Question

Identify the incorrect order against the stated property.

• A. Ge>Sn>Pb - Ionization enthalpy
• B. Ge>Pb>Sn - Melting point
• C. Pb>Sn>Ge - Density
• D. Ge>Pb>Sn - Electrical resistivity

Hint:

Periodic Trends for Group 14 (C, Si, Ge, Sn, Pb): - Ionization Enthalpy: Generally decreases down a group. Irregularity: IE(Pb)>IE(Sn) due to poor shielding by d and f electrons in Pb, leading to higher effective nuclear charge. Correct order: Ge>Pb>Sn. - Melting Point: Decreases from C to Ge, then Sn has a significantly lower MP, and Pb is slightly higher than Sn. C>Si>Ge>Pb>Sn (approx). - Density: Increases down the group: Pb>Sn>Ge. - Electrical Resistivity: Decreases down the group as metallic character increases (Ge is semiconductor, Sn and Pb are metals). So Ge >> (metals). For Sn and Pb (metals), Sn is a better conductor (lower resistivity) than Pb. So, Ge>Pb>Sn is generally true.

Answer 1

The Correct Option is A

The elements Ge, Sn, Pb are in Group 14 (Carbon family).
Order down the group: C, Si, Ge, Sn, Pb.

Option (1): Ionization Enthalpy (IE):
General trend: Ionization Enthalpy (IE) decreases down a group due to increasing atomic size and shielding effect.
Expected order: \( \text{Ge} > \text{Sn} > \text{Pb} \).
The stated order \( \text{Ge} > \text{Sn} > \text{Pb} \) is consistent with the general trend.
However, there can be irregularities.
The first IE values (kJ/mol) are:
Ge: 762, Sn: 709, Pb: 716.
So, the actual order is \( \text{Ge} > \text{Pb} > \text{Sn} \).
Therefore, the stated order \( \text{Ge} > \text{Sn} > \text{Pb} \) is incorrect.

Option (2): Melting Point:
Melting points generally decrease down this group for C, Si, Ge.
But then Sn and Pb have lower melting points due to weaker metallic bonding (d- and f-orbital effects, inert pair effect influences bonding character).
Melting points (in K):
C (diamond): ~3823 K, Si: 1687 K, Ge: 1211 K, Sn: 505 K, Pb: 601 K.
So the order of melting points is \( \text{Ge} > \text{Pb} > \text{Sn} \).
The stated order \( \text{Ge} > \text{Pb} > \text{Sn} \) is correct.

Option (3): Density:
Density generally increases down a group due to increasing atomic mass and similar packing efficiency or decreasing atomic volume change relative to mass change.
Densities (g/cm\(^3\)):
Ge: 5.32, Sn: 7.31 (white tin), Pb: 11.34.
So the order of density is \( \text{Pb} > \text{Sn} > \text{Ge} \).
The stated order \( \text{Pb} > \text{Sn} > \text{Ge} \) is correct.

Option (4): Electrical Resistivity:
Metals generally have low resistivity (high conductivity). Non-metals have high resistivity. Metalloids are in between.
Down Group 14, metallic character increases: Ge (metalloid/semiconductor) < Sn (metal) < Pb (metal).
Resistivity values (\( \Omega \cdot m \) at 20°C):
Ge: \( 0.46 \) (semiconductor), Sn: \( 1.09 \times 10^{-7} \), Pb: \( 2.08 \times 10^{-7} \).
So, for metals Sn and Pb, Pb has slightly higher resistivity than Sn.
Ge (semiconductor) has much higher resistivity than both.
The order of resistivity is \( \text{Ge} >> \text{Pb} > \text{Sn} \).
The stated order \( \text{Ge} > \text{Pb} > \text{Sn} \) is correct.

Conclusion:
The incorrect order is (1) \( \text{Ge} > \text{Sn} > \text{Pb} \) for Ionization Enthalpy.
The correct order is \( \text{Ge} > \text{Pb} > \text{Sn} \).
Therefore, the answer is \( \boxed{1} \).