Iridium (Ir) - Atomic Structure

Understanding Iridium: An Atomic Perspective

Iridium, symbolized as Ir, is a chemical element with an atomic number of 77. It is a dense, very hard, brittle, silvery-white transition metal of the platinum group. Known for its extreme resistance to corrosion even at high temperatures, Iridium is one of the densest naturally occurring elements. Its properties make it valuable in specialized applications, some of which are relevant to various advanced industries globally, including in India.

Atomic Structure of Iridium

The atomic structure of Iridium can be understood by examining its constituent subatomic particles and their arrangement.

Protons, Neutrons, and Electrons

1. Protons: The atomic number (Z) of an element defines the number of protons in its nucleus. For Iridium, the atomic number is 77. Therefore, an Iridium atom contains 77 protons.
2. Electrons: In a neutral atom, the number of electrons orbiting the nucleus is equal to the number of protons. Thus, a neutral Iridium atom possesses 77 electrons.
3. Neutrons: The number of neutrons in an atom can vary, leading to different isotopes of an element. The mass number (A) of an isotope represents the total number of protons and neutrons in the nucleus (A = protons + neutrons). For the most common isotope of Iridium, Iridium-192, the mass number is 192. Number of neutrons = Mass Number - Atomic Number Number of neutrons = 192 - 77 = 115 neutrons (for Iridium-192). Natural Iridium is a mixture of two stable isotopes, Iridium-191 and Iridium-193. The number of neutrons would be 114 for Ir-191 and 116 for Ir-193.

Electron Configuration

Electron configuration describes the distribution of electrons of an atom or molecule in atomic orbitals. For Iridium (Z=77), the electron configuration follows the Aufbau principle, Hund’s rule, and the Pauli exclusion principle.

The full electron configuration for Iridium is: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^7$

Using the noble gas notation for simplicity, the preceding noble gas is Xenon (Xe), which has an atomic number of 54. The electron configuration for Xenon is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6$.

Therefore, the condensed electron configuration for Iridium is: $[Xe] 4f^{14} 5d^7 6s^2$

This configuration indicates that after the stable electron core of Xenon, there are 14 electrons in the 4f subshell, 7 electrons in the 5d subshell, and 2 electrons in the 6s subshell. The 4f subshell, belonging to the f-block elements, is filled before the 5d subshell in transition metals like Iridium.

Valence Electrons

Valence electrons are the electrons located in the outermost shell of an atom that participate in chemical bonding. For transition metals like Iridium, valence electrons typically include the electrons in the outermost s subshell and the partially filled d subshell of the penultimate shell.

From Iridium’s electron configuration, $[Xe] 4f^{14} 5d^7 6s^2$:

• The outermost shell is the 6th shell, containing $6s^2$ electrons.
• The 5d subshell, although not the outermost principal energy level, is partially filled ($5d^7$) and these electrons can participate in bonding, a characteristic feature of transition metals.
• The 4f subshell is completely filled ($4f^{14}$) and is generally considered part of the core electrons, not typically participating in chemical reactions due to its deeper penetration and shielding effect.

Thus, the valence electrons for Iridium are the $6s^2$ electrons and the $5d^7$ electrons. Total number of valence electrons = 2 (from 6s) + 7 (from 5d) = 9 valence electrons. Iridium exhibits a variety of oxidation states, often including +3 and +4, which is consistent with the involvement of its 5d and 6s electrons in bonding.