Question

Cation and anion present (if any) in NaCl is :

• A. Anion: $C{l}^{2-}$ Cation: $N{a}^{+}$
• B. Anion: $C{l}^{-}$ Cation: $N{a}^{2+}$
• C. Cation: $C{l}^{-}$ Anion: $N{a}^{+}$
• D. Anion: $C{l}^{-}$ Cation: $N{a}^{+}$

Answer 1

The correct option is D Anion: $C{l}^{-}$ Cation: $N{a}^{+}$

Separate the compound into its cation and anion.

$NaCl\to N{a}^{+}+C{l}^{-}$
Cation$\to N{a}^{+}$
Anion $\to C{l}^{-}$