Question

Electronic configuration of O^-2 ion is

• A. 1s², 2s², 2p⁴
• B. 1s², 2s², 2p⁵
• C. 1s², 2s², 2p⁶
• D. 1s², 2s², 2p³

Answer 1

The Correct Option is C

The O²⁻ ion is formed when a neutral oxygen atom gains two additional electrons. Oxygen, with an atomic number of 8, has 8 protons in its nucleus and normally possesses 8 electrons in its neutral state. The electronic configuration of a neutral oxygen atom is: $$ 1s^2, 2s^2, 2p^4 $$ In this configuration, the 1s and 2s orbitals are completely filled, while the 2p subshell contains 4 electrons. To achieve a stable electron configuration resembling that of a noble gas (in this case, neon, Ne), oxygen tends to gain 2 more electrons. This addition fills the 2p subshell, which can hold a maximum of 6 electrons. When 2 electrons are added to the neutral oxygen atom, the total number of electrons in the O²⁻ ion becomes: $$ 8 + 2 = 10 \text{ electrons.} $$ These 10 electrons are distributed as follows: - The 1s orbital holds 2 electrons: $1s^2$, - The 2s orbital holds 2 electrons: $2s^2$, - The 2p orbital now holds 6 electrons: $2p^6$. Thus, the complete electronic configuration of the O²⁻ ion is: $$ 1s^2, 2s^2, 2p^6 $$ This configuration indicates that the O²⁻ ion has achieved a stable octet, with all of its outermost orbitals fully filled. Among the given options, the correct answer is: $$ \boxed{(3) \, 1s^2, 2s^2, 2p^6} $$ This configuration reflects the fact that the O²⁻ ion has the same electron arrangement as the noble gas neon (Ne), which also has 10 electrons.