Writing electron configurations for ions involves adjusting the electron count based on the ion's charge. For negative ions, such as Na-, one electron is added to the configuration, resulting in [Ne] 3s2. For positive ions, like Na+, one electron is subtracted, leading to [Ne]. Transition metals require careful consideration; for instance, titanium (Ti) has the configuration [Ar] 4s2 3d2, where Ti- becomes [Ar] 4s2 3d3, but Ti+ is [Ar] 4s1 3d2, indicating that electrons are removed from the higher energy 4s orbital before the 3d orbital.
PREREQUISITESChemistry students, educators, and anyone interested in mastering the principles of electron configurations and ion behavior in atomic theory.
kuahji said:If your ion has a negative charge, add one electron to the electron configuration. If your ion is positive, subtract one electron.
For example, take Na. The electron configuration for the atom is [Ne] 3s1. If you have Na- it becomes [Ne] 3s2 & Na+, just [Ne].
Another example, suppose you have Cl, the electron configuration is [Ne] 3s2 3p5. Cl- is [Ar] & Cl+ is [Ne] 3s2 3p4. Notice how you add & subtract from the orbital with the highest energy
You have to be careful with the transition metals. Take for example [B]Ti, its electron configuration is [Ar] 4s2 3d2. Ti- is [Ar] 4s2 3d3. But here is where it "can" be tricky. Ti+ is [Ar] 4s1 3d2. [/B] The d orbitals are written after the s, but the 4s orbitals are still of higher energy. So you remove from them first, before the d orbitals.