Write the condensed
electron configuration for each ion: Co2 N3− Ca2
The Correct Answer and Explanation is :
Here’s the condensed electron configuration for each ion you mentioned:
- Cobalt Ion (Co²⁺):
- Condensed Electron Configuration: ([Ar] 3d^7)
- Nitride Ion (N³⁻):
- Condensed Electron Configuration: ([He] 2s^2 2p^6)
- Calcium Ion (Ca²⁺):
- Condensed Electron Configuration: ([Ar])
Explanation:
1. Cobalt Ion (Co²⁺):
Cobalt (Co) has an atomic number of 27, meaning it has 27 electrons in its neutral state. The electron configuration of neutral cobalt is ([Ar] 4s^2 3d^7). When cobalt loses two electrons to form the Co²⁺ ion, the electrons are removed from the highest energy level, which is the 4s subshell. Therefore, the electron configuration for Co²⁺ becomes ([Ar] 3d^7), indicating that cobalt retains its 3d electrons but loses its 4s electrons.
2. Nitride Ion (N³⁻):
Nitrogen (N) has an atomic number of 7, with a neutral electron configuration of ([He] 2s^2 2p^3). To achieve a stable electron configuration, nitrogen can gain three electrons, resulting in the nitride ion (N³⁻). This addition fills the 2p subshell, leading to the configuration ([He] 2s^2 2p^6), which is the same as that of neon (Ne), a noble gas, indicating that the nitride ion is stable.
3. Calcium Ion (Ca²⁺):
Calcium (Ca) has an atomic number of 20. The electron configuration for neutral calcium is ([Ar] 4s^2). When calcium forms a Ca²⁺ ion, it loses two electrons from the 4s subshell, resulting in a configuration of ([Ar]). This shows that the Ca²⁺ ion has a full outer shell of electrons, making it more stable.
In summary, these electron configurations illustrate how atoms transition between their neutral states and their ionized forms, reflecting the changes in electron arrangement that correspond to their charge states.