\(H_2O_2\) behave like reducing agent in which of the following reactions:
\(Fe^{+2}+ H_2O_2 → Fe^{+3} + H_2O\)
\(H_2S+H_2O_2 → SO^{2-}_4+H_2O\)
\(HOCl + H_2O_2 → Cl^- + 2H_2O + O_2\)
\(Mn^{+2} + H_2O_2 → MnO_2 + H_2O\)
\(H_2O_2\) reduces \(HOCl\) to \(Cl^-\) and itself gets oxidised to \(O_2.\)
The correct option is (c ): \(HOCl + H_2O_2 → Cl^- + 2H_2O + O_2\)
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
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: