The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence
(a) GeX2

Q: What exactly is the inert pair effect?

The inert pair effect is the reluctance of the outermost $s$-orbital electrons ($ns^2$) to participate in chemical bonding, especially in the heavier elements of the p-block. This happens because these electrons are held tightly by the nucleus due to poor shielding by inner $d$ and $f$ electrons.

Q: Why does the inert pair effect increase down the group?

As you move down the group, the number of inner $d$ and $f$ electrons increases. Since $d$ and $f$ orbitals have poor shielding capacity, the effective nuclear charge pulling on the outermost $s$-electrons increases, making them more inert.

Q: Why are $d$ and $f$ orbitals considered poor shielders?

Shielding depends on the shape of the orbital. $s$ and $p$ orbitals are more compact and effectively block the nuclear charge. $d$ and $f$ orbitals are highly diffused and spread out, allowing the nuclear charge to easily pass through and attract the outermost electrons.

Q: Does Carbon show a $+2$ oxidation state?

Carbon primarily shows a $+4$ oxidation state. While it can form compounds like Carbon Monoxide ($CO$) where it is in a $+2$ state, the $+4$ state is vastly more stable. The inert pair effect is negligible in Carbon because it lacks $d$ and $f$ orbitals.

Q: Which is more stable: $SnCl_2$ or $SnCl_4$?

For Tin ($Sn$), the $+4$ oxidation state is still slightly more stable than the $+2$ state, so $SnCl_4$ is generally more stable than $SnCl_2$. However, $SnCl_2$ is a strong reducing agent because it wants to oxidize to the more stable $+4$ state. In contrast, for Lead ($Pb$), $PbCl_2$ is much more stable than $PbCl_4$.

Question

The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence

Options:

(A) $GeX_2 < SiX_2 < SnX_2 < PbX_2$

(B) $SiX_2 < GeX_2 < PbX_2 < SnX_2$

(C) $SiX_2 < GeX_2 < SnX_2 < PbX_2$

(D) $PbX_2 < SnX_2 < GeX_2 < SiX_2$

Sigma Percentile · Accepted Answer

### The Mystery of the Inert Pair Effect

When we look at the elements of Group 14—Silicon ($Si$), Germanium ($Ge$), Tin ($Sn$), and Lead ($Pb$)—we expect them to behave in a predictable manner. After all, they share the same general valence shell electronic configuration: $ns^2 np^2$. With four electrons in their outermost shell, it seems perfectly logical that they would readily form compounds in the $+4$ oxidation state by sharing or losing all four electrons.

However, nature loves to throw a curveball, and in the p-block of the periodic table, this curveball is known as the **Inert Pair Effect**.

### The Reluctant Electrons

As we travel down Group 14, from Silicon to Lead, the atoms get larger, and new inner electron shells are filled. Specifically, the intervening $d$ and $f$ orbitals begin to populate. Here is where the plot thickens: $d$ and $f$ orbitals are notoriously bad at shielding the outer electrons from the pull of the nucleus. Their shapes are diffuse, allowing the positive charge of the nucleus to "leak" through.

Because of this poor shielding, the effective nuclear charge ($Z_{eff}$) felt by the outermost $s$-electrons ($ns^2$) is surprisingly strong. The nucleus grips these two $s$-electrons tightly, pulling them closer and making them reluctant to participate in chemical bonding. They become "inert."

### The Shift in Stability

What does this mean for the oxidation states? For lighter elements like Silicon, the $+4$ state is highly stable. But as we move down to Germanium, Tin, and finally Lead, the inert pair effect becomes increasingly dominant. The two $s$-electrons refuse to bond, meaning only the two $p$-electrons are available for reactions.

Consequently, the stability of the $+4$ oxidation state decreases down the group, while the stability of the $+2$ oxidation state steadily increases. By the time we reach Lead ($Pb$), the $+2$ state is far more stable than the $+4$ state.

### Solving the Dihalide Puzzle

The question asks about the stability of dihalides ($MX_2$). In a dihalide, the central metal atom is in the $+2$ oxidation state. Since the stability of the $+2$ oxidation state increases as we move down the group due to the inert pair effect, the stability of the corresponding dihalides must follow the exact same trend.

Therefore, the stability order is:
**$SiX_2 < GeX_2 < SnX_2 < PbX_2$**

Silicon dihalides are highly unstable and tend to oxidize to the $+4$ state, whereas Lead dihalides are incredibly stable. Understanding the inert pair effect unlocks the secret to predicting the chemical behavior of heavy p-block elements!

Animated Solution for Chemistry - s and p-Block Elements: The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence

Select Answer:

Visualized Solution

Group $14$ Elements

Oxidation States

The Inert Pair Effect

Cause of Inert Pair Effect

Stability of $+ 2$ State

Stability of Dihalides

Final Answer

The Sigma Insight: Group 14 Elements

The Mystery of the Inert Pair Effect

The Reluctant Electrons

The Shift in Stability

Solving the Dihalide Puzzle

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