# Give reasons for the following :(i) Transition metals form alloys.(ii) $Mn_{2}O_{3}$ is basic whereas $Mn_{2}O_{7}$ is acidic.(iii) $Eu^{2+}$ is a strong reducing agent.

(i) The size of transition metals are similar size, Therefore, they get easily settled in the      crystal lattice of another metal, hence, they form alloys.

(ii) $Mn_{2}O_{3}$ is bnasic because $Mn$ is $+3$ and exist in $d^{+}$ electronic configuration.

$Mn$ can lose $4$ more electron to achieve $+7$ oxidation state of $d^{\circ}$ state.

$Mn_{2}O_{7}$ is acidic, because, it already exist in highest horrible oxidn state i.e, $+7$ and     has no more electrons left to lose. therefore, $Mn_{2}O_{7}$ is acidic

(iii) $Eu^{2+}$ get easily oxidised to $Eu^{3+}$,as $Eu^{3+}$ is more stable, hence $Eu^{2+}$ acts as         strong reducing agent.

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