The electron configuration of the U4+ ion can be determined by removing 4 electrons from the neutral uranium atom (U).
The electron configuration of neutral uranium (U) is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10 6p^6 7s^2 5f^3 6d^1.
To form the U4+ ion, four electrons are removed, starting from the outermost energy level.
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10 6p^6 7s^2 5f^3 6d^1 - 4e-
The resulting electron configuration of the U4+ ion is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 4d^10 5p^6 6s^2 4f^14 5d^10 6p^6 7s^2 5f^3.
The electron configuration of U4+ ion
3 answers
Use condensed electron configuration to write an equation for the formation of each transition metal ion, and predict whether it is paramagnetic;
a) U4+ (z=92)
b)Hg2+ (z=80)
c) La3+ (z=57)
d)Os3+ (z=76)
a) U4+ (z=92)
b)Hg2+ (z=80)
c) La3+ (z=57)
d)Os3+ (z=76)
a) The electron configuration of uranium (U) is [Rn]5f^3 6d^1 7s^2. To form U4+, 4 electrons are removed:
U: [Rn]5f^3 6d^1 7s^2
U4+: [Rn]5f^3 6d^0
The condensed electron configuration for U4+ is [Rn]5f^3. Since there are unpaired electrons in the 5f orbital, U4+ is paramagnetic.
b) The electron configuration of mercury (Hg) is [Xe]4f^14 5d^10 6s^2. To form Hg2+, 2 electrons are removed:
Hg: [Xe]4f^14 5d^10 6s^2
Hg2+: [Xe]4f^14 5d^10
The condensed electron configuration for Hg2+ is [Xe]4f^14 5d^10. Since all the orbitals are filled and there are no unpaired electrons, Hg2+ is diamagnetic (not paramagnetic).
c) The electron configuration of lanthanum (La) is [Xe]5d^1 6s^2. To form La3+, 3 electrons are removed:
La: [Xe]5d^1 6s^2
La3+: [Xe]5d^0 6s^0
The condensed electron configuration for La3+ is [Xe]5d^0 6s^0. Since all the orbitals are filled and there are no unpaired electrons, La3+ is diamagnetic.
d) The electron configuration of osmium (Os) is [Xe]4f^14 5d^6 6s^2. To form Os3+, 3 electrons are removed:
Os: [Xe]4f^14 5d^6 6s^2
Os3+: [Xe]4f^14 5d^5
The condensed electron configuration for Os3+ is [Xe]4f^14 5d^5. Since there are unpaired electrons in the 5d orbital, Os3+ is paramagnetic.
U: [Rn]5f^3 6d^1 7s^2
U4+: [Rn]5f^3 6d^0
The condensed electron configuration for U4+ is [Rn]5f^3. Since there are unpaired electrons in the 5f orbital, U4+ is paramagnetic.
b) The electron configuration of mercury (Hg) is [Xe]4f^14 5d^10 6s^2. To form Hg2+, 2 electrons are removed:
Hg: [Xe]4f^14 5d^10 6s^2
Hg2+: [Xe]4f^14 5d^10
The condensed electron configuration for Hg2+ is [Xe]4f^14 5d^10. Since all the orbitals are filled and there are no unpaired electrons, Hg2+ is diamagnetic (not paramagnetic).
c) The electron configuration of lanthanum (La) is [Xe]5d^1 6s^2. To form La3+, 3 electrons are removed:
La: [Xe]5d^1 6s^2
La3+: [Xe]5d^0 6s^0
The condensed electron configuration for La3+ is [Xe]5d^0 6s^0. Since all the orbitals are filled and there are no unpaired electrons, La3+ is diamagnetic.
d) The electron configuration of osmium (Os) is [Xe]4f^14 5d^6 6s^2. To form Os3+, 3 electrons are removed:
Os: [Xe]4f^14 5d^6 6s^2
Os3+: [Xe]4f^14 5d^5
The condensed electron configuration for Os3+ is [Xe]4f^14 5d^5. Since there are unpaired electrons in the 5d orbital, Os3+ is paramagnetic.
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