Match List-I (Compounds) with List-II (Oxidation states of nitrogen) and select answer using the codes given below List-I (Molecule) List-II (Oxidation state) (A) NaN₃ (P) +5 (B) N₂H₂ (Q) +2 (C) NO (R) −1/3 (D) N₂O₅ (S) −1

Oxidation number of N in NaN 3 , N 2 H 2 , NO , N 2 O 5 are − 3 1 , − 1 , + 2 , + 5 respectively.

Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsDetermination of Oxidation StateMedium2 minQB

ChemistryMediummatching list

Match List-I (Compounds) with List-II (Oxidation states of nitrogen) and select answer using the codes

given below

Answer:

Solution:

Oxidation number of N in $NaN_{3}, N_{2} H_{2}, NO, N_{2} O_{5}$ are $- \frac{1}{3}, - 1, + 2, + 5$ respectively.

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Determination of Oxidation State

⏱ 2mℹ️ Source: QB

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