The correct option(s) regarding $P_{4} O_{10}, C l_{2} O_{7}, I_{2} O_{5}$ is/are:

P_{4} O_{10}

is a more powerful dehydrating agent than

C l_{2} O_{7}

and

I_{2} O_{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C l_{2} O_{7}

is the product of dehydration of

H C l O

using

P_{4} O_{10}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

The dehydrating power will decrease as:

C l_{2} O_{7} > I_{2} O_{5} > P_{4} O_{10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{2} O_{5}

can be prepared by heating

H I O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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 $H C l O$ 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 $H C l O_{4}$ by $P_{4} O_{10}$ , the reverse reaction cannot be expected to proceed.
$12 H C l O_{4} + P_{4} O_{10} \to 4 H_{3} P O_{4} + 6 C l_{2} O_{7}$

Since, $I_{2} O_{5}$ can be prepared by heating $H I O_{3}$ , it cannot have too great an affinity for water.
$2 H I O_{3} Δ - \to I_{2} O_{5} + H_{2} O$

Hence, the correct order is: $P_{4} O_{10} > C l_{2} O_{7} > I_{2} O_{7}$

Suggest Corrections

Similar questions

P_{4} O_{10}, C O, S O_{2}, C l O_{2}, C l_{2} O, N_{2} O, N O_{2}, C l_{2} O_{6}, C O_{2}, C l_{2} O_{7}, S O_{3}

C l_{2} O_{7}

C l_{2} O_{7}

N a_{2} O, M g O, S O_{3}, C l_{2} O_{7}

C l

C l_{2} O_{7}

Join BYJU'S Learning Program