Total number of molecules/species from following which will be paramagnetic is ______
$$O_{2}$$, $$O_2^+$$, $$O_2^-$$, $$NO$$, $$NO_{2}$$, $$CO$$, $$K_{2}[NiCl_{4}]$$, $$[Co(NH_{3})_{6}]Cl_{3}$$, $$K_{2}[Ni(CN)_{4}]$$

We need to find the total number of paramagnetic species from the given list.

A species is paramagnetic if it has one or more unpaired electrons.

1. $$O_2$$ (Oxygen molecule):

MO configuration: $$(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^2$$. The two electrons in $$\pi_{2p}^*$$ are unpaired (one in each degenerate orbital, by Hund's rule). Paramagnetic (2 unpaired electrons).

2. $$O_2^+$$: Remove one electron from $$\pi_{2p}^*$$: 1 unpaired electron. Paramagnetic.

3. $$O_2^-$$: Add one electron to $$\pi_{2p}^*$$: 3 electrons in two degenerate orbitals gives 1 unpaired. Paramagnetic.

4. NO (Nitric oxide): Total 15 electrons. MO config has one unpaired electron in $$\pi_{2p}^*$$. Paramagnetic.

5. $$NO_2$$: Nitrogen dioxide has 23 electrons (odd number), so at least 1 unpaired electron. Paramagnetic.

6. CO: 14 electrons, all paired in MO configuration. Diamagnetic.

7. $$K_2[NiCl_4]$$: Contains $$[NiCl_4]^{2-}$$. $$Ni^{2+}$$ has $$d^8$$ configuration. $$Cl^-$$ is a weak-field ligand, so tetrahedral geometry. In tetrahedral field, $$d^8$$ has 2 unpaired electrons. Paramagnetic.

8. $$[Co(NH_3)_6]Cl_3$$: Contains $$[Co(NH_3)_6]^{3+}$$. $$Co^{3+}$$ has $$d^6$$. $$NH_3$$ is a strong-field ligand (octahedral), so low-spin: all 6 electrons fill $$t_{2g}$$ with 0 unpaired electrons. Diamagnetic.

9. $$K_2[Ni(CN)_4]$$: Contains $$[Ni(CN)_4]^{2-}$$. $$Ni^{2+}$$ has $$d^8$$. $$CN^-$$ is a strong-field ligand, forming square planar geometry. In square planar, $$d^8$$ has 0 unpaired electrons. Diamagnetic.

Count of paramagnetic species: $$O_2, O_2^+, O_2^-, NO, NO_2, K_2[NiCl_4]$$ = 6.

The answer is 6.