Q.The compound AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?

Concept: Electronic configuration and oxidation state stability

Silver typically exists in the +1 oxidation state with a stable $d^{10}$ configuration. In AgF₂, silver is forced into the unusual +2 state, giving Ag²⁺ a $d^9$ configuration.

The $d^9$ configuration is inherently unstable because:

• It has one unpaired electron in the $d$ orbitals, creating a high-energy state
• The filled $d^{10}$ configuration of Ag⁺ is far more stable (all orbitals paired, lower energy)
• Ag²⁺ has a very strong tendency to gain one electron and revert to the stable Ag⁺ state

Since Ag²⁺ desperately "wants" to accept an electron to achieve $d^{10}$, it readily oxidises other substances (taking their electrons). This electron-accepting tendency is precisely what defines a strong oxidising agent. …

The heart of this question lies in understanding why certain oxidation states are stable for transition metals and what happens when you force an element into an unfavorable state.

Silver normally exists as $\ce{Ag^{+}}$ with the electronic configuration $[\ce{Kr}]4d^{10}$—a completely filled $d$-subshell that is exceptionally stable. When you oxidise it further to $\ce{Ag^{2+}}$, you remove one more electron, leaving $[\ce{Kr}]4d^9$. This $d^9$ configuration is inherently less stable than $d^{10}$ for several reasons: it loses the exchange energy stabilisation of a filled subshell, introduces an unpaired electron, and disrupts the spherical symmetry of the electron cloud.

Now, why does $\ce{AgF2}$ form at all if $\ce{Ag^{2+}}$ is so unstable? Fluorine is the most electronegative element, and it can pull electrons away from silver more effectively than any other element. The high lattice energy of the resulting ionic compound provides just enough stabilisation to make $\ce{AgF2}$ barely stable enough to exist—but it remains thermodynamically poised to revert.

Let me walk through the oxidising behaviour step by step:

1. The driving force for reduction: $\ce{Ag^{2+}}$ has a very high reduction potential because the $d^{10}$ configuration is so much more stable than $d^9$. The half-reaction is:

This extraordinarily positive reduction potential means $\ce{Ag^{2+}}$ desperately wants to gain an electron.

2. What makes an oxidising agent strong?: A species is a strong oxidising agent if it readily accepts electrons (gets reduced). The more favorable its reduction, the stronger the oxidising agent. With $E^\circ \approx +1.98$ V, $\ce{Ag^{2+}}$ is one of the strongest oxidising agents among common metal ions.

3. The role of fluorine: Fluoride ion ($\ce{F^-}$) is extremely difficult to oxidise (it's the most reluctant to give up electrons), so it won't interfere by getting oxidised itself. This means all the oxidising power resides with the $\ce{Ag^{2+}}$ ion. In contrast, if you tried to make $\ce{AgCl2}$ or $\ce{AgBr2}$, the halide ions would be oxidised before you could stabilise $\ce{Ag^{2+}}$. …