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Sodium loses one electron, therefore Na+ Chlorine gains an electron, therefore Cl-
Na+ has 10 electrons and is isoelectronic with neon. Cl- has 18 electrons and is isoelectronic with argon.
Elements in the 1st group have 1 valence electron. So they are likely to donate 1 electron to get more stable. potassium, rubidium and cesium are likely to donate 1 electron.
Sodium chloride is neutral; only elements have an electronegativity.
The Elements Symbol Is : Na
In NaCl, there exists Na+ and Cl- ions and with the electron configuration of [He]2s22p6 (for Na+) and [Ne]3s23p6 (for Cl-)
The positive ion, the one that took the electrons, should have a full shell while, the negative ion, the one that lost the electrons, should have an empty shell.
Yes, Sodium chloride is an ionic compound composed of Na+ and Cl- ions.
Ionic bond is when a non-metal takes the metals valance electrons so both atoms can be balanced. Take NaCl for example. Na has one valance electron and Cl only needs one to get eight. So Cl takes Na's one valance electron...:Cl: Na. See the empty spot for Cl, that where Na's valance electron will go.'
No. This is an ionic compound and the electron of sodium is donated into the electron shell of chlorine. Na + ( the cation ) and Cl - ( the anion ) form the ionic compound NaCl, sodium chloride.
Sodium loses one electron, therefore Na+ Chlorine gains an electron, therefore Cl-
This is called an ionic bondExample:Na --> Na+ + 1 e-Cl + 1 e- --> Cl-Na+ + Cl- --> Na+Cl-
From greatest to least tendency to accept an electron, they are F, O, C, Li, and Be.
18. Cl=17 Na= 11
Na+ has 10 electrons and is isoelectronic with neon. Cl- has 18 electrons and is isoelectronic with argon.
Ionically. Sodium ionises its one outer most valence electon to form the cation Na^(+) Chlorine has electron affinity, wi thspace to accept one more electron to complete its outer energy shell octet. Cl^(-) Since the sodium cation and the chloride anion each have one opposite charge, they combine like the north and south poles of a magnet. Here are the equations Na => Na^(+) + e^- (ionisation) e^- + Cl => Cl^(-) (electron affinity) Na^(+) + Cl^(-) = (Na^+Cl^-) or just NaCl(s) (ionic attraction)
The sodium atom will donate an electron to the other atom which it is in compound with, leaving you with Na[+] and another, negatively charged ion (for example Cl[-]).