Animated Solution for Chemistry - Redox Reactions: The sum of oxidation states of two silver ions in $[Ag(N{H}_3{)}_2][Ag(CN{)}_2]$ complex is …… .

Unraveling the Oxidation States in Double Complex Salts

Have you ever looked at a massive coordination compound and felt a slight wave of intimidation? You are not alone. When we see a formula like $[Ag(N{H}_3{)}_2][Ag(CN{)}_2]$, it looks like a dense forest of brackets, symbols, and subscripts.

But here is the secret: chemistry is just a puzzle waiting to be taken apart.

In this article, we are going to break down this exact problem step-by-step. We will not just find the answer; we will understand the deep physical reality of what is happening inside this molecule. By the end of this journey, you will look at these double complexes and smile, knowing exactly how to conquer them.

The first step in solving any complex problem is to break it down into simpler, manageable pieces.

The compound $[Ag(N{H}_3{)}_2][Ag(CN{)}_2]$ is what we call a double complex salt. This means that both the positive part (the cation) and the negative part (the anion) are coordination complexes themselves.

Imagine dropping this salt into water. It doesn't just shatter into individual atoms. Instead, it gracefully splits into two distinct, stable entities.

Why does it split exactly like this? Silver ($Ag$) is a transition metal that has a very strong preference for the $+1$ oxidation state in aqueous chemistry.

When it binds with two neutral ammonia molecules, the overall charge remains $+1$. When it binds with two negatively charged cyanide ions, the overall charge becomes $-1$. These two perfectly balance each other out to form the neutral solid salt.

Before we dive into the calculations, we need a reliable tool.

The golden rule for finding the oxidation state of a central metal atom in a coordination complex is beautifully simple: The sum of the oxidation states of all the atoms (or ligands) in a complex ion must equal the net charge on that entire ion.

This is essentially a statement of charge conservation. The metal and the ligands pool their charges together to give the complex its overall identity.

Let's apply this master principle to our two separated ions.

Diamminesilver(I)

Let us focus our attention on the cationic part: $[Ag(N{H}_3{)}_2{]}^{+}$.

Inside this bracket, we have one central silver atom surrounded by two ammonia ($N{H}_3$) ligands.

To find the oxidation state of silver, we need to know the charge of the ammonia ligand. Ammonia is a stable, neutral molecule. It has no net charge. Therefore, its contribution to the oxidation state equation is exactly zero.

Let us assume the oxidation state of our silver atom is $x$.

Applying our master principle, we can set up the following equation:

This is a straightforward linear equation. The two ammonia molecules contribute nothing to the charge, so the entire $+1$ charge of the complex must come directly from the silver atom.

We have successfully decoded the first half of the puzzle. The silver atom in the cation is in a $+1$ oxidation state.

Dicyanoargentate(I)

Now, let us shift our focus to the anionic part: $[Ag(CN{)}_2{]}^{-}$.

Here, we have another silver atom, but this time it is bonded to two cyanide ($C{N}^{-}$) ligands.

Unlike ammonia, cyanide is not neutral. It is a polyatomic anion that carries a strict charge of $-1$. This is a crucial piece of chemical knowledge you must always keep in your toolkit.

Let us call the oxidation state of this second silver atom $y$.

We apply our master principle once again. The sum of the oxidation state of silver and the charges of the two cyanide ligands must equal the net charge of the complex, which is $-1$.

Now, we carefully solve this algebraic equation. Watch out for the minus signs!

We move the $-2$ to the other side of the equation by adding $2$ to both sides.

Fascinating, isn't it? Despite being in completely different chemical environments—one surrounded by neutral molecules and the other by negative ions—both silver atoms exist in the exact same $+1$ oxidation state.

Geometry and Hybridization

While the question only asked for the oxidation states, true mastery comes from understanding the physical reality of these molecules.

Both $[Ag(N{H}_3{)}_2{]}^{+}$ and $[Ag(CN{)}_2{]}^{-}$ feature a central silver atom bonded to exactly two ligands. In coordination chemistry, a coordination number of 2 almost universally leads to a linear geometry.

Imagine the silver atom sitting perfectly in the center, with the two ligands stretching out in opposite directions, forming a crisp $180^{\circ }$ angle.

This linear shape is a direct result of $sp$ hybridization. The silver atom mixes one $s$ orbital and one $p$ orbital to create two equivalent $sp$ hybrid orbitals, which then overlap with the orbitals of the ligands to form strong sigma bonds.

Understanding this spatial arrangement helps you visualize the molecule not just as a string of letters on a page, but as a real, three-dimensional object existing in space.

When tackling problems like this in high-stakes exams like JEE, students often fall into a few predictable traps.

First, misidentifying the ligand charge. If you mistakenly think cyanide is neutral, or ammonia has a charge, your entire calculation will collapse. Always maintain a mental catalog of common ligands and their charges.

Second, forgetting to split the complex. Some students try to calculate the oxidation state of silver by treating the entire double complex as one giant equation with two unknown silver atoms. While mathematically possible if you assume both silvers are identical, it is conceptually risky and can lead to errors if the metal centers are different. Always split the salt into its constituent ions first.

By staying disciplined and following the step-by-step dissociation method, you insulate yourself against these silly mistakes.

We have done the heavy lifting. Now it is time to answer the specific question asked by the examiner.

The question asks for the sum of the oxidation states of the two silver ions in the complex.

We found that the oxidation state of the first silver ion is $+1$, and the oxidation state of the second silver ion is also $+1$.

All that is left is a simple addition.

The final answer is $2$.

This problem is a beautiful reminder that complex structures are just combinations of simple rules. By understanding dissociation, knowing your ligand charges, and applying basic algebra, you can unravel even the most intimidating chemical formulas. Keep practicing, stay curious, and never let the brackets scare you!