Question

Among the species given below, the total number of diamagnetic species is

$H$ atom, $N{O}_2$ monomer, $O_2^{-}$ superoxide, dimeric sulphur in vapour phase, $M{n}_3{O}_4,\left(N{H}_4{)}_2\right[FeC{l}_4],(N{H}_4{)}_2\left[NiC{l}_4\right],K_{2}Mn{O}_4,K_{2}Cr{O}_4$

• A. 1
• B. 1.0
• C. 1.00

Answer 1

Answer: (A), (B), (C)

Among the given species only $K_{2}Cr{O}_4$ is diamagnetic as central metal atom $Cr$ in it has $\left[Ar\right]3d^0$ electronic configuration i.e., all paired electrons. The structure and oxidation state of central metal atom of this compound are as follows
Structure:



Oxidation state of $Cr=+6$

Rest all the compounds are paramagnetic. Reasons for their paramagnetism are given below
(i) H-atom have $1s^1$ electronic configuration, i.e. 1 unpaired electron.
(ii) $N{O}_2$ is itself is an odd electron species.


(iii) $O_2^{-}$ (Superoxide) has one unpaired electron in ${\pi }^{\ast }$ molecular orbital.
(iv) $S_2$ in vapour phase has $O_2$ like electronic configuration i.e., have 2 unpaired electrons in ${\pi }^{\ast }$ molecular orbital.
(v) $M{n}_3{O}_4$ has following structure


Thus, $Mn$ is showing $+2$ and $+4$ oxidation states. The outermost electronic configuration of elemental $Mn$ is $3d^{5}4s^2$. Hence, in both the above oxidation states it has unpaired electrons as



(vi) $(N{H}_4{)}_{2}FeC{l}_4$ has $Fe$ as central metal atom with $+2$ oxidation state. The electronic configuration of $F{e}^{2+}$ in the complex is



(vii) $(N{H}_4{)}_{2}NiC{l}_4$ has $Ni$ as central metal atom with $+2$ oxidation state. The electronic configuration of $N{i}^{2+}$ in the complex is:



(viii) In $K_{2}Mn{O}_4$ central metal atom $Mn$ has $+6$ oxidation state with following structure


Electronic configuration of $M{n}^{6+}$ is: