what is the electron configuration of lithium after making an octet

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I will lose a valance elcetron

3Li = 1s2 2s1. If you lose the outer electron it leaves 1s2 which is not an octet because there are only two electrons instead of 8 but it's a closed shell none-the-less.

To determine the electron configuration of lithium after making an octet, let's start with the original electron configuration of lithium.

The atomic number of lithium is 3, which means it has 3 electrons. The electron configuration of lithium is 1s² 2s¹.

Now, to make an octet, we need to have 8 electrons in the outermost energy level (also known as the valence shell). Since lithium is in the first period of the periodic table, it can only hold a maximum of 2 electrons in its valence shell.

To achieve an octet, lithium can either gain 7 electrons or lose its one valence electron. Let's consider both scenarios:

1. If lithium gains 7 electrons:
The resulting electron configuration would be the electron configuration of neon (the noble gas in the previous period), which is 1s² 2s² 2p⁶.

2. If lithium loses its one valence electron:
The resulting electron configuration would be the electron configuration of helium (the noble gas in the same period), which is 1s².

So, after making an octet, the possible electron configurations of lithium would be:

1s² 2s² 2p⁶ (like neon)
or
1s² (like helium)

It's important to note that by achieving an octet, lithium would no longer be lithium but rather a different element altogether.