Question

Write the cations and anions present in the following compounds:
i) NaCl,
ii) CH3COONH4.
iii) H2 iv) NH4NO3​

Answer 1

(a) CH

3

COONa

Cation - Na

+

Anion - CH

3

COO

−

(b) NaCl

Cation - Na

+

Anion - Cl

−

(c) H

2

Cation - None

Anion - None

(d) NH

4

NO

3

Cation - NH

4

+

Anion - NO

3

−

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Answer 2

Answer:

I) $NaCl$ - cation: $Na^{+}$ and anion: $Cl^{-}$.

II) $CH_{3}COONH_{4}$ - cation: $NH_{4}^{+}$ and anion: $CH_{3}COO^{-}$.

III) $H_{2}$ - no cation and no anion.

IV) $NH_{4}NO_{3}$ - cation: $NH_{4}^{+}$ and anion: $NO_{3}^{-}$.

Explanation:

Cations:

• They are positively charged ions.

Anions:

• They are negatively charged ions.

Now, in the given compounds, the cations and anions are:

I) $NaCl$:

• The dissociation of $NaCl$ occurs in the way given below:
• $NaCl \rightarrow Na^{+} + Cl^{-}$

Thus,

• The positively charged ion $Na^{+}$ is cation.
• The negatively charged ion $Cl^{-}$ is anion.

II) $CH_{3}COONH_{4}$

• The dissociation of $CH_{3}COONH_{4}$ occurs in the way given below:
• $CH_{3}COONH_{4} \rightarrow CH_{3}COO^{-} + NH_{4}^{+}$

Thus,

• The positively charged ion $NH_{4}^{+}$ is cation.
• The negatively charged ion $CH_{3}COO^{-}$ is anion.

III) $H_{2}$

• $H_{2}$ is a neutral molecule; it is formed by the covalent bond.
• $H_{2} \rightarrow 2H$

Thus,

• It does not contain cation and anion.

IV) $NH_{4}NO_{3}$

• The dissociation of $NH_{4}NO_{3}$ occurs in the way given below:
• $NH_{4}NO_{3} \rightarrow NH_{4}^{+} + NO_{3}^{-}$

Thus,

• The positively charged ion $NH_{4}^{+}$ is cation.
• The negatively charged ion $NO_{3}^{-}$ is anion.