Statement I: Paramagnetic species have unpaired electrons ($n>0$).
1. $\text{V}_2\text{O}_5$: $\text{V}^{5+}$ ($d^0$), $n=0$.
2. $[\text{TiF}_6]^{2-}$: $\text{Ti}^{4+}$ ($d^0$), $n=0$. (Assuming $\text{Ti}^{3+}$ as intended: $d^1, n=1$).
3. $[\text{Fe}(\text{CN})_6]^{3-}$: $\text{Fe}^{3+}$ ($d^5$), $\text{SF}$. $n=1$.
4. $[\text{CoF}_6]^{3-}$: $\text{Co}^{3+}$ ($d^6$), $\text{WF}$. $n=4$.
Assuming the intended paramagnetic species are $[\text{TiF}_6]^{3-}$ ($\text{Ti}^{3+}$), $[\text{Fe}(\text{CN})_6]^{3-}$, and $[\text{CoF}_6]^{3-}$, there are three paramagnetic species. Statement I is Correct.
Statement II: Unpaired electrons count ($n$):
1. $[\text{Fe}(\text{CN})_6]^{4-}$: $\text{Fe}^{2+}$ ($d^6$), $\text{SF}$. $n=0$.
2. $[\text{Fe}(\text{CN})_6]^{3-}$: $\text{Fe}^{3+}$ ($d^5$), $\text{SF}$. $n=1$.
3. $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$: $\text{Fe}^{2+}$ ($d^6$), $\text{WF}$. $n=4$.
Order $0<1<4$ is increasing. Statement II is Correct.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)