Oxidation number of Nitrogen in $NH_4NO_3$ is
- A$+3$ and $+5$
- B$-3$ and $+5$
- C$+3$ and $-3$
- D$+1$ and $+3$
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Similar Questions
$1 \ mol$ of compound $N_2H_4$ loses $10 \ mol$ of electrons to form a new compound $Y$. Assuming all nitrogen atoms are present in compound $Y$,what is the oxidation state of $N$ in compound $Y$?
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View SolutionWhat are the oxidation states of both the chlorine atoms in bleaching powder $(CaOCl_2)$?
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View SolutionIn thiosulphuric acid $(H_2S_2O_3)$,which of the following statements is correct?
In alkaline condition,$KMnO_4$ reacts as $2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + [O]$. The equivalent weight of $KMnO_4$ would be (Atomic mass of $K = 39$,$Mn = 55$,$O = 16$).
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View SolutionMatch the Column-$I$ with Column-$II$ for the oxidation states of the central atoms.
Column-$I$ Column-$II$ $A$. $Cr_2O_7^{2-}$ $1$. $+3$ $B$. $MnO_4^{-}$ $2$. $+4$ $C$. $VO_3^{-}$ $3$. $+5$ $D$. $FeF_6^{3-}$ $4$. $+6$ $5$. $+7$
| Column-$I$ | Column-$II$ |
| $A$. $Cr_2O_7^{2-}$ | $1$. $+3$ |
| $B$. $MnO_4^{-}$ | $2$. $+4$ |
| $C$. $VO_3^{-}$ | $3$. $+5$ |
| $D$. $FeF_6^{3-}$ | $4$. $+6$ |
| $5$. $+7$ |
Medium
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