How do I make an ionic bond?
I'm writing a chemistry paper tomorrow. One of the topics is ionic bonding. However, I don't understand it at all. Could someone give me an example? Using aluminum and sulfur? I would be very grateful, as I've been trying for hours 🙂
An example of ion bond is the sodium chloride NaCl. The sodium ion is positively charged – an electron is missing. Chlorion is negatively charged – there is also no electron. However, the chlorion has 7 electrons on the outer shell – there is an electron, so it is still a negatively charged ion.
LA
There’s something going on…
Would you just say how much electrons have an uncharged chlorine atom and how many a clorid ion?
A chlorine atom has 7 electrons on the outer shell. the chlorine atom is negatively charged.
(sodium) chloride is also negatively charged. Also owns 7 electrons on the outer shell ..
Thanks for the star 🌟
I corrected this to 8 – unfortunately my contingent has disappeared
How do you get 7?
I’m good that you start to address the points you haven’t understood so early;)
Hihi, that works very well. Don’t try at home!
Fun aside: In the case of an ion bond, the aim is to attract positively charged ions (cations) and negatively charged ions (anions). As the positive pole of a magnet and the negative pole of another magnet attract each other.
This attraction force acts in each case in all directions, for which reason ion compounds are often present as “endless gratings”, where a three-dimensional combination of the ions in question is continuously repeated. So not in molecules.
The simplest and most famous example is sodium chloride, NaCl. These are sodium cations Na(+) and chlorine anions Cl(-). They dress each other. Each cation is surrounded by anions on all sides and each anion is surrounded by cations on all sides. Looks like so out. And that just continues in all directions without interruption.
In the case of aluminum sulfide, it is somewhat more complicated because aluminum ions are three times positively charged, i.e. Al(3+)… and sulfur is twice negative at this point, i.e. S(2-). Since the charges have to compensate for each other, it requires different Al ions and S ions. Just come two Al(3+) on three S(2-). This also results in an endless grid, only that it more complicated is as if the number ratio was 1:1.