# How do you determine the electron configuration of "W"?

Feb 15, 2018

$\left[X e\right] 4 {f}^{14} \setminus 5 {d}^{4} \setminus 6 {s}^{2}$

#### Explanation:

We first need to find the closest noble gas to tungsten on the periodic table. From here, tungsten is number $74$ on the periodic table, so the closest noble gas to it would be xenon, which is number $54$. So, we start by writing the first part of tungsten's electron configuration which is

$\left[X e\right]$

Next, we have $74 - 54 = 20$ more electrons to fill.

We know that xenon's full electron configuration is $1 {s}^{2} \setminus 2 {s}^{2} \setminus 2 {p}^{6} \setminus 3 {s}^{2} \setminus 3 {p}^{6} \setminus 3 {d}^{10} \setminus 4 {s}^{2} \setminus 4 {p}^{6} \setminus 4 {d}^{10} \setminus 5 {s}^{2} \setminus 5 {p}^{6}$

From this picture, we see that the next shell we choose to fill will be the $6 s$ orbital. But don't take too much stock in it, as it doesn't work well for many transition metals. It can hold $2$ electrons, so we will continue with tungsten's electron configuration, and now have

$\left[X e\right] 6 {s}^{2}$

$20 - 2 = 18$ more electrons

Next comes the $4 f$ subshell. It can a maximum of $14$ electrons, so we can write it as $4 {f}^{14}$. Continuing with our original goal, we get

$\left[X e\right] 6 {s}^{2} 4 {f}^{14}$

$18 - 14 = 4$ more electrons

Since the $5 d$ subshell can hold $10$ electrons, but we only need $4$ electrons, it becomes $5 {d}^{4}$.

So, our final electron configuration for tungsten is:

$\left[X e\right] 6 {s}^{2} \setminus 4 {f}^{14} \setminus 5 {d}^{4}$

or we can rearrange it in order of $n$ to

$\left[X e\right] 4 {f}^{14} \setminus 5 {d}^{4} \setminus 6 {s}^{2}$