Given below are two statements:
Statement I: The number of paramagnetic species among $$[CoF_{6}]^{3-},[TiF_{6}]^{3-},V_{2}O_{5}\text{ and }[Fe(CN)_{6}]^{3-}$$ is 3.
Statement II: $$K_{4}[Fe(CN)_{6}] < K_{3}[Fe(CN)_{6}] < [Fe(H_{2}O)_{6}]SO_{4}.H_{2}O < [Fe(H_{2}O)_{6}]Cl_{3}$$ is the correct order in terms of number of unpaired electron(s) present in the complexes.
In the light of the above statements, choose the correct answer from the options given below

We need to evaluate two statements about paramagnetic species and unpaired electrons in coordination compounds.

Statement I: The number of paramagnetic species among $$[CoF_6]^{3-}$$, $$[TiF_6]^{3-}$$, $$V_2O_5$$, and $$[Fe(CN)_6]^{3-}$$ is 3.

- $$[CoF_6]^{3-}$$: Co³⁺ (3d⁶), F⁻ is weak field, so 4 unpaired electrons. Paramagnetic ✓

- $$[TiF_6]^{3-}$$: Ti³⁺ (3d¹), 1 unpaired electron. Paramagnetic ✓

- $$V_2O_5$$: V is in +5 state (3d⁰), no unpaired electrons. Diamagnetic ✗

- $$[Fe(CN)_6]^{3-}$$: Fe³⁺ (3d⁵), CN⁻ is strong field, pairing occurs: t₂g⁵ eg⁰, 1 unpaired electron. Paramagnetic ✓

Number of paramagnetic species = 3. Statement I is TRUE.

Statement II: $$K_4[Fe(CN)_6] < K_3[Fe(CN)_6] < [Fe(H_2O)_6]SO_4.H_2O < [Fe(H_2O)_6]Cl_3$$ is the correct order of unpaired electrons.

- $$K_4[Fe(CN)_6]$$: Fe²⁺ (3d⁶), CN⁻ strong field → t₂g⁶ eg⁰ → 0 unpaired electrons

- $$K_3[Fe(CN)_6]$$: Fe³⁺ (3d⁵), CN⁻ strong field → t₂g⁵ eg⁰ → 1 unpaired electron

- $$[Fe(H_2O)_6]SO_4.H_2O$$: Fe²⁺ (3d⁶), H₂O weak field → t₂g⁴ eg² → 4 unpaired electrons

- $$[Fe(H_2O)_6]Cl_3$$: Fe³⁺ (3d⁵), H₂O weak field → t₂g³ eg² → 5 unpaired electrons

Order: 0 < 1 < 4 < 5. Statement II is TRUE.

Both statements are true, corresponding to Option 1.