MCQ
Sulphur in $+3$ oxidation state is present in
- ASulphurous acid
- BPyrosulphuric acid
- ✓Dithionous acid
- DThiosulphuric acid
$HO - \mathop {\mathop S\limits^{||} }\limits^O - \mathop {\mathop S\limits^{||} }\limits^O - OH$ ;
$2( + 1) + 2x + 4( - 2) = 0$
$2x = 8 - 2 = 6$;
$x = +3$
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$X_2 + Y_2 \rightarrow 2XY,$ is given below :
$(i)\,\, X_2 \rightarrow X + X\, (fast)$
$(ii)\,\,X + Y_2 \rightleftharpoons XY + Y\, (slow)$
$(iii)\,\,X + Y \rightarrow XY\, (fast)$
The overall order of the reaction will be
takes places through the mechanism given below :
$NO + Br _2 \Leftrightarrow NOBr _2 \text { (fast) }$
$NOBr _2+ NO \rightarrow 2 NOBr \text { (slow) }$
The overall order of the reaction is $.....$.

