Q.Write the effect on valency when we go from top to bottom in a group.

Oxidation State Covalency: The First Meeting

Imagine you are looking at a molecule like $\text{SO}_3$. You know sulphur can show different oxidation states — here it is $+6$. But if you count the actual number of covalent bonds sulphur makes, it is also 6. That is not a coincidence. For many elements, especially in their highest oxidation states, the oxidation number equals the number of covalent bonds they form. That is the core intuition behind oxidation state covalency.

But this is not always true. Consider $\text{H}_2\text{O}_2$. Oxygen has an oxidation state of $-1$, but it forms two covalent bonds. Here the number of bonds (2) does not match the magnitude of the oxidation state (1). So the idea needs a precise boundary.

The Precise Statement

In other words:

But this equality breaks down when:

• The bond has significant ionic character (e.g., $\text{NaCl}$ — Na has oxidation state $+1$ but forms zero covalent bonds).
• The element uses d-orbitals to expand its octet (e.g., $\text{PCl}_5$ — P has oxidation state $+5$, covalency 5 — here it works).
• The molecule has coordinate bonds or back-bonding (e.g., $\text{CO}$ — C has oxidation state $+2$, but covalency is 3 due to a coordinate bond).

Why Does This Matter?

For main-group elements in their maximum oxidation state (equal to group number), oxidation state covalency is almost always obeyed. This is why:

• $\text{SF}_6$: S is $+6$, covalency 6.
• $\text{Cl}_2\text{O}_7$: Cl is $+7$, covalency 7 (each Cl is bonded to 3 O atoms and one O bridge — count carefully: 7 bonds per Cl).
• $\text{P}_4\text{O}_{10}$: P is $+5$, covalency 5.

But for transition metals, the concept is less useful because their oxidation states often exceed their covalency (e.g., $\text{MnO}_4^-$: Mn is $+7$, but covalency is only 4 — the extra oxidation state comes from oxygen's high electronegativity pulling electron density away).

A Common Mistake …