# quantum number - question about the ms-value

I've got the quantum number $\ce{(3,2,2,-1/2)}$, and the question is which electron belongs to this quantity. The answer is orbital $\ce{3d^1}$. But I don't understand why. I would have understood it, if the ms-value was $\ce{+1/2}$. But how can you have $\ce{-1/2}$?

• There is no difference between 1/2 and -1/2 in the problem you are solving. Independently of it there is one electron on d orbital – user43021 May 19 '18 at 16:10
• Okay, thank you. But how do I then know the answer is $\ce{3d^1}$ instead of $\ce{3d^2}$ – F.Sal May 19 '18 at 16:25
• The quantum numbers are $n,\;l,\;m_l$ for the orbital and $S$ and $m_s$ for the electron spin. The ranges are $l=0,\;1,\cdots n-1$, $m_l=0,\pm 1,\pm 2,\cdots \pm l$ and $S=1/2$ and $m_s=\pm 1/2$. You have $n=3,\ ; l=2,\; m_l=2$ making one of the $2l+1$ or five $3d$ orbitals. One electron makes $3d^1$. The $-1/2$ can only refer to the spin quantum number $m_s$ and is not usually included. – porphyrin May 19 '18 at 22:42

As porphyrin clearly explained, the first three quantum numbers usually are enough to determine the orbital of a given electron.

You probably know this but I'd like to let it here for everyone else.

Each different value of the principal quantum number $n$ yields a different wave function and a different state of the electron. In atoms, $n$ defines the order of all states according to energy.

The second and third quantum numbers $l$ and $m_{l}$, arise from the quantization of rotation where $l$ gives the shape of an orbital and its angular distribution and $m_{l}$ determines the number of orbitals and their orientations in a subshell.

The spin quantum number of an electron is denoted $s$ and it has the value of $s=1/2$. The spin magnetic quantum number is denoted $m_{s}$ gives the projection of the spin angular momentum on the z-axis. Therefore, one electron can have only one orientation: either $m_{s}=+1/2$ which we commonly denote up-spin or $\alpha$ and $m_{s}=-1/2$ denoted as down-spin or $\beta$.

What I see in the comments are partial answers because to associate an electron (not an orbital) to these numbers you need to use $m_{s}$. Then, you have a spin-down or $\beta$ electron in a $3d$ orbital.

Answering your question in the comments. To have a $3d^{2}$ you'd need two pairs of four quantum numbers: $(3,2,2,+1/2)$ and $(3,2,2,-1/2)$. Here, two electrons occupy the same region in space but they have different orientations with respect to their spin angular momenta.