Arrange in decreasing order the energy of 2s orbital in the following atoms H, Li, Na, K

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$E_{2 s} (H) > E_{2 s} (Na) > E_{2 s} (Li) > E_{2 s} (K)$

$E_{2 s} (H) > E_{2 s} (Na) = E_{2 s} (K) > E_{2 s} (Li)$

$E_{2 s} (K) > E_{2 s} (Na) > E_{2 s} (Li) > E_{2 s} (H)$

$E_{2 s} (H) > E_{2 s} (Li) > E_{2 s} (Na) > E_{2 s} (K)$

answer is A.

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Detailed Solution

As we go down the group IA, there is increase in the number of shells so size of atom increases and energy of 2s orbital decreases.

Hence the correct order is, $E_{2 s} (H) > E_{2 s} (Li) > E_{2 s} (Na) > E_{2 s} (K)$

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