आप निम्नलिखित को किस प्रकार से स्पष्ट करेंगे-

$(i)$ $d^{4}$ स्पीशीज में से $Cr ^{2+}$ प्रबल अपचायक है जबकि मैंगनीज $(III)$ प्रबल ओक्सीकरक है।

$(ii)$ जलीय विलयन में कोबाल्ट $(II)$ स्थायी है परंतु संकुलनकारी अभिकर्मकों की उपस्थिति में यह सरलतापूर्वक ओक्सीकृत हो जाता है।

$(iii)$ आयनों का $d^{1}$ विन्यास अत्यंत अस्थायी है।

$(i)$ $C r^{2+}$ 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.