- ✓Blue
- BRed
- CColourless
- DYellow
$Br _2+ KI \rightarrow KBr + I _2$
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$A.$ $Cl _{2} / Cl^{-}$ $B.$ $I _{2} / I^{-}$ $C.$ $Ag ^{+} / Ag$ $D.$ $Na ^{+} / Na$ $E.$ $Li ^{+} / Li$
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$\begin{array}{*{20}{c}}
{\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH = C{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow[{2.\,KOH - EtOH,\,heat}]{{1.\,HBr}}$
The major end product formed is :-
(Given : atomic number $Sm =62 ; Eu =63 ; Tb =65 ; Gd =64, Pm =61 \text { ) }$
$A.$ $Sm$ $B.$ $Eu$ $C.$ $Tb$ $D.$ $Gd$ $E.$ $Pm$
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