Compound $\to$ Oxidation state
- A$[Co{(N{H_3})_5}Cl]C{l_2}$ $→$ $0$
- B$N{H_2}OH$$ → $ $- 1$
- ✓${({N_2}{H_5})_2}S{O_4}$$ → $ $ + 2$
- D$M{g_3}{N_2}$$ → $ $ - 3$
Compound $\to$ Oxidation state
Let oxidation number of $N = x$
Oxidation number of $NH _3=0$
$x +1(3)=0$
$x =-3 \text { is correct. }$
$2.$ $NH _2 OH$
Let oxidation number of $N = x$
Oxidation number of $H =+1$
Oxidation number of $O =-2$
$x+3(+1)+(-2)=0$
$x =-1$ is correct
$\text { 3. }\left( N _2 H _5\right)_2 SO _4$
Let oxidation number of $N = x$
Oxidation number of $H =+1$
Oxidation number of $SO _4=-2$
$2(2( x )+5(+1))+(-2)=0$
$x =-2$ is incorrect
$4$.$Mg _3 N _2$
Let oxidation number of $N = x$
Oxidation number of $Mg =+2$
$2(x)+3(+2)=0$
$x =-3$ is correct
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Statement $(I)$ : $p-$nitrophenol is more acidic than $m-$nitrophenol and $o-$nitrophenol.
Statement $(II)$ : Ethanol will give immediate turbidity with Lucas reagent.
In the light of the above statements, choose the correct answer from the options given below :
$E^o_{Cl_2/Cl^-} = 1.36\,V,$ $E^o_{Cr^{3+}/Cr}= -0.74\,V,$
$E^o_{Cr_2/O_7^{2-}/Cr^{3+}}=1.33\,V,$ $E^o_{MnO^-_4/Mn^{2+}} = 1.51\,V$
Among the following, the strongest reducing agent is
