Question

The correct option(s) regarding $P_4{O}_{10},C{l}_2{O}_7,I_2{O}_5$ is/are:

• A. $P_4{O}_{10}$ is a more powerful dehydrating agent than $C{l}_2{O}_7$ and $I_2{O}_5$
• B. $C{l}_2{O}_7$ is the product of dehydration of $HClO$ using $P_4{O}_{10}$
• C. The dehydrating power will decrease as: $C{l}_2{O}_7>I_2{O}_5>P_4{O}_{10}$
• D. $I_2{O}_5$ can be prepared by heating $HI{O}_3$

Answer 1

Answer: (A), (B)

The correct options are
A $P_4{O}_{10}$ is a more powerful dehydrating agent than $C{l}_2{O}_7$ and $I_2{O}_5$
B $C{l}_2{O}_7$ is the product of dehydration of $HClO$ using $P_4{O}_{10}$

$P_4{O}_{10}$ is the most powerful dehydrating agent.

$C{l}_2{O}_7$ is the product of dehydration of $HCl{O}_4$ by $P_4{O}_{10}$, the reverse reaction cannot be expected to proceed.
$12HCl{O}_4+P_4{O}_{10}\to 4H_{3}P{O}_4+6C{l}_2{O}_7$

Since, $I_2{O}_5$ can be prepared by heating $HI{O}_3$, it cannot have too great an affinity for water.
$2HI{O}_3\stackrel{\Delta }{\to }{I}_2{O}_5+H_{2}O$

Hence, the correct order is: $P_4{O}_{10}>C{l}_2{O}_7>I_2{O}_7$