Assertion : The `1^(st)` IP of Be is greater than that of B
Reason : 2p orbital is lower in energy than 2s

Options

A. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).
B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A)
C. If (A) is correct, but (R) is incorrect.
D. If both (A) and (R) are incorrect.

Answer

C

Explanation

`IE_(1)` of Be is greater than `IE_(1)` of B because of fully filled configuration. 2p orbital has higher energy as it is at more distance from nucleus in comparison to 2s orbital.