Finding the g-value in the Boltzmann distribution

[JoJoQuinoa]

Hello,

I was wondering if someone could show me how to determine the number of orbitals available for a state and the number of electrons in that state. For calcium in the ground state, the electron config is 1s2 2s2 2p6 3s2 3p6 4s2. For the first excited state I assumed 1s2... 4s1 3d1.
From the solution for g_g, why the number of available orbitals is one when 4s is filled? Shouldn't it be zero?

Thanks!

 

[mjc123]

If there were zero available orbitals, where would the electrons go? It doesn't mean "available for other electrons". It means "available for the valence electrons". (All filled subshells have #e = 2#o, so g = 1; you only need to consider the valence electrons.) There is one 4s orbital into which 2 electrons can go to give the configuration 4s².

 

[JoJoQuinoa]

@mjc123, thank you for your explanation.

That is what I thought originally but when I applied it to the excited state, it didn't work. But I think my electron configuration for the first excited state is wrong. I think it should be $1s^2...4s^1 4p^1$ based on the selection rules. If that's true then #o is 3 in $4p^1$ and there are 2 valence electrons.

 

[DrClaude]

I think it should be $1s^2...4s^1 4p^1$ based on the selection rules.

Note that even without considering selection rules, $4s^1 4p^1$ is lower in energy than $4s^1 3d^1$. (The Aufbau rules when filling orbitals do not necessarily apply to finding excited states).

 

[mjc123]

I'm not sure what excited state is being referred to - certainly not the first excited state, as 226.51 nm is much higher energy. The expression for g_e would suggest a configuration like 4p², for which #o is 3 and #e is 2. (4s4p would have g = 12; 2 for the 4s electron x 6 for the 4p.) And the 15 microstates of 4p² don't all have the same energy - they are split into ¹S, ³P and ¹D states (and ³P is further split into J = 0, 1 and 2 states). So using the g value in the Boltzmann distribution would be problematic.

Can you provide more context for this problem?

 

[JoJoQuinoa]

I'm not sure what excited state is being referred to - certainly not the first excited state, as 226.51 nm is much higher energy. The expression for g_e would suggest a configuration like 4p², for which #o is 3 and #e is 2. (4s4p would have g = 12; 2 for the 4s electron x 6 for the 4p.) And the 15 microstates of 4p² don't all have the same energy - they are split into ¹S, ³P and ¹D states (and ³P is further split into J = 0, 1 and 2 states). So using the g value in the Boltzmann distribution would be problematic.

Can you provide more context for this problem?

Certainly

Problem: Use the Boltzmann distribution equation to calculate the percentage of calcium atoms that are in the first excited state in a hydrogen -air flame at 2,250K. The line associated with that transition has a wavelength of 226.51 nm.

Further Question: Is it not true that first excitation occurs when 'one' valence electron transition to a higher electronic state?

Thanks!

 

[JoJoQuinoa]

Note that even without considering selection rules, $4s^1 4p^1$ is lower in energy than $4s^1 3d^1$. (The Aufbau rules when filling orbitals do not necessarily apply to finding excited states).

Hi DrClaude,

Is this a common knowledge or is there a set of rules for this? I'm not a chemistry student so I've been using what I can vaguely remember from general chemistry 10 years ago

 

[DrClaude]

Is this a common knowledge or is there a set of rules for this?

There are no general rules that I am aware of.

You can see here the order of the excited energy levels for calcium:
https://physics.nist.gov/PhysRefData/Handbook/Tables/calciumtable5.htm