Iridium (Ir) is a dense, silvery-white metal belonging to the platinum group metals, highly valued for its resistance to corrosion and high melting point. The number of electrons an Iridium atom possesses is the foundation for its chemical identity. A neutral atom of Iridium contains exactly 77 electrons. This precise number governs the element’s physical and chemical behavior.
The Total Electron Count
The number of electrons in a neutral atom is directly determined by its atomic number (\(Z\)). For Iridium, the atomic number is 77. This means that the nucleus of every Iridium atom contains 77 protons, which carry a positive electrical charge.
In a neutral atom, the total positive charge from the protons must be perfectly balanced by an equal amount of negative charge from the electrons. Therefore, a neutral Iridium atom must possess 77 electrons to maintain this charge equilibrium. The number of electrons only changes when the Iridium atom loses or gains electrons to form an electrically charged particle known as an ion.
Arrangement of Iridium’s Electrons
The total count of 77 electrons is constant for a neutral Iridium atom, but their arrangement dictates the element’s distinct properties. These electrons are organized into distinct energy levels, or shells, surrounding the nucleus. For Iridium, these 77 electrons are distributed across six principal energy shells.
The electrons fill these shells in a pattern represented by the structure 2, 8, 18, 32, 15, 2, moving outward from the nucleus. The full ground-state electron configuration is written in a shorthand notation as \([Xe] 4f^{14} 5d^7 6s^2\), referencing the stable configuration of the noble gas Xenon.
This notation shows that 54 electrons are arranged like the Xenon atom, followed by 14 electrons in the \(4f\) subshell. The outermost arrangement consists of 9 electrons in the \(5d\) and \(6s\) subshells. Iridium is a \(d\)-block element, meaning that its characteristic chemical behavior is influenced by the filling of the \(5d\) orbitals.
Electrons and Iridium’s Chemical States
The chemical reactivity of Iridium is determined by its valence electrons, which are the electrons located in the outermost energy levels. Iridium has nine valence electrons: two in the \(6s\) orbital and seven in the \(5d\) orbital, totaling nine electrons available for chemical bonding. Chemical reactions occur when atoms share, gain, or lose these outermost electrons.
When Iridium forms compounds, it typically loses valence electrons to become a positively charged ion, or cation. The most common chemical states, known as oxidation states, are Iridium(III) (+3) and Iridium(IV) (+4).
In the +3 state, the Iridium atom has lost three electrons, leaving it with 74 electrons total. In the +4 state, it has lost four electrons, resulting in 73 electrons. Iridium can exhibit a wide range of chemical states, from as low as -3 to as high as +9.