7. Explain the formation of the bond in NaCl in terms of the octet rule.
Answers
Sodium (Na) has 1 electron in the 3s orbital and chlorine (Cl) has 7 electrons in the 3p orbital. Sodium gives away the one electron to Cl, leaving it with 8 electrons (octet) in the 2p orbital (like Neon). The chlorine takes that one electron giving it 8 electrons (octet) in the 3p orbital. The sodium then has a +1 charge, and the chloride ion now has a -1 charge. This is an ionic bond.
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".