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

.

.

.

.

.

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.

Fridrich Method: An Interesting Criticism

First of all let me just say it, I know all the cubers will laugh at this and all the speedcubers will know that I am not yet one of them, but still let me just say it- The Fridrich Method is a system of solving the Rubik's cube and a pretty good one while we are at it. So continue reading if you know what this is about, else feel free outrage at me for wasting your time.

The Fridrich method is the most used/popular system for solving the cube speedily. All the possible world records are held by practitioners of this system (why do I keep calling it a system ? vocabulary may be !). What we do in this method is beautiful in the sense that one is basically applying short hands (short cuts) to the "Beginners method". I mean in the beginners method- the first step is to build a cross as is in the Fridrich method. And than the beginner method takes forever to complete the first two layers which the Fridrich method does in one go (actually 4). Finally the Fridrich method offers two magical spells - OLL and PLL that together comprehend the last layer which the beginners method sucks at! 

Well these were the positives and mind it, these were really POSITIVE. Now what is about to follow is inspired from what Antony Snider said in one of his rare Youtube videos showing us (actually showing off) his own system. What he said was that when we are solving the cube using the CFOP (Fridrich) method and when we are doing the OLL, we are not being a genius in orienting all the top layer pieces all in one go but we are actually making things bad as in the next step we disturb everything we just did in order to have the pieces in the right places together with in the right orientation. Well he has a point. One can extend his way of looking at it to the whole CFOP and not just the OLL stage. How ? let us see.

Firstly we do the cross (A cliche according to me). Completing the cross is fine as there is nothing we are disturbing or redoing in order to achieve the cross due to this being the very first step. After the cross we do the F2L (the most beautiful part of this system and the one which stands tall intuitive, in a system which other wise has near a 100 algorithms). The F2L is beautiful, but notice while you are putting one of the four F2L pairs together we are constantly disturbing the cross and finally getting it to rest in the way it is meant to be but with a correctly placed F2L pair. My point is why build just only a cross when you know that it has to be disturbed in order to complete the  first two layers. After completing the First 2 Layers, we move on to the OLL stage. The OLL orients all the top layer pieces (more on that later). But notice when you are doing your OLL alg, you are actually disturbing some of your F2L pairs and manipulating them to circle back to their meant to be place but with that, orient everything in the last (top) layer. Let me repeat the question, why do just the F2L when you know that later on you have disturb (manipulate) it in order to achieve a solid color on top (Damn the vocabulary)? 

Now focusing more on what Snider said. Ones we have done the OLL we are on to the PLL stage. But before the implementation of the OLL alg, what we had was a top layer all messed up both in orientation and positioning (permutation). What OLL did was orient all the pieces (flip them correctly) caring less (rather nothing) about where each piece actually needed to go. And this little thing is taken care of by the PLL alg. But again, the PLL alg follows the same pattern i.e. disturb what was done just yet and manipulate it to achieve something extra together with the re-establishment of the previous step. 

So all I am asking is that why do something when you know that just after a quarter of a second (if you are Feilks Zemdegs) you will be disturbing it in order to achieve something extra. Its like writing a sentence in black plain ink (e.g. God I wish I were a speedcuber.) and than in order to highlight some portion of your sentence (say 'speedcuber') you erase that part (God I wish I were a __________) and then write it back in capitals (God I wish I were a SPEEDCUBER) when you could have easily written SPEEDCUBER (in blocks) to begin with. And its not just that, the whole repeating everything with little- little changes all gangs up together posing a serious  question to the efficacy of the whole system. I mean when You do the PLL, you think you are just disturbing the OLL right? but in reality you disturb the OLL which in turn disturbs the F2L which in turn disturbs the cross and you end up adjusting your cross for the third time + adjusting your F2L for the second time and adjusting your OLL for the first time,before finally solving the cube.

Its like after you have written -  "God I wish I were a SPEEDCUBER" , you again erase the "SPEEDCUBER" word cause you need it to be in some coloured font(SPEEDCUBER) and then after you have done that, you erase it again to rewrite it in italics(SPEEDCUBER) and then bold(SPEEDCUBER) and the pain goes on.

Its not like there no alternatives to this, its just that these alternatives takes a long time to master and get up to the same speed as the Fridrich method offers. Snider method it self is one of the alternatives but its not that quick. The Roux is another alternative and its quite fast too. In Roux in the last stage of the solve we only use the M and U slices which directly tells that we are not at all touching what we have acheived before, so there is that. But who am I to say anything, I use Fridrich alg while attempting to solve the cube using Roux!