To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
(N/A) The electronic configuration plays a crucial role in determining the stability of oxidation states in the first transition series.
$1$. $Sc$ $(3d^1 4s^2)$ forms a highly stable $Sc^{3+}$ ion because it achieves the stable noble gas configuration of $[Ar]$ by losing three electrons.
$2$. In the case of $Ti$ $(3d^2 4s^2)$ and $V$ $(3d^3 4s^2)$,the $+4$ and $+5$ oxidation states are stable as they attain the $[Ar]$ configuration.
$3$. For $Mn$ $(3d^5 4s^2)$,the $+2$ oxidation state is particularly stable because the $Mn^{2+}$ ion has a half-filled $d$-orbital $(3d^5)$,which provides extra stability.
$4$. Generally,the stability of the $+2$ oxidation state increases across the series as the $d$-orbitals are progressively filled.
$1$. $Sc$ $(3d^1 4s^2)$ forms a highly stable $Sc^{3+}$ ion because it achieves the stable noble gas configuration of $[Ar]$ by losing three electrons.
$2$. In the case of $Ti$ $(3d^2 4s^2)$ and $V$ $(3d^3 4s^2)$,the $+4$ and $+5$ oxidation states are stable as they attain the $[Ar]$ configuration.
$3$. For $Mn$ $(3d^5 4s^2)$,the $+2$ oxidation state is particularly stable because the $Mn^{2+}$ ion has a half-filled $d$-orbital $(3d^5)$,which provides extra stability.
$4$. Generally,the stability of the $+2$ oxidation state increases across the series as the $d$-orbitals are progressively filled.
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