Number of Diamagnetic Species Among Given Chemical Entities

To determine the number of diamagnetic species, we evaluate the electronic configuration and the presence of unpaired electrons in each of the given chemical entities:

1. $\text{H}$ atom: The electronic configuration is $1s^1$. It contains $1$ unpaired electron. $\implies \text{Paramagnetic}$

2. $\text{NO}_2$ monomer: The total number of valence electrons is $5 (\text{from N}) + 2 \times 6 (\text{from O}) = 17$ electrons. Being an odd-electron molecule, it contains an unpaired electron. $\implies \text{Paramagnetic}$

3. $\text{O}_2^-$ (superoxide ion): The total number of electrons is $17$. According to Molecular Orbital Theory (MOT), its electronic configuration is: \sigma_{1s}^2\, \sigma^*_{1s}{}^2\, \sigma_{2s}^2\, \sigma^*_{2s}{}^2\, \sigma_{2p_z}^2\, (\pi_{2p_x}^2 = \pi_{2p_y}^2)\, (\pi^*_{2p_x}^2 = \pi^*_{2p_y}^1) It possesses $1$ unpaired electron in its antibonding $\pi^*$ molecular orbital. $\implies \text{Paramagnetic}$

4. Dimeric sulphur in vapour phase ($\text{S}_2$): Analogous to the dioxygen molecule ($\text{O}_2$), $\text{S}_2$ contains $2$ unpaired electrons in its degenerate antibonding molecular orbitals: \pi^*_{3p_x}^1 = \pi^*_{3p_y}^1 $\implies \text{Paramagnetic}$

5. $\text{Mn}_3\text{O}_4$: This is a mixed oxide containing manganese in two different oxidation states, formulated as $\text{Mn}^{\text{II}}\text{Mn}^{\text{III}}_2\text{O}_4$.

   • $\text{Mn}^{2+}$ has a $3d^5$ configuration ($5$ unpaired electrons).
   • $\text{Mn}^{3+}$ has a $3d^4$ configuration ($4$ unpaired electrons). Since both ions have unpaired electrons, the compound is magnetic (paramagnetic at high temperatures). $\implies \text{Paramagnetic}$

6. $(\text{NH}_4)_2[\text{FeCl}_4]$: The complex contains the tetrahedral ion $[\text{FeCl}_4]^{2-}$.

   • The oxidation state of iron is $\text{Fe}^{2+}$ with electronic configuration $[\text{Ar}]\,3d^6$.
   • $\text{Cl}^-$ is a weak-field ligand, resulting in a high-spin tetrahedral configuration $e^3 t_2^3$, which has $4$ unpaired electrons. $\implies \text{Paramagnetic}$

7. $(\text{NH}_4)_2[\text{NiCl}_4]$: The complex contains the tetrahedral ion $[\text{NiCl}_4]^{2-}$.

   • The oxidation state of nickel is $\text{Ni}^{2+}$ with electronic configuration $[\text{Ar}]\,3d^8$.
   • In a tetrahedral geometry, its configuration is $e^4 t_2^4$, which has $2$ unpaired electrons. $\implies \text{Paramagnetic}$

8. $\text{K}_2\text{MnO}_4$: Contains the manganate ion $[\text{MnO}_4]^{2-}$.

   • The oxidation state of manganese is $+6$, giving an electronic configuration of $[\text{Ar}]\,3d^1$.
   • It possesses $1$ unpaired electron. $\implies \text{Paramagnetic}$

9. $\text{K}_2\text{CrO}_4$: Contains the chromate ion $[\text{CrO}_4]^{2-}$.

   • The oxidation state of chromium is $+6$, giving an electronic configuration of $[\text{Ar}]\,3d^0$.
   • It has no unpaired electrons. $\implies \text{Diamagnetic}$

Thus, only $\text{K}_2\text{CrO}_4$ is diamagnetic.

The total count of diamagnetic species is 1.