| The quantum number of four electrons are given below: |
| (I) \[n=4\], \[l=2\], \[{{m}_{1}}=1,\]\[{{m}_{s}}=-\frac{1}{2}\] |
| (II) \[n=3\], \[l=2\], \[{{m}_{1}}=1,\] \[{{m}_{s}}=+\frac{1}{2}\] |
| (III) \[n=4\], \[l=1\], \[{{m}_{1}}=0,\]\[{{m}_{s}}=+\frac{1}{2}\] |
| (IV) \[n=3\], \[l=1\], \[{{m}_{1}}=1,\] \[{{m}_{s}}=-\frac{1}{2}\] |
| The Correct order of their increasing energies will be: |
A) \[I<III<II<IV\]
B) \[IV<II<III<I\]
C) \[I<II<III<IV\]
D) \[IV<III<II<I\]
Correct Answer: B
Solution :
| Higher-the value of\[(n+l)\] | |
| Higher will be energy of orbital. If \[(n+l)\] are equal, then higher the value n higher will be energy | |
| I. \[n+l=6\] | \[n=4\] |
| II. \[n+l=5\] | \[n=3\] |
| III. \[n+l=5\] | \[n=4\] |
| \[IV<II<III<I.\] | |
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