Electronic Configuration?

Question

How do you wirte electonic configuration, how? method please for example NA, is 1s2 2s2 2p6 3s1 , (another think i dont get is 2s2)

Answer 1

The way I do it is by looking at the periodic table.

The table can be split into blocks with the first 2 groups being the "s" block. The main group in the middle of the table (groups 3 to 12) is the "p" block which includes all the metals. The "d" block is the groups from 13 onwards.

It's hard for me to explain in writing but if you are to obtain, for example, the electronic structure of bromine you would look and see that it was in the d block. Start at the beginning of the table and work it through from left to right.

The period n.o tells you what level is being filled - for H the period n.o is 1 and the electron config is 1s1. For He the electron config is 1s2 because it is in period 1 and there are 2 electrons. (remember as you go along from element to element a single electron is being added each time).

1s2 will always be filled (unless you are just loking at H).

Dropping to period 2, we know it is the 2nd level being filled (because it is period 2). Li and Be are in the 2s block and therefore we know this level is filled in Br so we can add 2s2 to the elctronic config.

So now we have 1s2 2s2 in our Br electronic config.

Moving right from Be, we come to the beginning of the p-block and reach B. You can see that there are 6 elements in period 2 of the p-block from B to Ne and these must all be full in Br so we know the next config is 2p6.

Now we have 1s2 2s2 2p6 in our Br config.

Do the same with period 3 to get 3s2 3p6.

When getting to the 4th period we know that the s shell is full so we have 4s2. Moving right from that we hit the d block NOT the p block. From Sc to Zn there are 10 elements and they are in the d block. They are all filled in Br, however, when you get to the d block, subtract 1 from the period number to give you 3d10, NOT 4d10.

So now we have 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10.

Keep moving right and you get the p block. Br is the 5th element along in this block of the period meaning that in Br, there are only 5 electrons in the 4p shell giving 4p5

1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p5

To double check you have the correct config, add up the electrons - in this case 2+2+6+2+6+2+10+5=35. The atomic n.o of Br is 35 and the atomic n.o = the n.o of electrons in an atom so we know the answer is correct.

Hope this makes sense!! Bit of a novel I've written lol

Answer 2

Aubau Principle: electrons enter the sublevel with the lowest energy first.

Order of the orbitals from lowest energy to highest for the known elements:

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

each letter represents a different orbital shape. The letter tells you how many electrons can fit in that orbital:

s = 2 electrons
p = 6 electrons
d = 10 electrons
f = 14 electrons

The number in front tells us the energy level, but it has NO bearing on how many electrons can fit in it.

For example, the 2p can hold 6 electrons max and the 4p can also only hold 6 electrons.

Using sodium as an example:

Sodium is number 11 on the Periodic Table. Since all atoms are neutral, with their proton count equaling their electron count, the atomic number (the number of protons) tells me how many electrons must be in my configuration.

So, we start with the first sublevel: 1s

1s can hold 2 electrons, so we put the first two in there.

1s^2

We still have 9 left, so we move to the next orbital: 2s. Just as before, the 2s can only hold 2 electrons as well, leaving us 7 more.

1s^2 2s^2

Next is the 2p. 2p can hold 6, so the next 6 go there, leaving us 1 left.

1s^2 2s^2 2p^6

The last sublevel we need is the 3s, which can hold 2 electrons. We only have one more to place, so the last on goes in that sublevel.

1s^2 2s^2 2p^6 3s^1

All 11 electrons are accounted for in the configuration.

Answer 3

Electrons have 4 basic quantum numbers. Energy, angular momentum, magnetic moment and spin. If you treat an electron as a standing wave in the potential electric field of the nucleus it has different modes of vibration in 3-D corresponding say to the modes of vibration of an acoustic wave in a pipe. The solution of the fundamental mode consists of a spherical shape(s-orbital) with the nucleus as centre. In this state the electron has zero ang momentum and magnetic moment (integer quantities allowed by quantum mechanics but +/- 'spin'). This orbital can hold two electrons only as every other quantum state is identical.(since the probability of observing a particle is given by multiplying its wave function by its complex conjugate (for fermions), identical particles yield zero probability). The next higher mode of vibration corresponds to a further spherically symnetrical one (2 electrons again) plus 3 'dumbell' shapes, symnetrically disposed in 3-D(p-orbitals), each capable of holding 2 electrons, making 6, or 2+6=8 in total for this energy level. In the next mode up we get an even more complicated new set of vibration shapes along with the p-orbitals and the s-orbital with higher energies for each. The d-orbitals have 4+1 symnetry capable of containing 5x2=10 electrons+6p electrons+2s electrons making 18 in total. These configurations go on to the 'f-level' with even more elaborate shapes, capable of containing even more electrons. The levels which correspond to different n-values or energies are called shells.
It is assumed that this solution for the simple hydrogen model with various avaliable energy levels also applies to progressively more complex atoms where the electrons occupy some or all of these shells. For Sodium then the config would be..2 (1s electrons) 8 (2(2s electrons)) and 1(3s electron) as the corresponding allowed value for n=2 is ang mom=1 and mag mom=0. Only at n=3 do you get full (permitted by quantum rules) expression of all the quantum numbers.