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CUET · CHEMISTRY · PYQ PAPER 2023

Answer the question on the basis of passage given below :
Nitrogen differs from rest of the members of group 15 due to its small size, high electronegativity, high ionization enthalpy and non-availability of d orbitals. Nitrogen has unique ability to form pa - pa multiple bonds with itself and with other elements having small size and high electronegativity. Heavier elements of this group do not form pa - pr bonds as their atomic orbitals are so large and diffuse that they cannot have effective overlapping.|
Match List I with List II.
LIST I (Nitrogen oxides)LIST II (Oxidation state)
A. \(N _2 O\)\(I. +3\)
B. \(N _2 O _3\)\(II. +5\)
C. \(N _2 O _4\)\(III. +1\)
D. \(N _2 O _5\)\(IV. +4\)

Choose the correct answer from the options given below:

  1. A \(A-III, B-IV, C-II, D-I\)
  2. B \(A-III, B-I, C-IV, D-II\)
  3. C \(A-IV, B-II, C-III, D-I\)
  4. D \(A-III, B-II, C-I, D-IV\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(A-III, B-I, C-IV, D-II\)

Step-by-step Solution

Detailed explanation

A. \(N_2O\): \(2x + (-2) = 0 \Rightarrow x = +1\) B. \(N_2O_3\): \(2x + 3(-2) = 0 \Rightarrow x = +3\) C. \(N_2O_4\): \(2x + 4(-2) = 0 \Rightarrow x = +4\) D. \(N_2O_5\): \(2x + 5(-2) = 0 \Rightarrow x = +5\) A-III, B-I, C-IV, D-II
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