Question

explain in detail formation of bond in sodium chloride

Answer 1

                               Formation of NaCl:

1. Sodium has atomic number 11, shows the electronic configuration 2,8,1 to become complete by it's octet state, Sodium looses it's 1 electron and forms the sodium ion Na+ (2,8).

Sodium (Na) (2,8,1)  -------------> Sodium ion (Na+) (2,8)

2. Chlorine has atomic number 17, shows electronic configuration 2,8,7. To become complete by it's octet state, it accepts 1 electron shared by Sodium and forms the Chlorine ion Cl- (2,8,8)

Chlorine (Cl) (2,8,7) --------------> Chloride ion (Cl-) (2,8,8)

3. Due to the opposite charges, there is a attraction between this two ions and one new chemical bond is formed.

                           Na₊ + Cl₋ -------------> NaCl

About Sodium Chloride:

Sodium Chloride is a common salt which is generally referred as "common salt" also. Sodium Chloride is colorless and crystalline ionic compound. By the further reactions on Sodium Chloride, Various salts like Na2Co3 and NaHCO3 are produced.

i) NaCl is generally salty in taste.

ii) It is Colourless , Neutral salt.

iii) There is no water of crystallization in it's crystalline structure.

iv) When the NaCl is heated at 800 degree Celsius, then it melts. This state of salt is called as "Fused state of salt".

Answer 2

❁ $\sf{ \pink{\Huge{Explanation}}}$❁

The electronic configuration of a sodium atom is 2,8,1. It has one electron in excess of the stable electronic configuration of the nearest noble gas i.e., neon, ( 2, 8 ). Therefore, an atom of sodium shows a tendency to give up the electron from its outermost shell, so as to acquire stable electronic configuration of neon.

$\sf{ Na \longrightarrow {Na}^{ + } + {1e}^{ - } }$

Here,

◉ Na - ( 2, 8, 1 ) atom

◉ $\sf{ {Na}^{ + } - (2, \: 8) \: cation}$

However, after giving up one electron, the sodium atom is no more electrically neutral. It has 11 protons in its nucleus but only 10 electrons revolving around it. Therefore, it has a net positive charge of +1. This positively charged atom is called sodium ion and is written as $\sf{ {Na}^{ + } }$ and its electronic configuration resembles that of the noble gas - neon.

The electronic configuration of chlorine is 2,8,7.

It has an electronic configuration with one electron less than that of the nearest noble gas i.e., argon ( 2, 8, 8 ). Therefore, the chlorine atom shows a tendency to acquire an electron to attain octet in its outermost shell.

$\sf{Cl + {1e}^{ - } \longrightarrow {Cl}^{ - }}$

Here,

◉ Cl - ( 2, 8, 7 ) atom

◉ $\sf {Cl}^{ - } - (2, \: 8, \: 8) \: anion$

An atom of chlorine is electrically neutral and it has contains 17 protons in its nucleus and 17 electrons revolving around nucleus. But, after acquiring an electron from the sodium atom, the chlorine atom does not remain electrically neutral. It has one electron more than the number of protons in its nucleus and therefore it has a charge of -1 represented as $\sf {cl}^{ - }$ i.e., chloride ion and its electronic configuration resembles that of the noble gas - argon.

Thus, when an atom of sodium combines with an atom of chlorine, "one" electron is transferred from the sodium atom to the chlorine atom, resulting in the formation of a sodium chloride molecule.

The cation $\sf {Na}^{ + }$ and anion $\sf {Cl}^{ - }$ are attached towards each other due to opposite electrical charges or coulomb force existing between them and form an ionic compound.

Refer the attachment above for orbit structure of electrovalent or ionic bonding in sodium chloride.

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