Give the electron configurations for the following ions: P5+, P3+, Sn4+, Se2+, I+, and Ni2+.
The Correct Answer and Explanation is:
Electron Configurations of Ions
- P⁵⁺ (Phosphorus 5+): [Ne]
- P³⁺ (Phosphorus 3+): [Ne] 3s²
- Sn⁴⁺ (Tin 4+): [Kr] 4d¹⁰
- Se²⁺ (Selenium 2+): [Ar] 3d¹⁰ 4s² 4p²
- I⁺ (Iodine 1+): [Kr] 4d¹⁰ 5s² 5p⁴
- Ni²⁺ (Nickel 2+): [Ar] 3d⁸
Explanation
Electron configuration refers to the arrangement of electrons in an atom or ion. When atoms become ions, they either lose or gain electrons to achieve greater stability, typically aiming for a noble gas configuration. The configurations of cations (positive ions) are formed by removing electrons from the outermost shells, while anions (negative ions) are formed by adding electrons to the outer orbitals.
- Phosphorus (P) has an atomic number of 15: [Ne] 3s² 3p³.
- In P⁵⁺, it loses all 5 valence electrons (3s² 3p³), leaving just the [Ne] core.
- In P³⁺, it loses three electrons (3p³), leaving the configuration [Ne] 3s².
- Tin (Sn) has an atomic number of 50: [Kr] 5s² 4d¹⁰ 5p².
- In Sn⁴⁺, it loses all four valence electrons (5s² 5p²), resulting in [Kr] 4d¹⁰.
- Selenium (Se) has atomic number 34: [Ar] 4s² 3d¹⁰ 4p⁴.
- In Se²⁺, it loses two 4p electrons, giving [Ar] 3d¹⁰ 4s² 4p².
- Iodine (I) has atomic number 53: [Kr] 5s² 4d¹⁰ 5p⁵.
- In I⁺, one electron is removed from the 5p orbital, yielding [Kr] 4d¹⁰ 5s² 5p⁴.
- Nickel (Ni) has atomic number 28: [Ar] 4s² 3d⁸.
- In Ni²⁺, two electrons are lost—typically from the 4s orbital first, resulting in [Ar] 3d⁸.
Note: Transition metals (like Ni and Sn) often lose s-electrons before d-electrons. Electron configuration helps in predicting chemical reactivity, bonding behavior, and magnetic properties of ions.
