Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
84 views
in Chemistry by (72.2k points)
closed by
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.

1 Answer

0 votes
by (70.4k points)
selected by
 
Best answer
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`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...