AP EAMCET · Chemistry · Structure of Atom
In an atom the order of increasing energy of electrons with quantum numbers
(i) \(n=4, l=1\)
(ii) \(n=4, l=0\)
(iii) \(n=3, l=2\) and (iv) \(n=3, l=1\) is
- A \(\mathrm{iii} < \) (i) \( < \mathrm{iv} < \) ii
- B \(\mathrm{ii} < \mathrm{iv} < \mathrm{i} < \mathrm{iii}\)
- C \(\mathrm{i} < \mathrm{iii} < \mathrm{ii} < \mathrm{iv}\)
- D iv \( < \mathrm{ii} < \mathrm{iii} < \mathrm{i}\)
Answer & Solution
Correct Answer
(D) iv \( < \mathrm{ii} < \mathrm{iii} < \mathrm{i}\)
Step-by-step Solution
Detailed explanation
The order of increase of energy can be calculated from \((n+I)\) rule. If two orbitals have same value of \((n+I)\), the orbital with lower value of \(n\) will be filled first. (i) For \(n=4, I=1,(n+I)=4+1=5\) (ii) For \(n=4, l=0,(n+l)=4+0=4\) (iii) For \(n=3, l=2,(n+l)=3+2=5\)…
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