Question

What is the ionic bond formation of $NaCl$?
A) $Na\; \to \;\;N{a^ + } + {\text{ }}{e^ - }$ and $Cl{\text{ }} + {\text{ }}{e^ - } \to \;\;C{l^ - }$
B) $Na\; \to \;\;N{a^ + } + {\text{ }}{e^ - }$ and $Cl{\text{ }} \to \;\;C{l^ - } + {e^ - }$
C) Bond formed between $N{a^ + }$ and $C{l^ - }$ ions
D) Both A and C

Answer 1

The sodium chloride is a neutral compound as it has no charge, the ions due to which $NaCl$ is formed is $N{a^ + }$ and $C{l^ - }$.

Complete step by step answer:
We have to remember that the $NaCl$ is an ionic compound which is commonly called as rock salt and has a chemical name as Sodium chloride.
It is formed by addition of two ions which are $N{a^ + }$ and $C{l^ - }$ .These two ions are present in equal ratio in this moiety.
The reaction for the formation of sodium chloride is as follows:
$N{a^ + } + C{l^ - } \to NaCl$
Electronic configuration of$Na$: $1{s^2}2{s^2}2{p^6}3{s^1}$
Electronic configuration of$Cl$: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}$
$Na$ has one electron in excess to attain stable electronic configuration whereas $Cl$ is deficient of one electron to get its octet filled.
So this is how an ionic bond is formed in sodium chloride, when an electron from $Na$ is transferred to $Cl$making both the atoms attaining stable octet configuration will lead to ionic bond formation.
Electronic configuration of$N{a^ + }$:$1{s^2}2{s^2}2{p^6}$
Electronic configuration of$C{l^ - }$: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
So by looking at this electronic configuration we can predict that $Na$ lose an electron to attain electronic configuration whereas $Cl$gains an electron and this is how an ionic bond is formed between $N{a^ + }$ and $C{l^ - }$.
Option A) This option can be correct $Na$lose an electron to form $N{a^ + }$ whereas $Cl$gain an electron to form $C{l^ - }$ and that how an ionic bond is formed i.e. $N{a^ + } + C{l^ - } \to NaCl$.
Option B) This is an incorrect option as when $Cl$will lose an electron it will not attain stable electronic configuration therefore no bond is as there is no electron transfer between $Na$ and $Cl$.
Option C) This option can be correct as an ionic bond is formed between $N{a^ + }$ and $C{l^ - }$.
Option D) This is a correct option as it has both the options which are correct i.e. option A and C.

Hence, the correct answer is, ‘Option D’.

Note: $NaCl$is an ionic bond that is formed with the exchange of ions between sodium and chloride ions. We also remember that the sodium chloride is a white crystalline solid having melting point of $801^\circ C$ and also know that the sodium chloride is readily soluble in water and other polar solvents.