- A$[Cr(H_2O)_6]^{3+}$
- ✓$[Fe(H_2O)_6]^{2+}$
- C$[Zn(H_2O)_6]^{2+}$
- D$[Cu(H_2O)_6]^{2+}$
The oxidation number of $Cr$ is $+3$
Atomic number $=24$
$Cr =[ Ar ] 3 d ^5 4 s ^1$
$Cr ^{3+}=[ Ar ] 3 d ^3 4 s ^0$
There are three unpaired electrons.
$\left[ Fe \left( H _2 O \right)_6\right]^{2+}$
The oxidation number of $Fe$ is $+2$.
Atomic number $=26$
$Fe =[ Ar ] 3 d ^6 4 s ^2$
$Fe e ^{2+}=[ Ar ] 3 d ^6 4 s ^0$
There are four unpaired electrons
$\left[ Cu \left( H _2 O \right)_6\right]^{2+}$
The oxidation number of $Cu$ is $+2$.
Atomic number $=29$
$Cu =[ Ar ] 3 d ^{10} 4 s ^1$
$Cu ^{2+}=[ Ar ] 3 d ^9 4 s ^0$
There is one unpaired electron.
$\left[ Zn \left( H _2 O \right)_6\right]^{2-}$
The oxidation number of $Zn$ is $+2$.
Atomic number $=30$
$Zn =[ Ar ] 3 d ^{10} 4 s ^2$
$Zn ^{2+}=[ Ar ] 3 d ^{10} 4 s ^0$
There are no unpaired electrons.
$\left[ Fe \left( H _2 O \right)_6\right]^{2+}$ has the maximum number of unpaired electrons; therefore, the highest degree of paramagnetism.
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$2N{H_3}(g)\xrightarrow[{Pt\,catalyst}]{{1130\,K}}{N_2}(g) + 3{H_2}(g)$
Choose the option which is not true with respect to the reaction, from the choices given below :-