| LIST I Elements | LIST II Colour imparted to the flame | ||
| A | K | 1 | Brick Red |
| B | Ca | 2 | Violet |
| C | Sr | 3 | Apple Green |
| D | Ba | 4 | Crimson Red |
Correct answer is (d) A-II, B-I, C-IV, D-III

The flame test is used to identify elements based on the characteristic colors they emit when heated. The correct matching is:
Thus, the correct matching corresponds to option (4).
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
For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:

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: