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# What are the quantum numbers for chlorine, iron, and tin?

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#### Explanation

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#### Explanation:

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35
Apr 15, 2016

$C l \to 3 , 1 , 0 , \pm \frac{1}{2}$
$F e \to 3 , 2 , - 2 , \pm \frac{1}{2}$
$S n \to 5 , 1 , 0 , \pm \frac{1}{2}$

#### Explanation:

The four quantum numbers describe the outermost or valence electron of an atom. They are:

$n$ is the distance of the orbital from the nucleus ($1 , 2 , 3 , 4. . .$),
$l$ is the shape of the orbital ($0 \to n - 1$),
${m}_{l}$ is the orientation of the orbital in space ($- l \to + l$), and
${m}_{s}$ is the spin of the electrons ($- \frac{1}{2}$ or $+ \frac{1}{2}$).

Begin each element by writing out the electronic configuration in terms of $s , p , d , f$ orbitals. From this you can directly get $n$ and find $l$ based on its equivalent orbital. From $l$ you can figure the range of ${m}_{l}$, and then count one up in the sequence for each electron in the outermost subshell. ${m}_{s}$ is either $- \frac{1}{2}$ or $+ \frac{1}{2}$.

Chlorine, $\left[N e\right] 3 {s}^{2} 3 {p}^{5}$, $n = 3$ straightaway because that is the highest energy level here, and since the outermost electron is in the $p$ orbital, this is the same as $l = 1$. ${m}_{l}$ can be $- 1 , 0 , + 1$, but since there are $5$ electrons, you count five, so $- 1 , 0 , + 1 , - 1 , 0$, which means the last electron has ${m}_{l} = 0$. ${m}_{s}$ can be either $- \frac{1}{2}$ or $+ \frac{1}{2}$, it doesn't really matter.

For iron, $\left[A r\right] 4 {s}^{2} 3 {d}^{6}$, then $n = 3$, $l = 2$ (because it is the $d$ orbital), ${m}_{l}$ is $- 2$ (because you count up one the series from $- l$ to $+ l$ six times, for the six electrons in the $d$ subshell) and ${m}_{s}$ is $\pm \frac{1}{2}$.

Tin has the electron configuration of $\left[K r\right] 5 {s}^{2} 4 {d}^{10} 5 {p}^{2}$, so $n = 5$, $l = 1$, ${m}_{l} = 0$ and ${m}_{s} = \pm \frac{1}{2}$

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