2
$\begingroup$

red-ox reaction

in the picture, the product is a neutral material, but in the illustration of the molecular structure, the compound is shown in its ions, why we can't show it in its ions even in the equation or if we can't, why in this molecular structure it is shown as ionic although it is neutral?

$\endgroup$
2
$\begingroup$

Your questions involve a lot of chemical concepts.

An oxydo-reduction is reaction where electrons are exchanged between species including or not molecular rearrangement.

You totally can write down any salt in the following way, emphasizing ion charges:

$\mathrm{NaCl}_{(s)} \equiv \mathrm{Na^+Cl^-}_{(s)}$

As long as you specify the phase, because when writing down an ionic form the chemist might assume that it is solvated into water. To underline it you may write the following:

$\mathrm{Na^+Cl^-}_{(aq)}$

Further more solid generally have numerous possible distinct phases, therefore you may need to specify which crystalline structure the salt have.

The following equation is explicit:

$2\mathrm{Na}^{(0)}_{(s)} + \mathrm{Cl_{2(g)}} = 2\mathrm{Na^+Cl^-}_{(s)}$

And underlines that an electron is withdrawn from metallic sodium in order to produce chlorine. Anyway, this equation is global balance this it hides many details of the complete and complex chemical process, especially the formation of the crystalline structure. And it does not provide any information on conditions of reaction.

$\endgroup$
2
$\begingroup$

Simply, Oxidation and reduction is the reaction in which one of the species donates eletron and another species accepts the electron .

As for the reaction involved in formation of Na Cl(s) , the reaction mechanism is as Na will release an electron from its outermost orbit to attain octet state forming Na+ cation.Likewise, the Cl atom will accept the electron forming Cl- anion to attain octet state. As a result, a ionic bond will be formed between them which is indeed a very strong chemical bond and finally after multiple bonding between them a crystalline structure will be formed.

Similarly even it is neutral it is presented in ionic form in its molecular structure mainly to specify that there is ionic linkage between these two atoms

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.