- ✓$6$
- B$5$
- C$7$
- D$10$
Hence the number of $s $ electron is $ 6$ in that element.
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$\mathrm{Pt}(s) \mid \mathrm{H}_2(g, 1 \text { bar })\left|\mathrm{H}^{+}(a q, 1 \mathrm{M}) \| \mathrm{M}^{4+}(a q), \mathrm{M}^{2+}(a q)\right| \mathrm{Pt}(s)$
$E_{\text {cell }}=0.092 \mathrm{~V} \text { when } \frac{\left[\mathrm{M}^{2+}(a q)\right]}{\left[\mathrm{M}^{4+}(a q)\right]}=10^x$
Given : $ E_{\mathrm{M}^4 / \mathrm{M}^{2+}}^0=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}$
The value of $x$ is
$(i)\,\,(CH_3)_2CH - CH_2Br \xrightarrow{{{C_2}{H_5}OH}}$ $ (CH_3)_2CH - CH_2OC_2H_5 + HBr$
$(ii)\,\,(CH_3)_2CH - CH_2Br \xrightarrow{{{C_2}{H_5}O^-}} $ $(CH_3)_2CH - CH_2OC_2H_5 + Br^-$
The mechanisms of reactions $(i)$ and $(ii)$ are respectively
(Excess)
$N H_{3}+3 C l_{2} \rightarrow Y$
(Excess)
What is $X$ and $Y$ in the above reaction?