Match List-I with List-II List-I
| List-I | List-I i |
| Reaction | Catalyst |
| \(4 NH _3( g )+5 O _2( g ) \rightarrow 4 NO ( g )+6 H _2 O ( g )\) | NO(g) |
| \(N _2( g )+3 H _2( g ) \rightarrow 2 NH _3( g )\) | H_2SO_4(l) |
| \(C _{12} H _{22} O _{11}( aq ) H _2 O (l) \rightarrow \text { (Glucose) }+ C _6 H _{12} O _6 C _6 H _{12} O _6\) | Pt(s) |
| \( SO _2( g )+ O _2( g ) \rightarrow 2 SO _3( g )\) | Fe(s) |
Choose the correct answer from the options given below:
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 symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in the figure. If the power of the original lens is 4D, then the power of a part of the divided lens is:
