$A$ compound contains atoms $X$,$Y$,and $Z$. The oxidation number of $X$ is $+2$,$Y$ is $+5$ and $Z$ is $-2$. Therefore,a possible formula of the compound is

  • A
    $XYZ_2$
  • B
    $X_2(YZ_3)_2$
  • C
    $X_3(YZ_4)_2$
  • D
    $X_3(Y_4Z)_2$

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$1 \ mol$ 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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Match the following Column-$I$ to Column-$II$ based on the oxidation state of the underlined nitrogen atom:
Column-$I$Column-$II$
$A$. $\underline{N}H_4 Cl$$P$. $+1$
$B$. $H\underline{N}O_3$$Q$. $+5$
$C$. $\underline{N}_2 O$$R$. $-3$
$D$. $\underline{N}H_2 OH$$S$. $-1$

The oxidation number of nitrogen in $NH_2OH$ is

When $K_2Cr_2O_7$ is converted to $K_2CrO_4$,the change in the oxidation state of chromium is

Oxidation number of oxygen in potassium superoxide $(KO_2)$ is

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