. It goes like this

1 s

2s 2p

3s 3p 3d

4s 4p 4d 4 f

5s, 5p, 5d, 5f

There is no e, g, nor h.

To write the configuration you have to do it putting the electrons in each orbital (s, p, d) but filling them in diagonal way (from 3p to 4 s, from 3 d to 4p, etc)

You cannot go 1 s, 2s, 2 p, 3s, 3p, 3d, 4s 4p, 4d, 4f (Wrong)

You have to do it like: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p

So for our case, arsenic has 33 protons so we need to have 33 electrons, we have to put those electrons in these levels, s,p,d (s can have 2 electros, p up to 6 and d up to 10) so

Arsenic 33

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³ . Now let me arrange the coeffients

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²  4p³

Now let´s bring back the electronic configuration diagram

1 s Here you can have 2 electrons

2s 2p Here you can have 2+6 = 8 electrons (Total electrons 2 _{from level 1}+ 8 = 10 )

3s 3p 3d Here you can have 2 + 6 + 10 = 18 electrons ( Total electrons 10 + 18 = 28)

4s 4p 4d 4 f Here you can have 2 + 6 + 10 + 14 = 32 electrons (We don´t need too many just five in order to reach 33 electrons, 28 + 5)

For our specific case we have

Level 1: 1s²

Level 2: 2s² 2p⁶

Level 3: 3s² 3p⁶ 3d¹⁰

Level 4: 4s²  4p³

You are asked to do a drawing, what we have to do is to draw 4 circles (4 levels) on inside the other, then you will put some dots (electrons) in every circle in order to match the 33 electrons.

In the first circle you will put 2, in the next you will put 8, in the next you will put 18 and in the last one you will put 5.

As

You repeat it for chlorine.

It has an atomic number of 17, which means it has 17 protons, therefore 17 electrons. the electronic configuration is

1s² 2 s² 2 p⁶ 3s² 3p⁵

Levels would go:

Level 1: 1s²

Level 2: 2s² 2p⁶

Level 3: 3s² 3p⁵

The drawing would be just 3 circles, one inside the other, the first one with 2 electrons, the third one with 8 electrons and the last one with 7 electrons

Cl