Question

Consider the following sequence of conversion. X, Y, and Z can be described as:
$\underset{\left(X\right)}{2NO+O_2}\rightleftharpoons \underset{\left(Y\right)}{2N{O}_2}\begin{array}{c}cool\\ \rightleftharpoons \\ heat\end{array}\underset{\left(Z\right)}{N_2{O}_4}$

• A. (X) Colourless (Y) Brown,paramagnetic (Z) Colourless,diamagnetic
• B. (X) Brown (Y) Colourless, diamagnetic (Z) Brown, paramagnetic
• C. (X) Colourless (Y) Colourless, paramagnetic (Z) Brown, diamagnetic
• D. (X) Brown (Y) Brown, paramagnetic(Z) Brown, diamagnetic

Answer 1

Answer: (A)

(X) Colourless (Y) Brown,paramagnetic (Z) Colourless,diamagnetic

$NO$ is colorless.

$N{O}_2$ is brown due to the presence of unpaired $e^{-}$ for which it is paramagnetic in nature. $N{O}_2$ is an odd-$e^{-}$ molecule.

Whereas $N_2{O}_4$ is colorless as it is diamagnetic in nature due to the presence of paired $e^{-}s$. Also, $N_2{O}_4$ is a dimer of $N{O}_2$.