“Ag dissolves in dilute $HN{O_3}$ yielding a colourless gas.” Answer whether the above statement is true or false. If true enter 1, else enter 0.

Hint: The silver reacts with nitric acid to give silver (+I) ions along with NO gas and two molecules of water. The NO gas has unpaired electrons but that electron can not be easily emitted such that it can produce a colour visible to eyes.

Complete step by step solution:
For this question, first, let us see the reaction that is taking place.
The reaction of silver with nitric acid ($HN{O_3}$) can be written as -
$3Ag + 4{H^ + } + NO_3^ - \to 3A{g^ + } + NO + 2{H_2}O$
The silver dissolves in the dilute nitric acid to give $NO$ gas. The $NO$ gas is colourless in nature.
So, the statement that “Ag dissolves in dilute $HN{O_3}$ yielding a colourless gas” is true.

So, here one needs to enter 1.

Note: The silver gives its one electron, that is it is being reduced and the nitric acid is being oxidised here because the nitric acid has given out hydrogen that forms water on reaction with oxygen. On seeing the Molecular orbital diagram of $NO$; one sees the unpaired electron. Thus, $NO$ should be coloured because we have studied that the paramagnetic substances give colour while the diamagnetic are the colourless ones. But here this is not the case. The MO diagram shows that the electron cannot move i.e. it can not go anywhere. So, it can not be emitted in the visible region. And we see the colours of emission only in the visible region. So, $NO$ will not produce colours in the visible region.