નીચે દર્શાવેલા વિધાનોને કેવી રીતે સ્પષ્ટ કરશો ?

$(i)$ $d^4$ સ્પિસીઝના $Cr^{2+}$ પ્રબળ રિડક્શનકર્તા છે જ્યારે મેંગેનીઝ $(II)$ પ્રબળ ઑક્સિડેશનકર્તા છે.

$(ii)$ કોબાલ્ટ $(II)$ જલીય દ્રાવણમાં સ્થાયી છે પરંતુ સંકીર્ણકર્તા પ્રક્રિયકોની હાજરીમાં તે સહેલાઈથી ઑક્સિડેશન પામે છે.

$(iii)$ આયનોમાં $d^1$ ઈલેક્ટ્રૉનીય રચના અત્યંત અસ્થાયી છે. 

$(i)$ $C r^{2+}$ is strongly reducing in nature. It has a $d^{4}$ configuration. While acting as a reducing agent, it gets oxidized to $Cr ^{3+}$ (electronic configuration, $d^{3}$ ). This $d^{3}$ configuration can be written as $t_{2 g }^{3}$ configuration, which is a more stable configuration. In the case of $Mn ^{3+}\left(d^{4}\right),$ it acts as an oxidizing agent and gets reduced to $Mn ^{2+}\left(d^{5}\right) .$ This has an exactly half-filled $d$ -orbital and is highly stable.

$(ii)$ $Co ( II )$ is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to $Co (III)$. Although the $3^{\text {rd }}$ ionization energy for $Co$ is high, but the higher amount of crystal field stabilization energy $(CFSE)$ released in the presence of strong field ligands overcomes this ionization energy.

$(iii)$ The ions in $d^{1}$ configuration tend to lose one more electron to get into stable $d^{0}$ configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the $d$ -orbital of these ions. Therefore, they act as reducing agents.