- A${[NiC{l_4}]^{2 - }}$
- B${[Co{F_6}]^{3 - }}$
- C${[Fe{({H_2}O)_6}]^{2 + }}$
- ✓${[Ni{(CN)_4}]^{2 - }}$
$Ni^{+2}$ in presence $C{N^ - } = [Ar]$
As ${[Ni{(CN)_4}]^{2 - }}$ has no unpaired electron. It is diamagnetic.
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$(i)$ $n\, = 4, l\, = 1$ $(ii)$ $n\, = 4, l\, = 0$
$(iii)$ $n\, = 3, l\, = 2$ $(iv)$ $n\, = 3, l\, = 1$
The sequence representing increasing order of energy, is
$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq}),(1 \mathrm{M}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$
The fraction of total iron present as $\mathrm{Fe}^{3+}$ ion at the cell potential of $1.500\, \mathrm{~V}$ is $\mathrm{X} \times 10^{-2}$. The value of $x$ is $.....$ (Nearest integer).
$\left(\right.$ Given $\left.E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=0.77\, \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \,\mathrm{~V}\right)$
$\begin{array}{*{20}{c}}
{\,{C_2}{H_5}} \\
{|\,\,\,\,\,} \\
{C{H_3} - C - C{H_2}Cl} \\
{|\,\,\,\,\,} \\
{\,{C_2}{H_5}}
\end{array}$
Reason : The only property that determines its aromatic behaviour is its planar structure.
Above reaction is known as