To write the equations for the formation of the transition metal ions along with their condensed electron configurations, we first need to determine the electron configuration of the neutral atoms. Then, we will show how they lose electrons to form ions.
a) U^(+4)
Atomic Number of Uranium (U): 92
Neutral Electron Configuration: \[ [Rn] 5f^3 6d^1 7s^2 \]
Formation of U^(+4):
- The formation of U^(+4) involves the loss of 4 electrons (typically from 7s and 6d orbitals): \[ [Rn] 5f^3 \]
- Paramagnetic Status: U^(+4) has 3 unpaired electrons in the 5f subshell, so it is paramagnetic.
b) Hg^(+2)
Atomic Number of Mercury (Hg): 80
Neutral Electron Configuration: \[ [Xe] 4f^{14} 5d^{10} 6s^2 \]
Formation of Hg^(+2):
- The formation of Hg^(+2) involves the loss of 2 electrons (typically from the 6s orbital): \[ [Xe] 4f^{14} 5d^{10} \]
- Paramagnetic Status: Hg^(+2) has no unpaired electrons (all electrons in 5d and 4f are paired), so it is diamagnetic.
c) La^(+3)
Atomic Number of Lanthanum (La): 57
Neutral Electron Configuration: \[ [Xe] 5d^1 6s^2 \]
Formation of La^(+3):
- The formation of La^(+3) involves the loss of 3 electrons, which would include both the 6s and the 5d electron: \[ [Xe] \]
- Paramagnetic Status: La^(+3) has no unpaired electrons, so it is diamagnetic.
d) Os^(+3)
Atomic Number of Osmium (Os): 76
Neutral Electron Configuration: \[ [Xe] 4f^{14} 5d^6 6s^2 \]
Formation of Os^(+3):
- The formation of Os^(+3) involves the loss of 3 electrons (generally 2 from the 6s and 1 from the 5d): \[ [Xe] 4f^{14} 5d^5 \]
- Paramagnetic Status: Os^(+3) has 5 unpaired electrons in the 5d subshell, so it is paramagnetic.
Summary
- U^(+4): \[ [Rn] 5f^3 \] - Paramagnetic
- Hg^(+2): \[ [Xe] 4f^{14} 5d^{10} \] - Diamagnetic
- La^(+3): \[ [Xe] \] - Diamagnetic
- Os^(+3): \[ [Xe] 4f^{14} 5d^5 \] - Paramagnetic
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