Question

When \(\mathrm{SnCl}_{4}\) is treated with excess of conc. \(\mathrm{HCl}\), the complex \(\mathrm{SnCl}_{6}{ }^{2}\) is formed. The oxidation state of \(\mathrm{Sn}\) in this complex is \((1)+6\) (2) \(+4\) (3) \(-2\) (4) \(+2\)

Short answer

The oxidation state of Sn in \(\text{SnCl}_6^{2-}\) is \(+4\).

Step-by-step solution

Identify the chemical reaction involved

When \(\text{SnCl}_4\) reacts with excess concentrated \(\text{HCl}\), the resulting complex is \(\text{SnCl}_6^{2-}\). The reaction can be represented as: \[ \text{SnCl}_4 + 2 \text{Cl}^- \rightarrow \text{SnCl}_6^{2-} \]

Determine the charge on the complex ion

The complex ion formed is \(\text{SnCl}_6^{2-}\), which has an overall charge of \(-2\).

Assign oxidation states to chlorine atoms

Each chlorine atom (Cl) in the complex has an oxidation state of \(-1\). Since there are 6 chlorine atoms, the total charge contributed by chlorine is \(-1 \times 6 = -6\).

Calculate the oxidation state of tin (Sn)

Let the oxidation state of tin (Sn) be \(x\). The sum of the oxidation states of all atoms in the complex ion must equal the overall charge of the ion. Therefore: \[ x + (-6) = -2 \] \[ x - 6 = -2 \] \[ x = -2 + 6 \] \[ x = 4 \]

Identify the oxidation state of Sn

The oxidation state of Sn in the complex \(\text{SnCl}_6^{2-}\) is \(+4\).

Key concepts

Chemical Reactions

In chemistry, reactions occur when substances interact to form new compounds. These reactions can often be represented through balanced chemical equations. Let's look into the reaction where \(\text{SnCl}_4\) reacts with concentrated \(\text{HCl}\). When \(\text{SnCl}_4\) is treated with excess \(\text{HCl}\), it forms a complex ion, \(\text{SnCl}_6^{2-}\). The chemical reaction can be written as:

• \(\text{SnCl}_4 + 2 \text{Cl}^- \rightarrow \text{SnCl}_6^{2-} \)

This is a simple representation of what's happening. The reactants, \(\text{SnCl}_4\) and \(\text{HCl}\), interact and transform into the product, \(\text{SnCl}_6^{2-}\).

Understanding these reactions helps explain how atoms and molecules rearrange themselves to form new substances. Notice that the equation balances: the number of chlorine atoms and the charges match on both sides of the equation. This balance is crucial in representing chemical reactions accurately. Chemical reactions are the foundation of understanding changes at the molecular level.

Complex Ions

Complex ions consist of a central metal ion bonded to one or more molecules or ions. These surrounding entities are called ligands.

In our example, the central metal ion is tin (\text{Sn}), and it forms a complex ion with six chlorine ligands resulting in \(\text{SnCl}_6^{2-}\). Here, chlorine ions (\(\text{Cl}^-\)) act as ligands that coordinate around the tin ion:

• Complex ion: \(\text{SnCl}_6^{2-}\)


• Central metal ion: tin (\(\text{Sn}\))


• Ligands: 6 chlorine ions (\(\text{Cl}^-\))

The formation of these complex ions involves donating or sharing of electrons between the central ion and the ligands. Complex ions are common in coordination chemistry and can be found in many naturally occurring substances and applications.

Understanding complex ions involves recognizing how ligands and central ions interact and why these structures form. Ligands can vary widely in their size and charge, affecting the properties and reactivity of the complex ion.

Oxidation States

Oxidation states, also known as oxidation numbers, indicate the degree of oxidation of an atom in a molecule. It is the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic.

Let's determine the oxidation state of tin (\text{Sn}) in \(\text{SnCl}_6^{2-}\).

1. Identify the charges for ligands: Chlorine (\(\text{Cl}^-\)) has an oxidation state of -1. With six chlorine atoms, the total charge contributed by the chlorine atoms is -6.

2. Set up the equation with the known charges and solve for the oxidation state of tin (x):
\[ x + (-6) = -2 \]
\[ x - 6 = -2 \]
\[ x = -2 + 6 \]
\[ x = 4 \]

Hence, the oxidation state of tin in \(\text{SnCl}_6^{2-}\) is +4.

Understanding oxidation states is crucial because it helps in balancing chemical equations and understanding redox reactions where electron transfer occurs. It provides insight into the chemical behavior and reactivity of elements within compounds and complex ions.