Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

The radius of \(\mathrm{La}^{3+}\) (atomic number \(=57\) ) is \(1.06 \AA\). Which one of the following given values will be closest to the radius of \(\mathrm{Lu}^{3+}\) (atomic number \(\left.=71\right)\) ? [2003] (a) \(0.85 \AA\) (b) \(1.06 \AA\) (c) \(1.40 \AA\) (d) \(1.60 \AA\)

Step-by-Step Solution

Answer

The radius of 7 ext{Lu}^{3+}7 is closest to (a) 0.85 7 ext{7}.

Step 1: Understanding Lanthanide Contraction

The lanthanide series in the periodic table consists of 15 elements, starting from lanthanum (La) to lutetium (Lu). In this series, there is a phenomenon called lanthanide contraction which results in the gradual decrease in atomic and ionic sizes with increasing atomic number.

Step 2: Analyzing the Trend

Due to the lanthanide contraction, despite having more electrons, the effective nuclear charge increases, which causes the atomic radii to decrease. Hence, as we move from La (7 ext{La}^{3+}7) to Lu (7 ext{Lu}^{3+}7), the radius is expected to decrease.

Step 3: Evaluating the Options

The given radius of 7 ext{La}^{3+}7 is 1.06 7 ext{7}. We expect 7 ext{Lu}^{3+}7 to have a smaller radius due to lanthanide contraction. Among the given options: - (a) 0.85 7 ext{7} is smaller than 1.06 7 ext{7} - (b) 1.06 7 ext{7} is equal - (c) 1.40 7 ext{7} is larger - (d) 1.60 7 ext{7} is much larger The closest and most reasonable answer considering the contraction is (a) 0.85 7 ext{7}.

Key Concepts

Ionic Radius

In chemistry, the ionic radius is the measure of an atom's ion in a crystal lattice. It is crucial in predicting the atom's behavior when forming compounds. As atoms lose or gain electrons to form ions, their radii can change significantly.
Specifically, when an atom loses electrons to become a positively charged ion (cation), the ionic radius generally decreases. This decrease happens because there are fewer electrons repelling each other, allowing the remaining electrons to be pulled closer to the nucleus. On the other hand, when an atom gains electrons, forming a negatively charged ion (anion), the ionic radius increases due to electron repulsion.In the case of lanthanides, such as the transition from Lanthanoides like La\(^{3+}\) and Lu\(^{3+}\), even though they have a positive charge, you would notice a decrease in ionic radius as the effective nuclear charge becomes stronger down the series. This is due to the unique lanthanide contraction phenomenon.

Periodic Trends

Periodic trends are consistent patterns or particular tendencies in the properties of elements that appear periodically throughout the periodic table. These trends can influence a variety of properties, such as atomic size, electronegativity, ionization energy, and more. One notable trend is the change in atomic or ionic size across periods and down groups. As you move across a period from left to right, ionic size usually decreases due to the increasing number of protons, which attracts the electrons more strongly, pulling them closer to the nucleus.
Additionally, as we move down a group in the periodic table, the ionic size tends to increase due to the addition of electron shells. However, there is a unique trend observed in the lanthanide series known as the lanthanide contraction. Despite adding more protons and electrons as we go from lanthanum (La) to lutetium (Lu), the elements show a decreasing trend in ionic radii. This decrease arises because the additional electrons fill inner f-orbitals, which are not efficient in shielding each other from the nucleus's pull.

Effective Nuclear Charge

The effective nuclear charge (Z\(_{eff}\)) refers to the net positive charge experienced by the outermost electrons of an atom. It is a crucial concept as it helps explain many periodic trends, including atomic and ionic sizes, ionization energies, and electronegativity.The effective nuclear charge is influenced by:

• Atom's proton count: More protons in the nucleus increase Z\(_{eff}\), as observed across a period.
• Electron shielding: Inner electrons can shield outer electrons from the full force of the nucleus's positive charge. In the lanthanides, the shielding by f-electrons is particularly poor, which causes a lack of effective shielding and increases Z\(_{eff}\).

The gradual decrease in lanthanide ionic radii from La\(^{3+}\) to Lu\(^{3+}\)is due to increasing effective nuclear charge. As Z\(_{eff}\) increases, electrons are more strongly attracted to the nucleus, ultimately resulting in the contraction of atomic and ionic sizes along the series.