What are the electron configurations of lithium, oxygen, nitrogen, and potassium?

Lithium:

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From its position, we know that it has `1` valence electron in the `2s` orbital series (because it's in the second period): `2s^1`.

We also know that its `1s` orbital is full, because to get to lithium in the periodic table, we have to pass `1s`. There are `2` electrons in an `s` orbital; this means that it has `2` electrons in its `1s` orbital: `1s^2`.

Putting it all together, we get `1s^2 2s^1`.

Oxygen:

PTable

From its position in the periodic table, we know that it has `4` valence electrons in the `2p` orbital series (because it's in the second period): `2p^4`.
We also know that:

• Its `1s` orbital is full. There are `2` electrons in an `s` orbital; this means that it has `2` electrons in its `1s` orbital: `1s^2`.
• Its `2s` orbital is full. There are `2` electrons in an `s` orbital; this means that it has `2` electrons in its `2s` orbital: `2s^2`.

Putting it all together, we get `1s^2 2s^2 2p^4`.

Nitrogen:

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Nitrogen is directly to the left of oxygen in the periodic table. This tells us that it has one less electron than oxygen—therefore, its electron configuration is the exact same as oxygen's, except with one less electron in the valence energy level.

Oxygen's electron configuration is `1s^2 2s^2 2p^4`.
After taking one electron from that, it becomes `1s^2 2s^2 2p^3`—nitrogen.

Potassium:

PTable

From its position in the periodic table, we know that it has `1` valence electron in the `4s` orbital series (because it's in the `s` block of the fourth period): `4s^1`.
We also know that:

• Its `1s` orbital is full. There are `2` electrons in an `s` orbital; this means that it has `2` electrons in its `1s` orbital: `1s^2`.

• Its `2s` orbital is full. There are `2` electrons in an `s` orbital; this means that it has `2` electrons in its `2s` orbital: `2s^2`.

• Its `2p` orbital is full. There are `6` electrons in an `s` orbital; this means that it has `6` electrons in its `2p` orbital: `2p^6`.

• Its `3s` orbital is full. There are `2` electrons in an `s` orbital; this means that it has `2` electrons in its `3s` orbital: `3s^2`.

• Its `3p` orbital is full. There are `6` electrons in an `s` orbital; this means that it has `6` electrons in its `3p` orbital: `3p^6`.

Putting it all together, we get `1s^2 2s^2 2p^6 3s^2 3p^6 4s^1`.