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JEE Main 202225 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Bonding and Molecular Structure

JEE Main 2022 · 25 July · Evening Shift · Chemistry · Chemical Bonding and Molecular Structure

Match List I with List II List-I List-II A XeO _ 3 I sp ^ 3 d ;...

Match List I with List II

List-I List-II
A XeO3{XeO}_{3} I sp3d{sp}^{3}d; linear
B XeF2{XeF}_{2} II sp3{sp}^{3}; pyramidal
C XeOF4{XeOF}_{4} III sp3d3{sp}^{3}{d}^{3}; distorted octahedral
D XeF6{XeF}_{6} IV sp3d2{sp}^{3}{d}^{2}; square pyramidal

Choose the correct answer from the options given below

Match List I with List II List-I List-II A XeO _ 3 I sp ^ 3 d ; linear B XeF _ 2 II sp ^ 3 ; pyramidal C XeOF _ 4 III

Options

  1. AA−II,B−IV,C−III,D−IA - II, B - IV, C - III, D - I
  2. BA−IV,B−II,C−III,D−IA - IV, B - II, C - III, D - I
  3. CA−II,B−I,C−IV,D−IIIA - II, B - I, C - IV, D - III
  4. DA−IV,B−II,C−I,D−IIIA - IV, B - II, C - I, D - III

Correct answer

C. A−II,B−I,C−IV,D−IIIA - II, B - I, C - IV, D - III

Step-by-step solution

Match List I with List II List-I List-II A XeO _ 3 I sp ^ 3 d ; linear B XeF _ 2 II sp ^ 3 ; pyramidal C XeOF _ 4 III

XeOF4{XeOF}_{4}

S.N=12(8+4)=6=sp3d2S . N = \dfrac{1}{2}\left(8 + 4\right) = 6 = {sp}^{3}{d}^{2} square pyramidal

XeO3{XeO}_{3}

12(8)=4=sp3\dfrac{1}{2}\left(8\right) = 4 = {sp}^{3} Pyramidal

XeF6{XeF}_{6}

12(8+6)=7=sp3d3\dfrac{1}{2}\left(8 + 6\right) = 7 = {sp}^{3}{d}^{3} distorted octahedral

XeF2{XeF}_{2}

12(8+2)=5=sp3d\dfrac{1}{2}\left(8 + 2\right) = 5 = {sp}^{3}d linear

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