Atomic radius regarding the bonding of Na and Cl

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Austin Qiu 1A
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Joined: Fri Sep 25, 2015 3:00 am

Atomic radius regarding the bonding of Na and Cl

Postby Austin Qiu 1A » Fri Oct 02, 2015 4:57 pm

When Na and Cl bond, I understand that since Cl is more electronegative than Na, Na will lose its valence electron to Cl, forming the ions Na+ and Cl-. Thus, since Na lost a electron in the 3s1 orbital, it follows that its atomic radius will decrease. Cl- however experiences an increase in atomic radius, even though electrons are not added to a higher shell. Why is this? Perhaps this is due to electrostatic repulsion between the electrons?

Annie Qing 2F
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am

Re: Atomic radius regarding the bonding of Na and Cl

Postby Annie Qing 2F » Fri Oct 02, 2015 11:21 pm

Yes, I think you're right with the electrostatic repulsion between electrons. Because the number of protons (and therefore positive charge) stays constant as chlorine gains an electron, the electrons aren't pulled closer in any way and instead spread out within the "shell" as an effect of the repulsion between their negative charges.

Chi-Yun 1B
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

Re: Atomic radius regarding the bonding of Na and Cl

Postby Chi-Yun 1B » Sat Oct 03, 2015 11:20 pm

I also believe you are correct.

Atomic radii decrease across a period due to the increase in protons and electrons; the increased attraction between the electrons and the nucleus creates a higher effective nuclear charge. Therefore, since no protons were added to chlorine, the extra electron creates more repulsion with the surrounding electrons, which will further expand the atomic radius of chloride.

My apologies if this is repetitive of the previous information stated.


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