Which of the following combination will produce `H_(2)` gas ?

Options

A. Cu metal and conc. `HNO_(3)`
B. Zn metal and NaOH (aq)
C. Au metal and NaCN (aq) in the presence of air
D. Fe metal and conc. `HNO_(3)`

Answer

B

Explanation

Nitric being a strong oxidising agent, oxidises initially Cu to `CuO` which further reacts with `HNO_(3)` to form `Cu(NO_(3))_(2)` with the evolution of `NO_(2)` gas. However , `H_(2)` gas is not produced.
Fe becomes passive in conc. `HNO_(3)` due to the formation of an impervious layer of iron oxide on its surface which prevents the further action. Au reacts with NaCN in presence of air to form soluble complex of dicyanoaurate (I)
`underset("Dicyanoaurate "(I))(4Au + 8 NaCN+ 2H_(2)O + O_(2) to 4Na [Au(CN)_(2)]+4NaOH`)
Only Zn reacts withe NaOH solution to evolve `H_(2)` gas
`Zn+2NaOH overset(Delta)to underset("Sod. zincate")(Na_(2)ZnO_(2))+H_(2)`