Question

How do you figure out the change of an element that is located in the d-block of the periodic table?

Answer 1

Hint: To answer this question we should know about the oxidation state. We should know what d-block elements are. The oxidation state is the number of electrons gained or loosed by an atom of an element in a compound. The atoms form ions. During the formation of ions, the charge gained by the ion shows its oxidation number.

Complete step by step answer:
The atoms have protons and neutrons in the nucleus and electrons in the outermost shell. The arrangement of electrons generates the electronic configuration of atoms. Neutral atoms have a fixed electronic configuration in which the outermost orbitals can be partial-filled, fully-filled, or can have one, or any number of electrons. It depends upon the atomic number of those atoms.

In the periodic table, the noble gases have a fully-filled electronic configuration. Fully-filled electronic configurations are stable, so every atom tends to achieve the fully-filled configuration. For this, atoms lose or gain electrons.

The outermost electronic configuration of d-block element is, $\left( {{\text{n}} - 1} \right){{\text{d}}^{1 - 10}}{\text{n}}{{\text{s}}^{\text{2}}}$ . The d-block element having configuration $\left( {{\text{n}} - 1} \right){{\text{d}}^1}{\text{n}}{{\text{s}}^{\text{2}}}$ (that is scandium) can lose a maximum of three electrons so, can show a $ + 3$ oxidation state. It is also possible that it loses only one d-electron and shows a $ + 1$ oxidation state.

The element having configuration $\left( {{\text{n}} - 1} \right){{\text{d}}^2}{\text{n}}{{\text{s}}^{\text{2}}}$ (that is titanium) can show, $ + 2$ , $ + 3$ and $ + 4$ oxidation state.

As we go left to right in d-block in periodic table, the configuration change from $\left( {{\text{n}} - 1} \right){{\text{d}}^1}{\text{n}}{{\text{s}}^{\text{2}}}$ to $\left( {{\text{n}} - 1} \right){{\text{d}}^{10}}{\text{n}}{{\text{s}}^{\text{2}}}$. So, the possibilities of showing various oxidation states increases.

So, if an element is showing a variable oxidation state during the formation of an ion we can it might be a d-block element.

Therefore, we figure out the change in the oxidation state of an element that is located in the d-block of the periodic table.

Note: A pure substance that cannot be broken down into smaller substances is known as an element. The d-block elements are the elements having outermost electrons in d-orbital. The d-block elements show variable oxidation states. The alkali metals show the $ + 1$ oxidation state only. The alkaline earth metals show $ + 2$ oxidation state only. The manganese can show the highest oxidation state in $3{\text{d}} - $block that is $ + 7$. Due to fully-filled electronic configuration zinc shows only $ + 2$ oxidation state.