Question

The metallic radii of \(\mathrm{Ni}\), \(\mathrm{Pd}\), and \(\mathrm{Pt}\) are \(125,138,\) and \(139 \mathrm{pm},\) respectively. Why is the difference in radius between Pt and Pd so much less than between Pd and Ni?

Short answer

The difference in atomic radius between Pt and Pd is much less than between Pd and Ni due to the lanthanide contraction, which results from an increase in effective nuclear charge and decrease in atomic size due to poor shielding of the nucleus by 4f electrons.

Step-by-step solution

Understand the Periodic Trends

The atomic radius of an element generally increases down a group and decreases across a period from left to right. This is due to the increase in the number of electron shells and effective nuclear charge respectively.

Analyze the Electron Configurations

The electron configurations are as follows: \n- Nickel (Ni) has the electron configuration [Ar] 3d8 4s2.\n- Palladium (Pd) has the unusual electron configuration [Kr] 4d10 5s0. Even though we would expect it to have 5s1, it fully occupies the d orbital instead.\n- Platinum (Pt) has the electron configuration [Xe] 4f14 5d9 6s1. It has one more d electron and one more f electron compared to Pd.

Relate the Trends with Configurations

The lesser increase in radius from Pd to Pt compared to Ni to Pd can be attributed to the lanthanide contraction, which is the greater-than-expected decrease in ionic radii of the elements in the lanthanide series (57–71), including the elements in period 6, like Pt. This is due to the poor shielding of the nucleus by 4f electrons, causing an increase in effective nuclear charge and decrease in atomic size.

Key concepts

Atomic Radius

The atomic radius is an essential concept in understanding the size of atoms and how they interact with each other. It refers to the distance from the center of an atom's nucleus to the outer edge of its electron cloud. However, this definition may vary slightly depending on the type of measurement, such as covalent or metallic radii.

The understanding of atomic radius is rooted in periodic trends. Generally, as you move down a group in the periodic table, the atomic radius increases. This is because each added row means a new electron shell is added, making the atom larger. Conversely, as you move across a period from left to right, the atomic radius decreases. This is due to the increase in effective nuclear charge – the nucleus's powerful pull – which pulls electrons closer to the nucleus, reducing the atomic size.

In the given problem, these trends explain why there are notable differences in the metallic radii of \(\mathrm{Ni}\), \(\mathrm{Pd},\) and \(\mathrm{Pt}.\) The change from \(\mathrm{Ni}\) to \(\mathrm{Pd}\) involves moving down the periodic table, leading to a more significant increase in atomic radius compared to the shift from \(\mathrm{Pd}\) to \(\mathrm{Pt}.\)

Electron Configuration

Electron configuration is a description of the number and arrangement of electrons in an atom's orbitals. It helps predict an element's chemical properties and behavior.

For nickel (Ni), the electron configuration is \[ \mathrm{[Ar] \ 3d^8 \ 4s^2} \], meaning it has electrons filling its 3d and 4s orbitals. Palladium (Pd) has a unique electron configuration \(\mathrm{[Kr] \ 4d^{10} \ 5s^0},\) where all the electrons are in the 4d orbital. This unusual configuration arises due to the stabilization provided by having a fully filled d orbital.

Platinum (Pt), with the configuration \[ \mathrm{[Xe] \ 4f^{14} \ 5d^9 \ 6s^1} \], has more electrons as it includes f orbitals. While it seems more complex, this configuration gives insights into its behavior. The presence of additional d and f electrons leads to increased electron-electron repulsion and effective nuclear charge, affecting the atomic radius.

Lanthanide Contraction

The lanthanide contraction is a phenomenon that has significant implications for the elements, particularly in the third transition series. It describes the unexpected decrease in ionic radii among the elements of the lanthanide series (atomic numbers 57–71).

This contraction is primarily due to the poor shielding effect of the 4f electrons. While electrons in these inner orbitals are added, they don't shield the outer electrons well from the nucleus's positive charge. Thus, the outer electrons experience a more substantial pull, leading to a smaller atomic radius than expected.

This phenomenon affects elements like platinum (Pt), which precede lanthanides but are in the same period. In the exercise, this contraction explains why the increase in radius from palladium (Pd) to platinum (Pt) is much less marked than expected. While Pt has additional electrons compared to Pd, these electrons don't expand the atomic size significantly, due to the strong nuclear attraction facilitated by the lanthanide contraction.