MCQ
Sulphur has lowest oxidation number in
- A${H_2}S{O_3}$
- B$S{O_2}$
- C${H_2}S{O_4}$
- ✓${H_2}S$
$\mathop {{H_2}S{O_4}}\limits^{\,\, * } = + 6$; $\mathop {{H_2}S}\limits^{\,\,\,\,\,\,\,\,\, * } = - 2$.
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$(A) i)$ Lindlar's catalyst, $H _2$; $ii$) $SnCl _2 / HCl$; iii) $NaBH _4$; iv) $H _3 O ^{+}$
$(B) i)$ Lindlar's catalyst, $H _2$; $ii$) $H _3 O ^{+}; iii)$ $SnCl _2 / HCl$; iv) $NaBH _4$
$(C) i)$ $NaBH _4; ii)$ $SnCl _2 / HCl; $iii$)$ $H _3 O ^{+}; iv)$ Lindlar's catalyst, $H _2$
$(D) i)$ Lindlar's catalyst, $H _2$; $ii$) $NaBH _4; iii)$ $SnCl _2 / HCl; iv)$ $H _3 O ^{+}$
The compound $A$ is a above reaction is
$KMn{O_4} + {H_2}S{O_4} + {H_2}{O_4} \to {K_2}S{O_4} + MnS{O_4} + {O_2} + {H_2}O$