New Method to find the Electronic Configuration of an Element

The most basic thing needed to solve a Chemistry problem is the knowledge of the Electronic structure of the molecules of the compound. And for that one must be able to write down the Electronic Configuration of each element (knowing its Atomic number). For example:

The EC of Carbon with atomic number 6 is 1s² 2s² 2p^2.

While the conventional method of writing down the EC of an element is not particularly complicated and doesn’t even require one to remember a lot of stuff, but still there is one flaw in it- Speed. Yes the traditional method is very slow. Every time we want an EC we have to draw a medium sized table sort of structure and draw zigzag lines upon it (U know what I mean don’t you?).

But I have got a brand new method that achieves the EC of an element from its Atomic number (Z) and is really fast. Let us get on with it.

The EC of any element consists of many terms with each term having 3 elements – 1s². First we are going to focus on the orbitals i.e. s, p, d, f. I am going to fix an order in which we are going to read these four orbitals for the rest of this article (or for the rest of your life if you’ll stick with me till the end). The order is: f, d, p, s. You are now going to divide this order into 4 segments. The first segment contains only the last element of the order i.e. s. The second segment contains the last two elements of the order – p, s. The third segment contains the last three elements of the order- d, p, s. The final segment contains all the elements in order- f, d, p, s. Write down each segment twice starting with the first and ending with the fourth to get your base EC:

s   s      p s   p s      d p s   d p s      f d p s   f d p s

The next part is to fill up the number of electrons in each of the orbitals. S can have max of 2, p can have max of 6, d can have max of 10 and f can have max of 14 electrons. So if I was to write the EC of an element with Z=32, I’ll start by writing the base of the EC as taught above and then filling each orbital with its maximum no. of electrons till all available 32 of them are properly consumed (you don’t have to write the whole base of the EC as it defeats the purpose of this method which is speed. So you can write a first few elements and fill them up with electrons and later if need be add some orbitals to the base to consume all the electrons) Remember you only to write that bit of the base which consumes all the electrons in the element. Now I presume that I’ll be needing only 6 first orbitals from the base so I write them and fill them with electrons-

s² s² p⁶ s² p⁶ s²

Using these 6 I have only consumed 2+2+6+2+6+2=20 electrons and I need to accommodate 12 more so I anticipate and write 3 more orbitals in the base and fill them up till 32 are put in.

S² s² p⁶ s² p⁶ s² d¹⁰ p².

So two out of three things are done till now. Now for the most complicated part- the orbit numbering of each orbital. Following are the thumb rules:

1) All orbits are numbered with numbers from 1 to ….. 8

2) Orbit number ‘1’ comes only ones, 2 comes twice, 3 comes thrice and …….. 8 comes 8 times at maximum in the base EC.

3) If suppose you are writing a number for the n^th  time then n-2 empty elements must be skipped before actually writing it in front of an element in the base EC. An already numbered element in the base EC is automatically skipped and it doesn’t count as skip for the above formula

4) Every time a new number is to be written for the first time, it must be written at the first available empty space from the beginning.

Let us follow these rules and number the entire base EC. Put 1 in front of the first element

1s  s  p  s  p  s  d  p  s  d  p  s  f  d   p  s  f   d  p  s

1’s are done

Put two at the first  empty element from the begining:

1s  2s  p  s  p  s  d  p  s  d  p  s  f  d   p  s  f   d  p  s

Put two for the second time with no skips(skips=n-2=2-2=0)

1s  2s  2p  s  p  s  d  p  s  d  p  s  f  d   p  s  f   d  p  s

2’s are done

3 for the first time at the first available empty space :

1s  2s  2p  3s  p  s  d  p  s  d  p  s  f  d   p  s  f   d  p  s

3 for the second time  (0 skips):

1s  2s  2p  3s  3p  s  d  p  s  d  p  s  f  d   p  s  f   d  p  s

3 for the third time (1 skip as we are writing 3 for the third time so skips=n-2=3-2=1)

1s  2s  2p  3s  3p  (SKIPPED) s  3d  p  s  d  p  s  f  d   p  s  f   d  p  s

3’s are done

4 for the first time at the first available empty space from the beginning:

1s  2s  2p  3s  3p  4s  3d  p  s  d  p  s  f  d   p  s  f   d  p  s

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1s  2s  2p  3s  3p  4s  3d  4p  5s  4d  5p  (SKIPPED)s  4f  5d   (SKIPPED)p  (SKIPPED)s  5f   d  p  s

Now I am about to write 5 for the fifth time(3 skips). But there is no element to number after skipping three empty elements. So we don’t write 5 for the fifth time at all (Imagine writing it in the air). In fact we ignore all those numbers for which the base EC runs out of elements. Following this consideration you will see that 8 can be written only ones in the Base EC.

So here is the final full EC :

1s  2s  2p  3s  3p  4s  3d  4p  5s  4d  5p  6s  4f  5d   6p  7s  5f   6d  7p  8s

While this process looked pretty daunting, it actually is quite simple and after one or two practices it comes out naturally. The thing to be kept in mind is that one can simply write the base EC sufficient enough for a given Atomic number without these orbit numbers and then after having the required segment of the base EC, one can proceed to number it. Hence obtaining the EC of the element very quickly. To demonstrate the quickness of this method, let us take an example:

Z= 65

STEP 1) Anticipate and write the base EC:

s s p s p s d p s d p s f (I guess this is enough to accommodate 65)

then upon the above base only we fill up the electrons till reaching 65

s²  s²  p⁶  s²  p⁶  s²  d¹⁰  p⁶  s²  d¹⁰  p⁶  s²  f¹¹

and finally we number it :

1s²  2s²  2p⁶  3s²  3p⁶  4s²  3d¹⁰  4p⁶  5s²  4d¹⁰  5p⁶  6s²  4f¹¹

And that is our Electronic Configuration.

This Electronic Configurations provided by this method are equally accurate as are the ones provided by the zigzag method and hence it is worth noting that the EC of some element should be excepted from this method as well.