Which of the following ions possess a noble gas electron configuration?
(a) K+
(d) O-
(b) He+
(e) F-
(c) H-
(f) Ca2+
The Correct Answer and Explanation is :
The ions that possess a noble gas electron configuration are: K⁺, F⁻, and Ca²⁺.
Explanation:
- What is a noble gas electron configuration?
Noble gases have a fully filled outer shell (also called a stable octet or duet for helium) which makes them highly stable. This configuration corresponds to the electron configuration of the noble gases like He, Ne, Ar, etc. - Analysis of each ion:
- (a) K⁺ (Potassium ion):
Potassium (K) has 19 electrons, and its configuration is ( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 ). When K loses one electron to form ( \text{K}^+ ), it has 18 electrons, matching the configuration of argon (( 1s^2 2s^2 2p^6 3s^2 3p^6 )). Thus, ( \text{K}^+ ) has a noble gas configuration. - (b) He⁺ (Helium ion):
Helium has 2 electrons (( 1s^2 )). Losing one electron gives ( \text{He}^+ ) with 1 electron, which does not match a noble gas configuration. So, ( \text{He}^+ ) does not have a noble gas configuration. - (c) H⁻ (Hydride ion):
Hydrogen (( H )) has 1 electron. When it gains one electron to form ( \text{H}^- ), its configuration becomes ( 1s^2 ), which matches helium. Thus, ( \text{H}^- ) has a noble gas configuration. - (d) O⁻ (Oxide ion):
Oxygen has 8 electrons (( 1s^2 2s^2 2p^4 )). Adding one electron gives ( \text{O}^- ) (( 1s^2 2s^2 2p^5 )), which does not match any noble gas configuration. - (e) F⁻ (Fluoride ion):
Fluorine has 9 electrons (( 1s^2 2s^2 2p^5 )). Adding one electron to form ( \text{F}^- ) gives 10 electrons (( 1s^2 2s^2 2p^6 )), matching neon’s configuration. Thus, ( \text{F}^- ) has a noble gas configuration. - (f) Ca²⁺ (Calcium ion):
Calcium has 20 electrons (( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 )). Losing two electrons to form ( \text{Ca}^{2+} ) leaves 18 electrons (( 1s^2 2s^2 2p^6 3s^2 3p^6 )), matching argon’s configuration. Thus, ( \text{Ca}^{2+} ) has a noble gas configuration.
Correct Answer:
(a) K⁺, (e) F⁻, and (f) Ca²⁺.