Question

$2N{O_2}\overset {} \leftrightarrows {N_2}{O_4}$. The dimerization of $N{{O}_{2}}$ is accompanied with:
(A) Decrease in paramagnetism
(B) Change in colour
(C) Increase in temperature
(D) Increase in paramagnetism

Answer 1

To answer this question we should know to predict whether compound is paramagnetic or diamagnetic. We should know what will be the consequence if a compound is paramagnetic or diamagnetic.

Complete answer:
Firstly, let's write the atomic number and electronic configuration of nitrogen and oxygen:
The atomic number of nitrogen is 7.
The electronic configuration of nitrogen is $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$.
The atomic number of Oxygen is 8
The electronic configuration of oxygen is $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$
We can easily know whether a compound is paramagnetic or diamagnetic by calculating .
Let's see how,
We know that only valence electrons participate in chemical bonding.
-$N{{O}_{2}}$
Valence electrons of nitrogen makes chemical bonding with the electrons of oxygen.
Let's sum the number of valence electrons of nitrogen and oxygen.
5 + 6 = 11
11 is an odd number, this means 1 electron in $N{{O}_{2}}$ is unpaired. The compound which has unpaired electrons is said to be paramagnetic in nature.
-${{N}_{2}}{{O}_{4}}$
Valence electrons of nitrogen makes chemical bonding with the electrons of oxygen.
Let's sum the number of valence electrons of nitrogen and oxygen.
5 $\times$2 + 6 $\times$4 = 34
34 is an even number, this means no electron in ${{N}_{2}}{{O}_{4}}$ is unpaired. The compound which has no unpaired electron is said to be diamagnetic in nature. We can observe that there is a decrease in paramagnetism from $N{{O}_{2}}$ to ${{N}_{2}}{{O}_{4}}$.
Therefore, the option A is the correct answer.
In the compounds which are paramagnetic in nature, the unpaired electron absorbs light of certain radiation and emits back which imparts some colour. The compound which is diamagnetic in nature does not possess this property as all the electrons are bonded. We can observe that there is will be change from some colour to colourless when it changes from $N{{O}_{2}}$ to ${{N}_{2}}{{O}_{4}}$.
Therefore, it is option B is the correct answer.

So, from the above discussion we can conclude that option A and B is the correct answer.

Note:
This will be a very easy method from an exam point of view as time will be conserved if the logic behind this method is understood. Only valence electrons take part in chemical bonding. That means according to our case the valance electrons of nitrogen and valence electrons of oxygen are involved in chemical bonding.