MCQ
$Fe(OH)_2$ is precipitated from $Fe(II)$ solutions as a while solid but turns dark green and then brown due to the formation of
- A$Fe(OH)_2$ and $Fe(OH)_3$
- Bonly $Fe(OH)_3$
- ✓$Fe_2O_3. (H_2O)_n$
- D$Fe_2O_3 ·2H_2O$
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(figure) $\xrightarrow[CC{{l}_{4}}]{B{{r}_{2}}}(A)\xrightarrow[(ii)\,NaN{{H}_{2}}]{(i)\,alc.\,KOH}(B)\xrightarrow[(ii)\,C{{H}_{3}}-Cl]{(i)\,NaN{{H}_{2}}}(C)$
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,;$ $K_1$
$N_2 + O_2 \rightleftharpoons 2NO\,;$ $K_2$
$H_2 + 2 O_2 \rightleftharpoons H_2O\,;$ $K_3$
The equilibrium constant $(K)$ of the reaction :
$2NH_3 + \frac{5}{2} \overset K \leftrightarrows 2NO + 3H_2O$