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Which type of bonding is present in ${{\left( N{{H}_{4}} \right)}_{2}}C{{O}_{3}}$
1. Ionic
2. Covalent
3. Co-ordinate (Dative)

A.) 1,2 and 3 are correct
B.) 1 and 2 are correct
C.) 2 and 3 are correct
D.) 1 is correct

Answer
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Hint: Ionic bonding involves force of attraction between opposite charges.
- Covalent bond is formed from mutual sharing of electrons to form a bond.
- Co-ordinate bond is formed when there is a lone pair of electrons which is being donated to the acceptor atom to form a bond.
- In ammonium carbonate ${{\left( N{{H}_{4}} \right)}_{2}}C{{O}_{3}}$, all three types of bonding are present.

Complete Solution :
-vAs we can see the formula for ammonium carbonate it is formed by the attraction of $N{{H}_{4}}^{+}$ ammonium ion and $C{{O}_{3}}^{2-}$ by ionic bonding as the charge on carbonate ion is $-2$ so to balance it we need two ammonium ions to form ammonium carbonate.
- It also involves covalent bonding the formation of ammonium ion in which the bond present between nitrogen and hydrogen is a covalent bond and similarly in carbonate ion the bond between carbon and oxygen is a covalent bond.
- Ammonium carbonate also has a co-ordinate bond. The ammonium ion is formed when ammonia with its lone pair on nitrogen donates it to a proton $\left( {{H}^{+}} \right)$ and forms ammonium ion.
Hence ammonium carbonate has all the three bonds present.
So, the correct answer is “Option A”.

Note: Ammonium carbonate readily degrades to gaseous ammonia and carbon dioxide upon heating.
- It is also known as baker's ammonia and was a predecessor to the more modern leavening agents baking soda and baking powder. It is also used as a leavening agent.