Question
How is ethylamine prepared from (a) Nitro-alkane (b) Oxime?
Explain basic nature of amines.

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Fluorine shows only $-1$ oxidation state while other halogens show $-1, +1, +3, +5$ and $+7$ oxidation states. Explain.
(a) $Sn _{( aq )}^{2+}+2 AgCl _{( s )} \rightarrow Sn _{( aq )}^{4+}+2 Ag ( s )+2 Cl _{( aq )}^{-}$
The overall reaction takes place into two steps:
(i) $Sn _{( aq )}^{2+} \rightarrow Sn ^{4+}+2 e ^{-}$(Oxidation half reaction)
(ii) $2 AgCl _{( s )}+2 e ^{-} \rightarrow 2 Ag _{( s )}+2 Cl _{\text {(aq) }}^{-}$(Reduction half reaction)
Hence the cell is,
$Pt \left| Sn _{\text {(aq }}^{2+}, Sn _{\text {(aq) }}^{4+} \| Cl _{\text {(aq) }}^{-}(1 M )\right| AgCl _{( s )} \mid Ag _{( s )}$

(b) $Mg _{( s )}+ Br _{2(l)} \rightarrow Mg _{( aq )}^{2+}+2 Br _{( aq )}^{-}$
The overall reaction takes place into two steps:
(i) $Mg _{( s )} \longrightarrow Mg _{\text {(aq) }}^{2+}+2 e ^{-}$
(Oxidation half reaction)
(ii) $Br _{2()}+2 e ^{-} \longrightarrow 2 Br _{\text {(aq) }}^{-}$
(Reduction half reaction)
Hence the cell is,
$Mg _{( s )}\left| Mg _{( aq )}^{2+}(1 M ) \| Br _{( aq )}^{-}(1 M )\right| Br _{2( l )} \mid Pt$
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