Match List I with List II 
Choose the correct answer from the options given below:
Step 1: Determine the Lone Pairs for Each Molecule/Ion
The number of lone pairs on the central atom can be determined using the following steps:
Count the total valence electrons of the central atom.
Subtract the electrons used for bonding with surrounding atoms.
Divide the remaining electrons by 2 to get the number of lone pairs.
Step 2: Analyze Each Molecule/Ion
\(\text{IF}_7\): Iodine has 7 valence electrons. All are used for bonding with 7 fluorine atoms. Therefore, 0 lone pairs (IV).
\(\text{ICl}_4^-\): Iodine has 7 valence electrons and gains 1 due to the negative charge. Four are used for bonding with chlorine atoms, leaving 4 electrons (2 lone pairs). Therefore, 2 lone pairs (III).
\(\text{XeF}_6\): Xenon has 8 valence electrons. Six are used for bonding with fluorine atoms, leaving 2 electrons (1 lone pair). Therefore, 1 lone pair (II).
\(\text{XeF}_2\): Xenon has 8 valence electrons. Two are used for bonding with fluorine atoms, leaving 6 electrons (3 lone pairs). Therefore, 3 lone pairs (I)}
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 :
