Easy
An electron having the quantum numbers $n = 4, l = 3, m = 0, s = -\frac{1}{2}$ would be in the orbital
- A$3s$
- B$3p$
- C$4d$
- D$4f$
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Write the electronic configuration of the following ions:
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$(b) Na^+$
$(c) O^{2-}$
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$(a) H^-$
$(b) Na^+$
$(c) O^{2-}$
$(d) F^-$
Easy
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View SolutionThe electronic configuration of $Cu^{2+}$ ion is
For a multi-electron atom,the highest energy level among the following is
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