Match List - I with List - II : 
Choose the most appropriate answer from the options given below :
Step 1: Understanding the Question:
We need to determine the number of lone pairs of electrons on the central Xenon (Xe) atom in four different compounds and match them accordingly.
Step 2: Key Formula or Approach:
We can use the VSEPR theory formula to find the number of electron pairs and then the number of lone pairs. Number of electron pairs (Steric Number, SN) = \(\frac{1}{2}\) [ (Valence electrons of central atom) + (No. of monovalent atoms) - (Charge on cation) + (Charge on anion) ] Number of lone pairs (LP) = SN - (Number of surrounding atoms)
Alternatively, we can directly count the electrons. Xenon (a noble gas) has 8 valence electrons.
Step 3: Detailed Explanation:
(a) XeF\(_2\): - Valence electrons of Xe = 8. - Electrons used in bonding with 2 Fluorine atoms = 2 \(\times\) 1 = 2. - Remaining non-bonding electrons = 8 - 2 = 6. - Number of lone pairs = 6 / 2 = 3. - Match: (a)-(iv).
(b) XeO\(_2\)F\(_2\): - Valence electrons of Xe = 8. - Electrons used in bonding with 2 Oxygen atoms (double bonds) = 2 \(\times\) 2 = 4. - Electrons used in bonding with 2 Fluorine atoms (single bonds) = 2 \(\times\) 1 = 2. - Total electrons used in bonding = 4 + 2 = 6. - Remaining non-bonding electrons = 8 - 6 = 2. - Number of lone pairs = 2 / 2 = 1. - Match: (b)-(ii).
(c) XeO\(_3\)F\(_2\): - Valence electrons of Xe = 8. - Electrons used in bonding with 3 Oxygen atoms (double bonds) = 3 \(\times\) 2 = 6. - Electrons used in bonding with 2 Fluorine atoms (single bonds) = 2 \(\times\) 1 = 2. - Total electrons used in bonding = 6 + 2 = 8. - Remaining non-bonding electrons = 8 - 8 = 0. - Number of lone pairs = 0. - Match: (c)-(i).
(d) XeF\(_4\): - Valence electrons of Xe = 8. - Electrons used in bonding with 4 Fluorine atoms = 4 \(\times\) 1 = 4. - Remaining non-bonding electrons = 8 - 4 = 4. - Number of lone pairs = 4 / 2 = 2. - Match: (d)-(iii).
Step 4: Final Answer:
The correct set of matches is (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii). This corresponds to option (B).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)