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.
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
