- ✓$C{H_3} - CH = C{H_2}\mathop {\xrightarrow{{{B_2}{H_6}}}}\limits_{O{H^ - }} $
- B$C{H_3}C{H_2}C{H_2}I\mathop {\xrightarrow{{HI}}}\limits_P $
- C$C{H_3}C{H_2}C{H_2}Cl\xrightarrow{{Na}}$
- DNone of these
$R - CH = C{H_2}\xrightarrow{{{B_2}{H_6}}}{(R - C{H_2} - C{H_2})_3} - B$ $\xrightarrow{{O{H^ - }}}R - C{H_2} - C{H_2}OH$
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$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \mathrm{E}^{\circ}=1.33 \mathrm{~V}$
$\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \mathrm{E}^{\circ}=-0.04 \mathrm{~V}$
$\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} \mathrm{E}^{\circ}=-0.25 \mathrm{~V}$
$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \mathrm{E}^{\circ}=0.80 \mathrm{~V}$
$\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \mathrm{E}^{\circ}=1.40 \mathrm{~V}$
Consider the given electrochemical reactions, The number of metal$(s)$ which will be oxidized be $\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}$, in aqueous solution is. . . . . .
Here $W, Y$ and $Z$ are left, up and right elements with respect to the element $'X'$ and $'X'$ belongs to $16^{th}$ group and $3^{rd}$ period. Then according to given information the incorrect statement regarding given elements is