A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements abou

Q: Why does a higher positive charge make the ion smaller?

When an atom loses more electrons to form a higher oxidation state, the remaining electrons are pulled more strongly by the nucleus because the effective nuclear charge per electron increases. This shrinks the electron cloud, making $M^{4+}$ significantly smaller than $M^{2+}$.

Q: What exactly is polarizing power?

Think of polarizing power as the cation's ability to distort the electron cloud of a neighboring anion. A small, highly charged cation acts like a concentrated point of positive charge, aggressively pulling the anion's negative electron cloud towards itself.

Q: Why does polarization lead to covalent character?

In a purely ionic bond, electrons are completely transferred. But when the cation strongly polarizes the anion, it pulls the electron cloud into the space between the two nuclei. This sharing of electron density is the very definition of a covalent bond!

Q: If $MCl_4$ is covalent, why is it more volatile?

Covalent compounds exist as discrete molecules held together by relatively weak intermolecular forces (like Van der Waals forces). Ionic compounds form massive, tightly bound crystal lattices. It takes much less heat energy to separate molecules than to break an ionic lattice, making $MCl_4$ more volatile.

Q: Why is $MCl_4$ more easily hydrolysed than $MCl_2$?

Covalent halides with a central atom in a high oxidation state (like $M^{4+}$) often have empty d-orbitals and a high positive charge density, making them highly susceptible to nucleophilic attack by water molecules. Ionic chlorides are generally more stable against such hydrolysis.

Question

A metal, $M$ forms chlorides in its $+2$ and $+4$ oxidation states. Which of the following statements about these chlorides is correct?

Options:

(A) $MCl_2$ is more volatile than $MCl_4$

(B) $MCl_2$ is more soluble in anhydrous ethanol than $MCl_4$

(C) $MCl_2$ is more ionic than $MCl_4$

(D) $MCl_2$ is more easily hydrolysed than $MCl_4$

Sigma Percentile · Accepted Answer

### The Battle of Oxidation States

When a metal $M$ forms chlorides in multiple oxidation states, such as $+2$ and $+4$, it sets the stage for a fascinating comparison of chemical bonding. We are looking at two distinct compounds: $MCl_2$ and $MCl_4$. To understand which statement among the options is correct, we must dive into the microscopic world of ions and electron clouds.

### Enter Fajan's Rule

To predict the nature of the bond between the metal and chlorine, we rely on **Fajan's Rule**. This fundamental principle helps us determine the partial covalent character in an ionic compound. According to Fajan's Rule, the covalent character increases when:
1. The cation has a **high positive charge**.
2. The cation has a **small atomic radius**.
3. The anion has a large atomic radius.

In our scenario, the anion (chloride, $Cl^-$) remains constant. The deciding factor is the metal cation.

### The Tug of War: Polarizing Power

Let's compare the two cations: $M^{2+}$ and $M^{4+}$.

The $M^{4+}$ ion has double the positive charge of the $M^{2+}$ ion. Furthermore, because it has lost more electrons, the effective nuclear charge pulls the remaining electron shells closer, making $M^{4+}$ significantly smaller in size than $M^{2+}$.

This combination of high charge and small size gives $M^{4+}$ an immense **polarizing power**. It acts like a powerful magnet, aggressively distorting the spherical electron cloud of the neighboring chloride ions and pulling that electron density into the space between the nuclei. This sharing of electrons is the very essence of a covalent bond. Therefore, $MCl_4$ exhibits a high degree of covalent character.

Conversely, the $M^{2+}$ ion, being larger and carrying less charge, has a much lower polarizing power. It cannot distort the chloride's electron cloud as effectively, meaning the electrons remain largely localized on the chloride ions. Thus, $MCl_2$ retains its predominantly **ionic character**.

### The Verdict on Bonding

With this understanding, we can confidently evaluate the given options. Because $MCl_4$ is highly covalent, it exists as discrete molecules held together by weak intermolecular forces, making it highly volatile and soluble in non-polar organic solvents.

On the other hand, $MCl_2$ is an ionic compound, forming a robust crystal lattice that requires significant energy to break, making it less volatile and possessing a higher melting point.

Therefore, the statement that **$MCl_2$ is more ionic than $MCl_4$** is the only scientifically accurate conclusion. The microscopic tug-of-war perfectly explains the macroscopic physical properties!

Animated Solution for Chemistry - s and p-Block Elements: A metal, $M$ forms chlorides in its $+ 2$ and $+ 4$ oxidation states. Which of the following statements about these chlorides is correct?

Select Answer:

Visualized Solution

$M C l_{2} vs M C l_{4}$

\text{Fajan's Rule}

\text{Comparing Polarising Power}

\text{Covalent vs Ionic Character}

\text{Conclusion}

\text{Physical Properties}

The Sigma Insight: Group 14 Elements

The Battle of Oxidation States

Enter Fajan's Rule

The Tug of War

The Verdict on Bonding

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