show the formation pf Na2O by transfer of electron
Answers
Answer:
It is two electrons short of octet configuration, thus it will gain two electrons. During the bond formation between Sodium and Oxygen, two Sodium atoms loses one electron each to Oxygen atom.
Explanation:
Sodium has a tendency to lose the valence electron and oxygen has a tendency to gain the electron lost by sodium. Since sodium can lose only one electron of the valence shell and oxygen atom needs two electrons to complete its octet in the valence shell, two atoms of sodium combine with one atom of oxygen. By losing valence electron, sodium is changed into Na+ and by gaining two electrons lost by two sodium atoms, oxygen atom is changed into an oxide anion, O2-. In this process, both the atoms, sodium and oxygen, obtain the stable electronic configuration of the noble gas neon.The oppositely charged sodium ion, Na+ and oxide ion, O2- are now held together by electrostatic forces of attraction or by ionic or electrovalent bond. Na2O is, therefore, an ionic or electrovalent compound.
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If N2O4 decompise to60% into NO2 at 90'Cel then vapour denisty of equilibrium mixture wiill be 57.5 28.5 115 49
