Answer: The atoms whose ions have electronic configuration as \(nd^5\) are Iridium (Ir), Iron (Fe) and Rhenium (Re)
Explanation:
An ion is formed when a neutral atom looses or gains electrons.
- When an atom looses electrons, it results in the formation of positive ion known as cation.
- When an atom gains electrons, it results in the formation of negative ion known as anion.
For the given elements:
Iridium: The electronic configuration of this element is \([Xe]6s^25d^7\). The stable oxidation state of iridium is +4.
The electronic configuration of \(Ir^{4+}\) is \([Xe]5d^5\)
Radon: The electronic configuration of this element is \([Xe]6s^25d^{10}6p^6\). This is a stable element.
Iron: The electronic configuration of this element is \([Ar]4s^23d^6\). The stable oxidation state of iron is +3.
The electronic configuration of \(Fe^{3+}\) is \([Ar]3d^5\)
Protactinium: The electronic configuration of this element is \([Rn]7s^26d^15f^2\). The stable oxidation state of protactinium is +3.
The electronic configuration of \(Pa^{3+}\) is \([Rn]7s^2\)
Ytterium: The electronic configuration of this element is \([Kr]5s^24d^1\). The stable oxidation state of ytterium is +3.
The electronic configuration of \(Y^{3+}\) is \([Ar]4s^23d^{10}4p^6\)
Rhenium: The electronic configuration of this element is \([Xe]6s^25d^5\). The stable oxidation state of rhenium is +2.
The electronic configuration of \(Re^{2+}\) is \([Xe]5d^5\)
Hence, the atoms whose ions have electronic configuration as \(nd^5\) are Iridium (Ir), Iron (Fe) and Rhenium (Re)
