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The following sets of quantum numbers represent four electrons in an atom.
(i) $n=4, l=1$
(ii) $n=4, l=0$
(iii) $n=3, l=2$
(iv) $n=3, l=1$
The sequence representing increasing order of energy, is
Options:
(i) $n=4, l=1$
(ii) $n=4, l=0$
(iii) $n=3, l=2$
(iv) $n=3, l=1$
The sequence representing increasing order of energy, is
Solution:
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Verified Answer
The correct answer is:
(iv) < (ii) < (iii) < (i)
(iv) < (ii) < (iii) < (i)
(i) $4 p$
(ii) $4 \mathrm{~s}$
(iii) $3 d$
(iv) $3 p$
According to Bohr Bury's $(n+\ell)$ rule, increasing order of energy will be (iv) < (ii) $ < $ (iii) $ < $ (i).
Note: If the two orbitals have same value of $(n+\ell)$ then the orbital with lower value of $n$ will be filled first.
(ii) $4 \mathrm{~s}$
(iii) $3 d$
(iv) $3 p$
According to Bohr Bury's $(n+\ell)$ rule, increasing order of energy will be (iv) < (ii) $ < $ (iii) $ < $ (i).
Note: If the two orbitals have same value of $(n+\ell)$ then the orbital with lower value of $n$ will be filled first.
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