Match List I with List II
| LIST I | LIST II | ||
| A | XeF4 | I | See-saw |
| B | SF4 | II | Square planar |
| C | NH4+ | III | Bent T-shaped |
| D | BrF3 | IV | Tetrahedral |
Choose the correct answer from the options given below :

Correct answer is (C): A-II, B-I, C-IV, D-III
A-I, B-I, C-IV, D-III.
In this matching, the structures of various molecules are being analyzed based on the number of bonding pairs and lone pairs. The tetrahedral geometry of \( \text{NH}_4^+ \) is typical for molecules with four bonding pairs and no lone pairs. \( \text{BrF}_3 \), with its lone pairs, adopts a bent T-shaped structure, while \( \text{XeF}_4 \) adopts a square planar structure due to the presence of two lone pairs and four bonding pairs.
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
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 