Question

Answer the following questions on this basis.
Which is correct matching of salt X?

A.$NiB{r_2}$
B.$CuB{r_2}$
C.$Cr{\left( {N{O_3}} \right)_2}$
D.$Cu{\left( {N{O_3}} \right)_2}$

Answer 1

Copper is a metal and can easily lose electrons and exist as a cation. Copper ion treated with sodium hydroxide solution forms a blue color precipitate. Further addition of ammonia leads to deep blue color. The nitrite or nitrate group when treated with sulphuric acid forms fumes.

Complete answer:
Chemical elements may lose electrons or gain electron to get nearest noble gas configuration. When an element loses electron, it becomes a positive ion called as cation. This cation involves in the bond formation with a negative ion called anion and form molecules or compounds.
Copper can exist as cation and forms a compound copper nitrate with molecular formula of $Cu{\left( {N{O_3}} \right)_2}$ by bonding with nitrate group.
Copper ion when treated with sodium hydroxide gives an insoluble precipitate and the addition of ammonia forms a deep blue coloured precipitate which is soluble in sodium hydroxide.
The nitrate group is existing an anion, when treated with concentrated sulphuric acid brown fumes were evolved. When these brown fumes were got into contact with starch paper blue spots will appear instantaneously on starch paper.
Thus, the cation is copper and anion is nitrate.
Thus, the compound X is copper nitrate $\left( {Cu{{\left( {N{O_3}} \right)}_2}} \right)$.

Option D is the correct one.

Note:
The copper can itself does not react with sodium hydroxide but the presence of ammonia or water can be able to soluble copper in sodium hydroxide. The nitrogen dioxide gas will be released in the presence of copper and sulphuric acid. This can be done when the compound has nitrate or nitrite groups.