To determine the number of diamagnetic species, we need to identify which of the given species have no unpaired electrons. A species is diamagnetic if all its electrons are paired. Let's analyze each species step-by-step.

1. H atom: The electronic configuration of a hydrogen atom is $1s^1$. It has one unpaired electron in the 1s orbital. Therefore, the H atom is paramagnetic.

2. NO₂ monomer: To determine if NO₂ is paramagnetic, we can count its total number of valence electrons. Nitrogen has 5 valence electrons, and each oxygen atom has 6. Total valence electrons = $5 + 2 \times 6 = 17$. Since the total number of electrons is odd, NO₂ must have at least one unpaired electron. Thus, NO₂ is paramagnetic.

3. O₂⁻ (superoxide): The superoxide ion has a total of $2 \times 8 + 1 = 17$ electrons. An odd number of electrons means the species must be paramagnetic. According to Molecular Orbital Theory (MOT), the electronic configuration of O₂⁻ ends in $(\pi^*_{2p_y})^2 (\pi^*_{2p_z})^1$, showing one unpaired electron. Thus, O₂⁻ is paramagnetic.

4. Dimeric sulphur in vapor phase (S₂): Sulphur is in the same group as oxygen. The $S_2$ molecule is analogous to the $O_2$ molecule. According to MOT, just like $O_2$, the $S_2$ molecule has two unpaired electrons in its antibonding $\pi^*$ molecular orbitals. The valence molecular orbital configuration is $(\sigma_{3s})^2 (\sigma^*_{3s})^2 (\sigma_{3p_z})^2 (\pi_{3p_x})^2 (\pi_{3p_y})^2 (\pi^*_{3p_x})^1 (\pi^*_{3p_y})^1$. Therefore, $S_2$ is paramagnetic.

5. Mn₃O₄: This is a mixed oxide, with the formula $MnO \cdot Mn_2O_3$. It contains manganese in two different oxidation states: $Mn^{2+}$ and $Mn^{3+}$.

   • $Mn$ (Z=25) configuration is $[Ar] 3d^5 4s^2$.
   • $Mn^{2+}$ configuration is $[Ar] 3d^5$, which has 5 unpaired electrons.
   • $Mn^{3+}$ configuration is $[Ar] 3d^4$, which has 4 unpaired electrons. Since both manganese ions present are paramagnetic, the compound $Mn_3O_4$ is paramagnetic.

6. $(NH_4)_2[FeCl_4]$: This compound contains the complex anion $[FeCl_4]^{2-}$.

   • The oxidation state of Fe is +2. ($x + 4(-1) = -2 \implies x = +2$).
   • $Fe^{2+}$ has the electronic configuration $[Ar] 3d^6$.
   • $Cl^-$ is a weak-field ligand, forming a high-spin tetrahedral complex.
   • In a tetrahedral field, the $d^6$ configuration is $e^3t_2^3$. This is incorrect. The correct filling for $d^6$ in a high spin tetrahedral field ($e$ orbitals are lower in energy) is $e(↑↓) e(↑) t_2(↑) t_2(↑) t_2(↑)$.
   • There are 4 unpaired electrons. Thus, the complex is paramagnetic.

7. $(NH_4)_2[NiCl_4]$: This compound contains the complex anion $[NiCl_4]^{2-}$.

   • The oxidation state of Ni is +2. ($x + 4(-1) = -2 \implies x = +2$).
   • $Ni^{2+}$ has the electronic configuration $[Ar] 3d^8$.
   • With $Cl^-$ (a weak-field ligand), it forms a tetrahedral complex.
   • The $d^8$ configuration in a tetrahedral field is $e^4t_2^4$, which corresponds to the filling $e(↑↓) e(↑↓) t_2(↑↓) t_2(↑) t_2(↑)$.
   • There are 2 unpaired electrons. Thus, the complex is paramagnetic.

8. K₂MnO₄: This is potassium manganate, containing the manganate ion $[MnO_4]^{2-}$.

   • The oxidation state of Mn is +6. ($x + 4(-2) = -2 \implies x = +6$).
   • $Mn^{6+}$ has the electronic configuration $[Ar] 3d^1$.
   • It has one unpaired electron. Thus, $K_2MnO_4$ is paramagnetic.

9. K₂CrO₄: This is potassium chromate, containing the chromate ion $[CrO_4]^{2-}$.

   • The oxidation state of Cr is +6. ($x + 4(-2) = -2 \implies x = +6$).
   • $Cr$ (Z=24) has the configuration $[Ar] 3d^5 4s^1$.
   • $Cr^{6+}$ has the configuration $[Ar] 3d^0$.
   • Since there are no valence electrons, there are no unpaired electrons. Thus, $K_2CrO_4$ is diamagnetic.

Conclusion: Out of all the given species, only $K_2CrO_4$ is diamagnetic. Therefore, the total number of diamagnetic species is 1.