Question

${\mathrm{Fex}}_2$and${\mathrm{Fey}}_3$ are known when $\mathrm{x}$ and $\mathrm{y}$ are:

• A. $\mathrm{x}=\mathrm{Cl},\mathrm{Br},\mathrm{I}\mathrm{and}\mathrm{y}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}$
• B. $\mathrm{x}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}\mathrm{and}\mathrm{y}=\mathrm{F},\mathrm{Cl},\mathrm{Br}$
• C. $\mathrm{x}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}\mathrm{and}\mathrm{y}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}$
• D. $\mathrm{x}=\mathrm{F},\mathrm{Cl},\mathrm{Br}\mathrm{and}\mathrm{y}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}$

Answer 1

The correct option is B

$\mathrm{x}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}\mathrm{and}\mathrm{y}=\mathrm{F},\mathrm{Cl},\mathrm{Br}$

B is the correct option.

Explanation for the correct option:

${\mathrm{FeI}}_3$, does not form because of${\mathrm{I}}^{–}$ being a very good reducing agent.

${\mathrm{FeI}}_3$ does not exist as ${\mathrm{I}}^{-}$ reduces ${\mathrm{Fe}}^{3+}$to ${\mathrm{Fe}}^{2+}$.

$2{\mathrm{Fe}}^{3+}+2{\mathrm{I}}^{-}\to 2{\mathrm{Fe}}^{2+}+{\mathrm{I}}_2$

Hence, in option B $\mathrm{y}$ doesn't have $\mathrm{I}$ as an option.

Therefore, B is the correct option.

A, C and D are incorrect options.

Explanation for the incorrect options :

In options A, C and D $\mathrm{y}$ has $\mathrm{I}$ as an option. And we know that ${\mathrm{FeI}}_3$does not exist as ${\mathrm{I}}^{-}$. Therefore, ${\mathrm{Fey}}_3$is not known when $\mathrm{y}\mathrm{is}\mathrm{I}.$

Hence, A, C and D are incorrect options.