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