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Topic: Writing Quantum Numbers for energy levels?  (Read 2159 times)

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Offline henrybar

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Writing Quantum Numbers for energy levels?
« on: September 08, 2014, 08:56:13 PM »
According to my teacher, a unique set of quantum numbers for the last energy level of Titanium is as follows:

How can this be? Isn't Ti in row 4 meaning n should equal 4.

3d^2 is the final sub orbital of Ti, so shouldn't that mean "l"equals 2? (s=0, p=1, d=2, f=3)

IT IS POSSIBLE I made an error in copying the board in regards to the value of "l." But I'm 100% certain he wrote n=3....Is this because of the "3" in "3d^2?" If so, does that mean elements from scandium to zinc are technically in row 3?? Or does this mean I should ignore row # when determining the value of "n" and instead look for the # of the energy shell?

Please spare the confusion and only answer if you know for a fact. An explanation would be nice too.

Offline AdiDex

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Re: Writing Quantum Numbers for energy levels?
« Reply #1 on: September 09, 2014, 01:27:43 AM »
Mate ,
first of all you wrote down another mistake l should be equal to 2
Because It is D subshell....
Nope your teacher was not wrong....
Actually he is talking about last entering electron in the subshell....
Because 4s subshell has lower energy than 3d ( Use  n+l aufbau rule)
So 4s subshell get filled before the 3d shubshell....

Sc , Ti , V , Cr .. Undergoes in 4th period of perodic table.

Order of sucshell in increasing order of energy .

Offline rwiew

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Re: Writing Quantum Numbers for energy levels?
« Reply #2 on: September 09, 2014, 02:07:11 AM »
Ok, so this is one of these problems where I don't know what to write because this is a high school chemistry forum. What adidx has told you is great for high school and will get you the full marks on a test. I believe that such situations are bulls&$# and if you're going to have to lie to students, then better do not ask such questions. You see, the problem is this is not correct.

The 3d level gets filled after 4s, because until after potassium the energy levels indeed go 3d > 4s. However, as the nuclear charge, and hence "effective" nuclear charge felt by electrons increases, the 3d becomes more stabilized, and between potassium and calcium crosses through the 4s level (when I say between I mean if you plot the two energies on a graph, of course there is no actual thing between the two). For Ca 3d is energetically lower than 4s, and so it is for all the subsequent transition metals. Why don't the electrons drop from the 4s level into 3d and we always have two electrons in the 4s levels? This is because energy of the orbital (by which the potential energy of electron - nucleus interaction is meant) is not everything, you have to consider factors such as electron - electron repulsion - this is lower in the bigger, more diffuse 4s orbital. Also, the maximization of unpaired spins will play a role - hence you will occassionally get the "funny" configurations such as s1 d5.

In conclusion, the actual scientific situation - the 4s level is the energetically highest level in a Ti atom (if this is what is meant by "last"). If you want an easy life - go with what adidx told you. If you want truth and are absolutely confident in that you understand this and can have a lengthy argument with your teacher (oh how many of those have I had in the past) - go for it. But if you don't understand what I said and are confused about it, just forget it, might not be worth the trouble.

As a side note - noting the m and s numbers for a "last" energy level is pointless. They all have the same energy for a given level.

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