The Correct Answer and Explanation is:
Here are the electron configurations for the requested ions.
Correct Answer:
(a) Rb⁺: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶ or [Kr]
(b) Br⁻: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶ or [Kr]
(c) S²⁻: 1s²2s²2p⁶3s²3p⁶ or [Ar]
(d) Ba²⁺: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶ or [Xe]
(e) Al³⁺: 1s²2s²2p⁶ or [Ne]
Explanation
Writing an electron configuration for an ion involves first understanding the configuration of the neutral atom and then adjusting the number of electrons based on the ion’s charge. Cations, which are positive ions, have lost electrons, while anions, which are negative ions, have gained electrons. Electrons are always removed from or added to the orbitals in the outermost shell first.
(a) Rb⁺: A neutral rubidium (Rb) atom has the atomic number 37, meaning it possesses 37 electrons. Its configuration is [Kr] 5s¹. To form the Rb⁺ ion, the atom loses one electron. This electron is removed from the outermost orbital, the 5s orbital. The resulting ion has 36 electrons, giving it the electron configuration of 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶. This is the same configuration as the noble gas krypton (Kr), making Rb⁺ very stable.
(b) Br⁻: A neutral bromine (Br) atom has an atomic number of 35, giving it 35 electrons with a configuration of [Ar] 4s²3d¹⁰4p⁵. The bromide ion, Br⁻, has a negative one charge, indicating it has gained one electron. This additional electron fills the last available spot in its outermost shell, the 4p orbital. The Br⁻ ion now has 36 electrons and an electron configuration of 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶, which is also isoelectronic with krypton.
(c) S²⁻: A neutral sulfur (S) atom has 16 electrons and a configuration of [Ne] 3s²3p⁴. To form the sulfide ion, S²⁻, the atom gains two electrons. These electrons are added to the 3p orbital, filling it completely. The resulting ion has 18 electrons and a configuration of 1s²2s²2p⁶3s²3p⁶. This is the stable electron configuration of the noble gas argon (Ar).
(d) Ba²⁺: A neutral barium (Ba) atom, with atomic number 56, has 56 electrons and a configuration of [Xe] 6s². The Ba²⁺ ion has a positive two charge because it has lost two electrons. These electrons are removed from its outermost 6s orbital. The resulting ion has 54 electrons, giving it the configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶. This is identical to the configuration of the noble gas xenon (Xe).
(e) Al³⁺: A neutral aluminum (Al) atom has 13 electrons and a configuration of [Ne] 3s²3p¹. The Al³⁺ ion is formed by losing three electrons. One is lost from the 3p orbital, and the remaining two are lost from the 3s orbital. The ion is left with 10 electrons and a configuration of 1s²2s²2p⁶, which is isoelectronic with the noble gas neon (Ne).
