The correct order of increasing energy of atomic orbitals is
- $5 \mathrm{p} < 4 \mathrm{f} < 6 \mathrm{~s} < 5 \mathrm{~d}$
- $5 \mathrm{p} < 6 \mathrm{~s} < 4 \mathrm{f} < 5 \mathrm{~d}$
- $4 \mathrm{f} < 5 \mathrm{p} < 5 \mathrm{~d} < 6 \mathrm{~s}$
- $5 p < 5 d < 4 f < 6 s$
Solution
increasing order of energy: $1 \mathrm{~s} < 2 \mathrm{~s} < 2 \mathrm{p} < 3 \mathrm{~s} < 3 \mathrm{p} < 4 \mathrm{~s} < 3 \mathrm{~d} < 4 \mathrm{p} < 5 \mathrm{~s} < 4 \mathrm{~d} < 5 \mathrm{p} < 6 \mathrm{~s} < 4 \mathrm{f}$ $ < 5 \mathrm{~d} < 6 \mathrm{p} < 7 \mathrm{~s}$ /
Asked in: JEE-TOPICTESTS-CHEMISTRY