Assign $A, B, C, D$ from the given type of reaction.
$K_4[Fe(CN)_6] + 2CuSO_4 \longrightarrow Cu_2[Fe(CN)_6] \downarrow + 2K_2SO_4$
$K_4[Fe(CN)_6] + 2CuSO_4 \longrightarrow Cu_2[Fe(CN)_6] \downarrow + 2K_2SO_4$
- AFor precipitate formation reaction
- BFor precipitate dissolution reaction
- CFor precipitate exchange reaction
- DFor no reaction
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View Solution$A$ solution of $FeCl_3$ when treated with $K_4[Fe(CN)_6]$ gives a Prussian blue precipitate due to the formation of:
Match List-$I$ with List-$II$:
$A$. Haber process $I$. Fe catalyst $B$. Wacker oxidation $II$. $PdCl_2$ $C$. Wilkinson catalyst $III$. $[(PPh_3)_3RhCl]$ $D$. Ziegler catalyst $IV$. $TiCl_4$ with $Al(CH_3)_3$
Choose the correct answer from the options given below:
| $A$. Haber process | $I$. Fe catalyst |
| $B$. Wacker oxidation | $II$. $PdCl_2$ |
| $C$. Wilkinson catalyst | $III$. $[(PPh_3)_3RhCl]$ |
| $D$. Ziegler catalyst | $IV$. $TiCl_4$ with $Al(CH_3)_3$ |
Choose the correct answer from the options given below:
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