Question:

Match List - I with List - II : 

Choose the most appropriate answer from the options given below : 
 

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A quick way to find lone pairs on a central atom: Start with its valence electrons (e.g., 8 for Xe). Subtract the number of electrons required by the surrounding atoms (1 for halogens, 2 for oxygen). Divide the remainder by 2 to get the number of lone pairs.
Updated On: Feb 3, 2026
  • (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
  • (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
  • (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
  • (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
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The Correct Option is B

Solution and Explanation

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). 
 

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