Show the formation of sodium oxide?
Answers
Answer:
Answer:
The ions present in sodium oxide compound (Na2O) are sodium ions (2Na+) and oxide ions (O2-).
The atomic number of Sodium is 11
The electronic configuration of Sodium is
Na(Z=11) = 1s22s22p63s1
In order to obtain an octet configuration it will lose one electron.
The atomic number of oxygen is 8
The electronic configuration of oxygen is
O(Z = 8) = 1s22s22p4
In order to obtain an octet configuration it has to gain two electrons.
Two sodium atoms transfer their 2 outermost electrons to an oxygen atom. By losing two electrons, the two sodium atoms form tow sodium ions (2Na+). And by gaining two electrons, the oxygen atom forms an oxide ion (O2-)
2 Na+ + O → Na2O
Answer:
The ions present in sodium oxide compound (Na2O) are sodium ions (2Na+) and oxide ions (O2-).
The atomic number of Sodium is 11
The electronic configuration of Sodium is
Na(Z=11) = 1s22s22p63s1
In order to obtain an octet configuration it will lose one electron.
The atomic number of oxygen is 8
The electronic configuration of oxygen is
O(Z = 8) = 1s22s22p4
In order to obtain an octet configuration it has to gain two electrons.
Two sodium atoms transfer their 2 outermost electrons to an oxygen atom. By losing two electrons, the two sodium atoms form tow sodium ions (2Na+). And by gaining two electrons, the oxygen atom forms an oxide ion (O2-)
2 Na+ + O → Na2O