[Chemistry] Formation of ions from the elements?

Hi.

I’ll try to answer your questions.

At the bottom, “Na” was written under the sodium atom and “Cl” (with the representation of 7 valence electrons) under the chlorine atom. I understand. “Na” thus emits a valence electron to “Cl”, both of which are then transformed into an ion (sodium ion and chloride ion) which fulfills the noble gas configuration or the octet rule.

Correct, you’ve already understood the most important thing.

Unfortunately, I don’t understand the name on the right side below. Why is the sodium ion shown here only as “Na^+”? (####I understand the high-level plus, that’s what you write, since an electron has now been given and there is an excess of a proton. So there’s a proton more than electrons. Therefore, the high-level plus comes to “Na^+”.###### But why are the valence electrons not shown here at “Na^+”?

When atoms emit or absorb electrons, they become ions:

• Onthe electrons onare therefore negatively charged (maybe this helps as an Eselsbrücke).
• Cat.ions have emitted the electrons, so are positively charged.

Sodium has as atom a valence electron (shown on the left as a point). It has 11 protons and 11 electrons. When the valence electron is released, it still has ten electrons and eleven protons. So there is a proton surplus from a proton, which is why the cation then has a simply positive charge (Na+). It also has no more valence electrons because it has delivered them to the reactant.

So why are there no 4 lines around “Na^+”? Is that what’s going on? Does that have to be?

It has an electron octet, but no eight valence electrons (not at all). This looks like the cation has eight valence electrons, but that’s not the case. In the case of main group elements, only the electrons of the outer shell (which are absent from sodium now) participate in the reaction.

Is this because Na (sodium) usually has 3 shells and is now located on the third shell 0 valence electrons?

Exactly.

Is a bowl with 0 electromen, like here, never drawn?

No, why should they be drawn? It’s empty. Look at the atomic and ion radius of sodium or the cation. In sodium, the atomic radius is 180 pm, the cation is an ion radius of 97 pm – the last shell is no longer there.

So if you imagine the third shell in your head to draw the valence electrons? Is this the reason that only under the sodium ion on the right-hand side below is “Na^+” and therefore no valence electrons were shown?

I don’t understand the question. The third shell is present at the sodium atom, in which is the valence electron. However, if the valence electron has been released, there is no need to draw the third shell – if you see Na+ and the two shells that are fully occupied, you know that the third shell is empty because Valenzelectronic has been released. This can be “ablesen” to the ion if one knows in which period and main group the corresponding element is in its atomic form.

LG