MCQ
Prussian blue is due to the formation of
- ✓$F{e_4}{\left[ {Fe{{(CN)}_6}} \right]_3}$
- B$F{e_2}\left[ {Fe{{(CN)}_6}} \right]$
- C$F{e_3}\left[ {Fe{{(CN)}_6}} \right]$
- D$Fe{\left[ {Fe{{(CN)}_6}} \right]_3}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Observation | $[A]$ | $[B]$ | Rate of reaction |
| $1$ | $0.1$ | $0.1$ | $2 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$ |
| $2$ | $0.4$ | $0.1$ | $0.4 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$ |
| $3$ | $0.1$ | $0.2$ | $1.4 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$ |
