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प्रश्न
Explain, why lanthanum (Z = 57) forms La3+ ion, while cerium (Z = 58) forms Ce4+ ion?
उत्तर
57La -Electronic configuration is [Xe]4f0 5d1 6s2
After losing 3 electrons La forms La3+ ion which is stable due to empty 4f-orbitals.
58Ce - Electronic configuration is [Xe]4f2 5d0 6s2
After losing 4 electrons Ce forms Ce4+ ion which is stable due to empty 4f-orbitals.
Thus, lanthanum (Z = 57) forms La3+ ion, while cerium (Z = 58) forms Ce4+ ion.
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संबंधित प्रश्न
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(i) \[\ce{Ce}\]
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| Column I | Column II | |
| (i) | Oxidation state of Mn in MnO2 is | (a) + 2 |
| (ii) | Most stable oxidation state of Mn is | (b) + 3 |
| (iii) | Most stable oxidation state of | (c) + 4 |
| Mn in oxides is | (d) + 5 | |
| (iv) | Characteristic oxidation state of lanthanoids is | (e) + 7 |
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| Column I (Property) | Column II (Element) | |
| (i) | Lanthanoid which shows +4 oxidation state |
(a) Pm |
| (ii) | Lanthanoid which can show +2 oxidation state |
(b) Ce |
| (iii) | Radioactive lanthanoid | (c) Lu |
| (iv) | Lanthanoid which has 4f7 electronic configuration in +3 oxidation state |
(d) Eu |
| (v) | Lanthanoid which has 4f14 electronic configuration in +3 oxidation state |
(e) Gd |
| (f) Dy |
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