Animated Solution for Chemistry - d and f-Block Elements: Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them ?

Have you ever wondered why transition metals are the true chameleons of the periodic table? Unlike the predictable alkali metals that always show a $+1$ oxidation state, transition metals can exhibit a dazzling array of oxidation states. This unique ability stems from their electronic structure.

In transition metals, the energy difference between the outermost $ns$ orbital and the inner $(n-1)d$ orbitals is remarkably small. Because these energy levels are so close, electrons from both subshells can participate in chemical bonding. This means that the maximum oxidation state an element can achieve is generally the sum of its $ns$ and $(n-1)d$ electrons.

Let's put on our detective hats and analyze the given electronic configurations to see which one holds the potential for the highest oxidation state. The rule of thumb is simple: count the valence electrons, but beware of the pairing trap!

First, we look at $(n-1)d^{8}n{s}^2$. You might quickly add $8$ and $2$ to get $10$, but nature isn't that simple. In this configuration (like Nickel), the $d$ orbitals have paired electrons. Paired electrons are deeply stabilized and require an immense amount of energy to be removed. Therefore, the maximum oxidation state is restricted, typically reaching only $+4$.

Next is $(n-1)d^{5}n{s}^1$, the classic configuration of Chromium. Here, we have $5$ unpaired $d$ electrons and $1$ unpaired $s$ electron. Since none of them are paired, all $6$ can be lost, giving a maximum oxidation state of $+6$.

Then we have $(n-1)d^{3}n{s}^2$, which corresponds to Vanadium. With $3$ unpaired $d$ electrons and $2$ $s$ electrons, the math is straightforward: $3+2=5$. The maximum oxidation state here is $+5$.

Finally, we arrive at $(n-1)d^{5}n{s}^2$. This is the electronic signature of Manganese. Let's break it down: the $d$ subshell is exactly half-filled with $5$ unpaired electrons, and the $s$ subshell has $2$ electrons.

Because there is absolutely no pairing in the $d$ subshell yet, all $5$ of these $d$ electrons, along with the $2$ $s$ electrons, are available for bonding. When Manganese goes all out, it can lose all $7$ of these electrons, achieving a spectacular $+7$ oxidation state! A famous example of this is the deep purple Potassium Permanganate ($KMn{O}_4$), where Manganese proudly wears its $+7$ crown.

As we move further right across the periodic table (like Iron with $d^6{s}^2$), electrons in the $d$ subshell begin to pair up. This pairing increases the energy required to remove them, which is why the maximum oxidation state drops after group 7. Always remember: unpaired electrons are the true currency of high oxidation states!