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JEE Main 202228 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
28 July
Shift
Morning Shift
Subject
Chemistry
Chapter
p Block Elements (Group 15, 16, 17 & 18)

JEE Main 2022 · 28 July · Morning Shift · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

Match List-I with List-II. List-I List-II A 4 NH _ 3 (g) + 5 O _ 2...

Match List-I with List-II.

List-I List-II
A 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4{NH}_{3}\left(g\right) + 5{O}_{2}\left(g\right) \to 4NO\left(g\right) + 6{H}_{2}O\left(g\right) I NO(g)NO\left(g\right)
B N2(g)+3H2(g)→2NH3(g){N}_{2}\left(g\right) + 3{H}_{2}\left(g\right) \to 2{NH}_{3}\left(g\right) II H2SO4(l){H}_{2}{SO}_{4}\left(l\right)
C C12H22O11(aq)+H2O(l)→C6H12O6(Glucose)+C6H12O6(Fructose){C}_{12}{H}_{22}{O}_{11}\left(aq\right) + {H}_{2}O\left(l\right) \to {C}_{6}{H}_{12}{O}_{6}\left(Glucose\right) + {C}_{6}{H}_{12}{O}_{6}\left(Fructose\right) III Pt(s)Pt\left(s\right)
D 2SO2(g)+O2(g)→2SO3(g)2{SO}_{2}\left(g\right) + {O}_{2}\left(g\right) \to 2{SO}_{3}\left(g\right) IV Fe(s)Fe\left(s\right)

Choose the correct answer from the options given below

Options

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

Correct answer

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

Step-by-step solution

(a)
4NH3(g)+5O2(g)⟶Pt(s)4NO(g)+6H2O(g)4{NH}_{3}\left(g\right) + 5{O}_{2}\left(g\right)\overset{Pt\left(s\right)}{⟶}4NO\left(g\right) + 6{H}_{2}O\left(g\right)
Ostwald process 500K500K

(b) N2+3H2⟶Fe(s)2NH3(g){N}_{2} + 3{H}_{2}\overset{Fe\left(s\right)}{⟶}2{NH}_{3}\left(g\right)
Haber's process

(c)C12H22O11{C}_{12}{H}_{22}{O}_{11}\left(aq\right) + {H}_{2}O(l)⟶H+C6H12O6(glucose)+C6H12O6(fructose)\left(l\right)\overset{{H}^{+}}{⟶}\mathop{{C}_{6}{H}_{12}{O}_{6}}_{\left(glucose\right)} + \mathop{{C}_{6}{H}_{12}{O}_{6}}_{\left(fructose\right)}
Inversion of sugar cane

(d) 2SO2(g)+O2(g)⟶NO(g)2SO3(g)2{SO}_{2}\left(g\right) + {O}_{2}\left(g\right)\overset{NO\left(g\right)}{⟶}2{SO}_{3}\left(g\right)

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