| List-I (Oxoacids of Sulphur) | List-II (Bonds) | ||
| A | Peroxodisulphuric acid | I | Two S–OH, Four S=O, One S–O–S |
| B | Sulphuric acid | II | Two S–OH, One S=O |
| C | Pyrosulphuric acid | III | Two S–OH, Four S=O, One S–O–O–S |
| D | Sulphurous acid | IV | Two S–OH, Two S=O |
The task is to match oxoacids of sulphur in List-I with their corresponding bond structures in List-II. We will analyze each oxoacid and bond description to find matches.
| List-I (Oxoacids of Sulphur) | List-II (Bonds) | ||
| A | Peroxodisulphuric acid | I | Two S–OH, Four S=O, One S–O–S |
| B | Sulphuric acid | II | Two S–OH, One S=O |
| C | Pyrosulphuric acid | III | Two S–OH, Four S=O, One S–O–O–S |
| D | Sulphurous acid | IV | Two S–OH, Two S=O |
Let's match them:
Therefore, the correct option is:
Given below are two statements. 
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements:
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
In the light of the above statements, choose the correct answer from the options given below:
Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $(\mathrm{B}<\mathrm{Al}),(\mathrm{Al}<\mathrm{Ga}),(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $\left(\mathrm{M}^{3+}\right)$ than the other one. The atomic number of the element (X) is
