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# How does electron configuration affect electronegativity?

Last updated date: 20th Jun 2024
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Hint: In the event that you ask enough to pose this inquiry you know the fundamentals of electron configuration. You know, for instance, that the $n = 1$ electron shell has just two electrons in it and that it is extremely near the nucleus.

We should take a look at the $n = 2$ electron shell. That shell covers the components from lithium to neon. At the point when you open another electron shell, you in reality just do it once and it opens an entire bundle of potential places for electrons to hang out. so with lithium, you have $2$ electrons in the $n = 1$ shell ($1{s^2}$ ) and one electron in the $n = 2$ shell (2s1).
It is the attraction to the positive charge of the three protons in the lithium nucleus. Nonetheless, the single $2s$ electron is additionally repulsed by the two electrons in the $n = 1$ shell. They, basically, block the $2s$ electron from feeling the attraction of every one of the three protons. All things considered, in light of the fact that the two $n = 1$ electrons are nearer to the nucleus than the $2s$ electron, it possibly feels as though it were attracted to one proton!
This idea is known as the effective nuclear charge. So for lithium, the effective atomic charge experienced by the $2s$ electron is $+ 1$ . Presently, when you go on to beryllium, you are adding another electron to the $n = 2$ shell, so it is neither further from the nucleus nor nearer to the nucleus than the first $n = 2$ electron. Consequently, it doesn't hinder the attraction of the nucleus. Every one of the $n = 2$ electrons feels an attraction of $n = 2$ , so both are held more unequivocally in beryllium than the one electron in lithium. The effective nuclear charge of beryllium for the $n = 2$ electrons is $+ 2.$
Presently how about we take a gander at fluorine! By a similar thinking, fluorine has an effective nuclear charge for $n = 2$ electrons of $+ 7$ .Electrons truly like (feel an extremely solid attraction for) being in the $n = 2$ shell of fluorine!
Every electron shell is further away from the nucleus than the one preceding it. What's more, the attraction between the effective nuclear charge and the electrons weakens with distance, along these lines, for instance, oxygen, with an effective atomic charge of $6$ has almost similar attraction for electrons as chlorine, with an effective nuclear charge of $7$ on the grounds that the chlorine n=3 shell is further away than the oxygen $n = 2$ shell.
We realize that there are really eight spaces for electrons in the $n = 2$ shell. So that implies there is space for another electron. What's more, that additional electron will feel a similar attraction for the fluorine nucleus as the initial seven $n = 2$ electrons! It is neither farther away nor nearer to the nucleus than the other seven.