Among the following substances, select the one(s) presenting a ft (pi) bonding NOTE TO STUDENTS: This question may or ma
Posted: Wed Apr 27, 2022 6:28 am
Among the following substances, select the one(s) presenting a ft (pi) bonding NOTE TO STUDENTS: This question may or may not have more than one correct answer. If you select an incorrect answer, points will be deducted from your score, however the total mark for the question cannot be negative. O a. SO3 O b. HCIO O c. NC13 O d. SF6 Oe. P023- are
a ALL BUT ONE of the following sentences provide reasonable explanations for the overall trend and the irregularities observed in the first ionization energies (.e., E) of the neutral atoms of the elements shown in the table. Identity the one statement that is INCORRECT. Na Mg AL Si Р S 09 СІ Ar VE (kJ/mol) 496 738 577 787 1012 1000 1251 1520 O a. When one electron is removed from a neutral sulfur atom. It adopts a configuration where all 3p orbitals are equally populated with one unpaired electron, which helps stabilizing the resulting Son That is why the IE of sulfur has a slightly lower than expected ionization energy Exam Support O b. The higher the energy level is, the more energy will be required to remove electrons from the atom. The same apply to the orbitals in the same energy level: electrons in higher-energy orbitals will have a higher ionization energy than those in lower-energy orbitals. O c. For elements in a same row, the effective nuclear charge is directly proportional to the atomic number. That explains the overall trend we observe when lonization energy increases as we move to the right in the periodic system. O d. Removing a paired electron from a 3s subshell consumes more energy than removing an unpaired electron from a 3p subshell. This could explain why the ionization energy of Mg is significantly higher than that of Al. Finis Byl page
a ALL BUT ONE of the following sentences provide reasonable explanations for the overall trend and the irregularities observed in the first ionization energies (.e., E) of the neutral atoms of the elements shown in the table. Identity the one statement that is INCORRECT. Na Mg AL Si Р S 09 СІ Ar VE (kJ/mol) 496 738 577 787 1012 1000 1251 1520 O a. When one electron is removed from a neutral sulfur atom. It adopts a configuration where all 3p orbitals are equally populated with one unpaired electron, which helps stabilizing the resulting Son That is why the IE of sulfur has a slightly lower than expected ionization energy Exam Support O b. The higher the energy level is, the more energy will be required to remove electrons from the atom. The same apply to the orbitals in the same energy level: electrons in higher-energy orbitals will have a higher ionization energy than those in lower-energy orbitals. O c. For elements in a same row, the effective nuclear charge is directly proportional to the atomic number. That explains the overall trend we observe when lonization energy increases as we move to the right in the periodic system. O d. Removing a paired electron from a 3s subshell consumes more energy than removing an unpaired electron from a 3p subshell. This could explain why the ionization energy of Mg is significantly higher than that of Al. Finis Byl page