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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${E^o}_{\left( {{I_2}/{I^ - }} \right)},{E^ \circ }_{\left( {Br^-/B{r_2}} \right)}$ and ${E^ o }_{\left( {{Fe}/{Fe^{2+} }} \right)}$
are respectively $+ 0.54\ V$, $-1.09\ V$ and $0.44\ V$. On the basis of above data which of the following process is non-spontaneous
$(I)\,[Ni(CO)_4]$ $(II)\, [Mn(CN)_6]^{4-}$ $(III)\, [Cr(NH_3)_6]^{3+}$ $(IV) \,[CoF_6]^{3 -}$
Given $: \frac{2.303 RT }{ F }=0.06 V$
$Pd _{( aq )}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\circ}=0.83\,V$
$PdCl _4^{2-}( aq )+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-}( aq )$
$E ^{\circ}=0.65\,V$

