Chemical Bonding - Result Question 115 — JEE PYQ Chapterwise
14. Among the species given below, the total number of diamagnetic species is ______
$H$ atom, $NO _2$ monomer, $O _2^ {-}$ (superoxide), dimeric sulphur in vapour phase, $Mn _3 O _4,\left(NH _4\right) _2\left[FeCl _4\right],\left(NH _4\right) _2\left[NiCl _4\right]$, $K _2 MnO _4, K _2 CrO _4$
(2018 Adv.)
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Answer:
Correct Answer: 14. (1)
Solution:
- Among the given species only $K _2 CrO _4$ is diamagnetic as central metal atom $Cr$ in it has $[Ar] 3 d^{0}$ electronic configuration i.e., all paired electrons. The structure and oxidation state of central metal atom of this compound are as follows
[Image]
Rest all the compounds are paramagnetic. Reasons for their paramagnetism are given below
(i) H-atom have $1 s^{1}$ electronic configuration, i.e. 1 unpaired electron.
(ii) $NO _2$, i.e. in itself is an odd electron species.
(iii) $O ^{-}_2 $ (Superoxide) has one unpaired electron in $\pi *$ molecular orbital.
(iv) $S _2$ in vapour phase has $O _2$ like electronic configuration i.e., have 2 unpaired electrons in $\pi *$ molecular orbitals.
(v) $Mn _3 O _4$ has following structure
[Image]
Thus, $Mn$ is showing +2 and +4 oxidation states. The outermost electronic configuration of elemental $Mn$ is $3 d^{5} 4 s^{2}$. Hence, in both the above oxidation states it has unpaired electrons as
(vi) $\left(NH _4\right) _2 FeCl _4$ has $Fe$ as central metal atom with +2 oxidation state. The electronic configuration of $Fe^{2+}$ in the complex is
(vii) $\left(NH _4\right) _2 NiCl _4$ has $Ni$ as central metal atom with +2 oxidation state. The electronic configuration of $Ni^{2+}$ in the complex is
(viii) In $K _2 MnO _4$ central metal atom $Mn$ has +6 oxidation state with following structure
[Image]
Electronic configuration of $Mn^{6+}$ is
BETA
