Asked by jena
The ionization energies (kJ/mol) of hydrogen (H) , nitrogen (N) , potassium (K) , and oxygen (O) are 1,312, 1,402, 419, and 1,314, respectively.
Which element combination is least likely?
Responses
OH−
uppercase O H superscript minus end superscript
KO−
uppercase K uppercase O superscript minus end superscript
NO3 −
uppercase N O subscript 3 end subscript superscript minus end superscript
NH4 +
All Answers 1
Answered by
GPT-5 mini
AI
KO−
Reason: K has a very low ionization energy (419 kJ·mol−1) and will form K+; O is a nonmetal (high IE ≈1314) and typically forms O2− (or O2−/O2− peroxides/superoxides), so the normal stoichiometry is K2O (K+ / O2−). OH−, NO3− and NH4+ are common, stable ions; KO− (a single K with an O carrying −1 overall) is not a reasonable, commonly formed species.
Reason: K has a very low ionization energy (419 kJ·mol−1) and will form K+; O is a nonmetal (high IE ≈1314) and typically forms O2− (or O2−/O2− peroxides/superoxides), so the normal stoichiometry is K2O (K+ / O2−). OH−, NO3− and NH4+ are common, stable ions; KO− (a single K with an O carrying −1 overall) is not a reasonable, commonly formed species.
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