Here’s how I understand electron configuration.

You will need to create your own electron configuration chart. Your chart will have 4 columns.

In the first column, write the following:

1s 2s 3s 4s 5s 6s 7s

In the second column, write the following (but be sure to align the numbers with the numbers of the first column):

2p 3p 4p 5p 6p 7p

(thus, 2p should line up with 2s, 3p should line up with 3s, etc)

In the third column, write the following (again aligning the numbers):

3d 4d 5d 6d

In the fourth column, write the following (again aligning the numbers)

4f 5f

The final thing that you need to do is draw diagonal lines (preferably with a ruler ).

You will end up with 8 lines as follows:

1s

2s

2p 3s

3p 4s

3d 4p 5s

4d 5p 6s

4f 5d 6p 7s

6d 7p

Now, using your electron configuration table, just imagine that you are dropping electrons down each of the diagonal lines. Remember that the s orbital can hold 2 electrons, the p orbital can hold 6 electrons, the d orbital can hold 10 electrons, and the f can hold 14 electrons.

So, say you want to write the electron configuration for Chlorine (Cl = element 17). There are 17 electrons in the Chlorine Atom.

The first 2 go into 1s and you write 1s2.

The next 2 go into 2s and you write 2s2.

The next 6 go into 2p and you write 2p6.

(At this point, you have used up 10 electrons, and you have 7 left.)

The next 2 go into 3s and you write 3s2

The remaining 5 go into 3p and you write 3p5

Chlorine electron configuration = 1s2 2s2 2p6 3s2 3p5

If you add up the “exponents” (2 + 2 + 6 + 2 + 5), you get 17.

digital configuration of Carbon (C), Nitrogen(N), Oxygen(O) are as follows: C(6)- 1s2 2s2 2Px1 2Py1 N(7)- 1s2 2s2 2Px1 2Py1 2Pz1 O(8)- 1s2 2s2 2Px2 2Py1 2Pz1 right here, Carbon has 2 electrons in its 2p subshell. So, there is an threat to make sturdy 0.5-crammed P orbital interior the final shell of Carbon by way of gaining a million electron. So, to benefit stability Carbon can certainly income a million electron. on the otherhand, Nitrogen has already 0.5-crammed P orbital interior the final shell. So, it won’t settle for further electrons using fact it is sturdy sufficient. to benefit extra electrons it has to break its stability. it is the reason, Nitrogen is way less constructive(extra valuable) than that of Carbon In case of Oxygen, there r 4 electrons in P orbital. it is neither 0.5-crammed nor finished-crammed. So it is not sturdy sufficient to benefit extra electrons. So, oxygen additionally can income extra electrons in its outermost shell ensuing in a extra constructive eelectronic affinity than Nitrogen. hence, digital affinity of Nitrogen> Carbon & Oxygen

Here’s the order:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6…..

now it depends on the element: Group 1 always ends in (period #) s1, Group 2: (period #) s2, Group 3A: (period #) p1, Group 4A: (period #) p2, Group 5A: (period #) p3,Group 6A: (period #) p4, Group 7A: (period #) p5, Group 8A: (period #) p6

ex:) carbon: 1s2 2s2 2p2

ex:) potassium: 1s2 2s2 2p6 3s2 3p6 4s1

the transitional elements are tricky, but they always end in the d orbital: (period #) d_