A diamagnetic metal burns in air to form a dioxide. This dioxide can reduce iodine to hydrogen iodide. When hydrogen gas is bubbled through this metal
A diamagnetic metal burns in air to form a dioxide. This dioxide can reduce iodine to hydrogen iodide. When hydrogen gas is bubbled through this metal, it forms a hydride. It was found that the reaction of the dioxide and hydride of this metal produces the same metal again. The metal is
A. S
B. Si
C. Te
D. Po.
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Correct Answer - A
The given metal is S. It is the dioxide of S ie.. `SO_2` which can act both a reducing agent and oxidising agent. The reactions involved are
`S + O_2 to SO_2`(Oxide)
`S+H_2 to H_2S` (Hydride)
`SO_2 + H_2S to 2H_2O + 3S`
`SO_2 +I_2 +2H_2O overset(OH^-)toH_2SO_4+HI`
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