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 : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,