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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$Z{n^{2 + }} + 2e \to Zn,{E^o} = - 0.762\,V$
$M{g^{2 + }} + 2e \to Mg,{E^o} = - 2.37\,V$
When zinc dust is added to the solution of $MgC{l_2}$
Spin only magnetic moment of $Fe$ in $\left[ Fe \left( H _2 O \right)_6\right]^{3+}$ and $\left[ Fe ( CN )_6\right]^{3-}$ complexes respectively is:
$_H^{Br} > \mathop C\limits_1 = \mathop C\limits_2 < _H^{Br}$ $\xrightarrow[{Catalyst}]{{{H_2}}}Br\mathop {C{H_2}}\limits_3 \mathop {C{H_2}}\limits_4 Br$

In a closed vessel follows first order kinetics. If starting with pure $N_2O(g)$ total pressure after time $'t'$ is $P_t$ and after a very long time is $P_\infty$ then which of the following expressions is correct