Chemistry - PERIODIC TABLE Question with Solution | TestHub
If the energy of 2p atomic orbitals and 2s atomic orbitals in an oxygen atom are –15.85 eV and –32.38 eV, respectively.
Number of bonding electrons in O₂ molecules which have higher energy than –32.38 eV.
Number of antibonding electrons in O₂ molecules which have lower energy than –32.38 eV.
Find .
Number of species from given below which have higher ionization energy as compared to molecular oxygen.
Atomic oxygen,
Atomic nitrogen,
Molecular nitrogen,
Atomic fluorine,
Neon,
Helium ,
Superoxide ion
Find the value of Z ?
Answer:
Solution:
Question Explanation: Comparing the Ionization Energy (IE) of various species with .
Concept: Ionization Energy Trends.
Solution: The IE of (12.1 eV) is lower than atomic oxygen because the electron is removed from a higher energy antibonding orbital. Atomic nitrogen, fluorine, noble gases, and all have higher IE due to stable configurations or smaller sizes.
Species with IE (12.1 eV):
Atomic O (13.6 eV),
Atomic N (14.5 eV),
Molecular (15.6 eV),
Atomic F (17.4 eV),
Neon (21.6 eV),
Helium (24.6 eV).
has a much lower IE as it is an anion.
Total count .
Final Answer: 6.00