Fe(III) Reaction with KSCN and NaF

The question describes a series of reactions involving an aqueous solution of a metal ion (A), leading to a colored product (B) upon reaction with KSCN, which then turns colorless (C) upon addition of NaF.

Let's analyze the reactions step-by-step:

Metal Ion (A) Reaction with KSCN

• We are told that the metal ion solution (A) gives a blood-red colored product (B) when reacted with KSCN (potassium thiocyanate).
• A common qualitative test in chemistry is the reaction of ferric ions ($\text{Fe}^{3+}$) with thiocyanate ions ($\text{SCN}^{-}$), which produces a distinctive blood-red or reddish-brown complex.
• Ferrous ions ($\text{Fe}^{2+}$), on the other hand, do not give this intense blood-red color with $\text{SCN}^{-}$ under normal conditions.
• Therefore, the metal ion (A) is likely $\text{Fe}^{3+}$ in aqueous solution. This corresponds to aq. Fe(III).

Formation of Product (B)

• When aq. $\text{Fe}^{3+}$ reacts with $\text{SCN}^{-}$ ions from KSCN, it forms a complex. The blood-red color is attributed to the formation of thiocyanatoiron(III) complexes.
• The most common species responsible for the intense red color, especially at low $\text{SCN}^{-}$ concentration, is the pentaaquathiocyanatoiron(III) ion.
• The formula for this complex is $[\text{Fe(SCN)(H}_2\text{O)}_5]^{2+}$. The overall charge is calculated as $(+3 \text{ from Fe}) + (-1 \text{ from SCN}) + (0 \text{ from } 5 \text{ H}_2\text{O}) = +2$.
• So, the blood-red colored product (B) is likely $[\text{Fe(SCN)(H}_2\text{O)}_5]^{2+}$.

Reaction of Product (B) with NaF

• Upon dropwise addition of NaF (sodium fluoride), which introduces fluoride ions ($\text{F}^{-}$), the blood-red complex (B) turns into a colorless compound (C).
• This indicates a ligand exchange reaction is occurring. Fluoride ions ($\text{F}^{-}$) are known to be strong ligands for $\text{Fe}^{3+}$ and can displace weaker ligands like water and thiocyanate from the coordination sphere of the iron ion.
• Fluoride forms very stable complexes with $\text{Fe}^{3+}$, such as the hexafluoridoferrate(III) ion.
• The reaction involves the replacement of $\text{SCN}^{-}$ and $\text{H}_2\text{O}$ ligands by $\text{F}^{-}$ ligands: $$\text{[Fe(SCN)(H}_2\text{O)}_5]^{2+} + 6\text{F}^{-} \rightarrow \text{[FeF}_6]^{3-} + \text{SCN}^{-} + 5\text{H}_2\text{O}$$ (This is a simplified representation; the exchange likely happens stepwise).
• The complex $[\text{FeF}_6]^{3-}$ is known to be colorless. This ligand exchange reaction with a strong ligand like $\text{F}^{-}$ disrupting the color-producing electronic transitions is consistent with the observation.
• Therefore, the colorless compound (C) is likely $[\text{FeF}_6]^{3-}$.

Summary of Identification

• Metal ion (A): aq. Fe(III)
• Blood-red product (B): $[\text{Fe(SCN)(H}_2\text{O)}_5]^{2+}$
• Colorless compound (C): $[\text{FeF}_6]^{3-}$

Comparing this identification with the given options, Option 2 matches our findings:

• aq. Fe(III) for A
• $[\text{Fe(SCN)(H}_2\text{O)}_5]^{2+}$ for B
• $[\text{FeF}_6]^{3-}$ for C

The other options either start with the wrong metal ion ($\text{Fe}^{2+}$) or propose incorrect formulas/charges for the complexes or incorrect final products (simple salts instead of complexes in solution).