Question

(6) Show the formation of Na2O and MgO by the transfer of electrons

Answer 1

Answer: The formation of both the compounds are given below.

Explanation:

The given compounds are ionic compounds.

Ionic compounds are defined as the compounds which are formed by the complete transfer of electrons from one atom to another atom. In these compounds, oppositely charged ions are attracted towards each other to form a compound.

The atom which looses electrons is known as electropositive atom and it attains positive charge. The atom which gains electrons is known as electronegative atom and it attains negative charge.

These bonds are usually formed between a non-metal and a metal.

• For $Na_2O$

Sodium is the 11th element of the periodic table having 11 electrons. The electronic configuration of this element is $1s^22s^22p^63s^1$

This will loose 1 electron to form $Na^+$ ion.

Oxygen is the 8th element of the periodic table having 8 electrons. The electronic configuration of this element is $1s^22s^22p^4$

This will gain 2 electrons to form $O^{2-}$ ion.

To form $Na_2O$ compound, 2 sodium ions are needed to neutralize the charge on oxygen ion.

• For $MgO$

Magnesium is the 12th element of the periodic table having 12 electrons. The electronic configuration of this element is $1s^22s^22p^63s^2$

This will loose 2 electron to form $Mg^{2+}$ ion.

Oxygen is the 8th element of the periodic table having 8 electrons. The electronic configuration of this element is $1s^22s^22p^4$

This will gain 2 electrons to form $O^{2-}$ ion.

To form $MgO$ compound, 1 magnesium ion is needed to neutralize the charge on oxygen ion.

The formation of both the compounds are given below.

Answer 2

For the formation of Na2O: It is an ionic compound and sodium has 11 atomic number and oxygen has 8 atomic number. Sodium has one valance electron whereas oxygen has 6. Sodium will lose the one valance electron to full fill his octet whereas oxygen will gain this one electron but to complete the octet of oxygen it needs two electrons. Therefore two atom of sodium will combine with one atom of oxygen and each sodium will lose one electron to oxygen and therefore completing its octet.

$\mathrm{O}+2 \mathrm{e}-\rightarrow \mathrm{O}^{2-} \text { and } \mathrm{Na} \rightarrow \mathrm{Na}^{+}+\mathrm{e}$

Therefore, $2 \mathrm{Na}+\mathrm{O}^{2-} \rightarrow 2 \mathrm{Na}^{+} \mathrm{O}^{2-}$

Formation of MgO: Magnesium has two electrons in its outer shell so it will lose two electron to complete its octet and oxygen has six electron in the outer shell so it will take those two electrons to complete its octate.  

$\mathrm{Mg} \rightarrow \mathrm{Mg} 2++2 \mathrm{e}-\text { and } \mathrm{O}+2 \mathrm{e}-\rightarrow \mathrm{O}^{2-}$

Therefore,  

$\mathrm{Mg}^{+2}+\mathrm{O}^{2-} \rightarrow \mathrm{MgO}$