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प्रश्न
Nickel atom can lose two electrons to form \[\ce{Ni^{2+}}\] ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons.
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उत्तर
\[\ce{Ni}\] (28): 1s2 2s2 2p6 3s2 3p6 3d8 4s2
\[\ce{Ni^{2+}}\] (26): 1s2 2s2 2p6 3s2 3p6 3d8
Hence, to form \[\ce{Ni^{2+}}\] ion, it will lose 2 electrons from 4s orbital.
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| (i) | 1 | 1 | +2 |
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| Atom / Ion | Electronic configuration |
| (i) \[\ce{Cu}\] | (a) 1s2 2s2 2p6 3s2 3p6 3d10 |
| (ii) \[\ce{Cu^{2+}}\] | (b) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 |
| (iii) \[\ce{Zn^{2+}}\] | (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 |
| (iv) \[\ce{Cr^{3+}}\] | (d) 1s2 2s2 2p6 3s2 3p6 3d9 |
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