- A$K_4[Fe^{+2}(CN)_5(O_2 )]$
- ✓$[NiF_6]^{-2}$
- C$[Fe(H_2O)_5(NO)]^{+2}$
- DAll of these
$[NiF_6]^{2-} \to d^2sp^3$ hybridisation, diamagnetic
$[Fe(H_2O)_5(NO)]^{+2} \to $ paramagnetic
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Major final product $(Z)$ is
A current was passed for $10$ hours. $10.8\ g Ag$ was deposited at cathode in $(III)$ electrolytic cell during electrolysis if current efficiency is $50\%.$
If $Z_1,$ $Z_2$ and $Z_3$ be the electrochemical equivalent in the formation of $H_2, Cu$ and $Ag$ respectively, hence $Z_1 : Z_2 : Z_3$ would be
$C{H_3} - CH = O + {C_6}{H_5} - CH =O \xrightarrow[\Delta ]{{\mathop O\limits^\Theta H\,\left( {aq} \right)}}\mathop X\limits_{{\text{(}}{{\text{C}}_4}{{\text{H}}_6}{\text{O)}}} + \mathop Y\limits_{{\text{(}}{{\text{C}}_9}{{\text{H}}_8}{\text{O)}}} $

gly $=$ glycinato; bpy $=2,2$ '-bipyridine