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An inorganic iodide (A) on heating with ...

An inorganic iodide (A) on heating with a solution of KOH gives a gas (B) and a solution of a compound. The gas (B) on ignition in air gives a compound (C ) and water. Copper sulphate is finally reduced to the methal on passing (B) through its solution.
What is true about gas (B) and compound (C ) ?

A

The oxidation number of central atom of gas (B) is `+ IV`

B

The gas (B) produces a black precipitate of methallic sliver with sliver nitrate solution.

C

Compound (C ) dissolves in water forming an acid which with solution hydroxide forms three series of salts.

D

(B) and (C ) both

Text Solution

Verified by Experts

`(A) is PH_(4)I`. "The given changes are" :
(i) `underset((A))PH_(4)I+KOH to KI+underset((B))(PH_(3))+H_(2)O`
(ii) `underset((B))(4PH_(3))+8O_(2) to underset((C))(P_(4))O_(10)+6H_(2)O`
(iii) `4Cu^(+)+PH_(3)+4H_(2)O to H_(3)PO_(4)+4Cudarr+8H^(+)`
`3AgNO_(3)+PH_(3) to Ag_(3)Pdarr+3HNO_(3)`
`Ag_(3)P+3agNO_(3)+3H_(2)O to 6Agdarr("black")+3HNO_(3)+H_(3)PO_(3)`
`6Ag^(+)+PH_(3)+3H_(2)O to 3H_(3)PO_(3)+6Agdarr("black")+6H^(+)`
In `P_(4)O_(10)` there are `16 sigma-`bonds and all central atoms have `sp^(3)` hybridisation.
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