Write electronic configurations for the following ions: a. Al3+ (Z = 13) b. O2- (Z = 8) c. Fe3+ (Z = 26) d. Cu2+ (Z = 29) e. Cu+ (Z = 29)
The Correct Answer and Explanation is:
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Electronic Configurations of Ions
a. Al³⁺ (Z = 13)
Neutral Al: 1s² 2s² 2p⁶ 3s² 3p¹
Al³⁺: 1s² 2s² 2p⁶
b. O²⁻ (Z = 8)
Neutral O: 1s² 2s² 2p⁴
O²⁻: 1s² 2s² 2p⁶
c. Fe³⁺ (Z = 26)
Neutral Fe: [Ar] 4s² 3d⁶
Fe³⁺: [Ar] 3d⁵
d. Cu²⁺ (Z = 29)
Neutral Cu: [Ar] 4s¹ 3d¹⁰
Cu²⁺: [Ar] 3d⁹
e. Cu⁺ (Z = 29)
Neutral Cu: [Ar] 4s¹ 3d¹⁰
Cu⁺: [Ar] 3d¹⁰
Explanation
The electronic configuration of an ion depends on the number of electrons it possesses, which differs from its neutral atom depending on whether it loses or gains electrons.
For Al³⁺, aluminum has an atomic number of 13, with a neutral configuration of 1s² 2s² 2p⁶ 3s² 3p¹. It loses three electrons to become Al³⁺. The electrons are lost from the outermost shell first (3p and 3s orbitals), leaving the stable noble gas configuration of neon (1s² 2s² 2p⁶).
In the case of O²⁻, oxygen has 8 electrons in its neutral state (1s² 2s² 2p⁴). Gaining two electrons gives it a full 2p orbital (2p⁶), resulting in a noble gas configuration like neon. Negative ions (anions) gain electrons to achieve full outer shells.
For Fe³⁺, iron has 26 electrons, and its neutral configuration is [Ar] 4s² 3d⁶. Electrons are removed first from the 4s orbital (higher energy in ions) and then from the 3d. Removing three electrons gives [Ar] 3d⁵, which is a half-filled d-subshell and relatively stable.
Cu²⁺ and Cu⁺ are exceptions due to copper’s unusual electron configuration. Neutral copper has a configuration of [Ar] 4s¹ 3d¹⁰, favoring a full d-subshell over a filled 4s. Cu⁺ loses the 4s¹ electron, resulting in [Ar] 3d¹⁰. Cu²⁺ loses both the 4s¹ and one 3d electron, resulting in [Ar] 3d⁹.
These examples illustrate the principles of electron configuration in ions, where stability and energy levels govern which electrons are lost or gained.
