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Chemistry Question on p -Block Elements

Reduction potential of ions are given below:\textbf{Reduction potential of ions are given below:}
ClO4−IO4−BrO4− E∘=1.19 VE∘=1.65 VE∘=1.74 V \begin{array}{ccc} \text{ClO}_4^- & \text{IO}_4^- & \text{BrO}_4^- \\\ E^\circ = 1.19 \, \text{V} & E^\circ = 1.65 \, \text{V} & E^\circ = 1.74 \, \text{V} \\\ \end{array}
The correct order of their oxidising power is:

A

ClO4−>IO4−>BrO4−\text{ClO}_4^- > \text{IO}_4^- > \text{BrO}_4^-

B

BrO4−>IO4−>ClO4−\text{BrO}_4^- > \text{IO}_4^- > \text{ClO}_4^-

C

BrO4−>ClO4−>IO4−\text{BrO}_4^- > \text{ClO}_4^- > \text{IO}_4^-

D

IO4−>BrO4−>ClO4−\text{IO}_4^- > \text{BrO}_4^- > \text{ClO}_4^-

Answer

BrO4−>IO4−>ClO4−\text{BrO}_4^- > \text{IO}_4^- > \text{ClO}_4^-

Explanation

Solution

Solution: The standard reduction potentials (SRP) provide insight into the tendency of each ion to undergo reduction:
Understanding Standard Reduction Potential: The higher the value of the standard reduction potential, the greater the tendency to gain electrons (undergo reduction). Hence, ions with higher reduction potentials act as better oxidizing agents.

Comparative Analysis: From the given data: BrO4− has the highest Eo value (1.74 V). IO4− follows with 1.65 V. ClO4− has the lowest at 1.19 V. Thus, the correct order of oxidizing power based on SRP is:

BrO4−>IO4−>ClO4−\text{BrO}_4^- > \text{IO}_4^- > \text{ClO}_4^-.