STD 12 - 5. Co-ordination chemistry — JEE STD 11 Science — Question

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.