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In which of the following, orbitals are correctly arranged in the increasing order of their energies?
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Verified Answer
The correct answer is:
5 p $ < 6$ s $ < 4$ f $ < 5$ d
For multielectron atoms, the energy of the orbitals increases with an increase in their $(\mathrm{n}+l)$ value.
Thus, the order will be :-
$$
5 \mathrm{p} < 6 \mathrm{~s} < 4 \mathrm{f} < 5 \mathrm{~d}
$$
Thus, the order will be :-
$$
5 \mathrm{p} < 6 \mathrm{~s} < 4 \mathrm{f} < 5 \mathrm{~d}
$$
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