Electronic configuration of 4Ве = 2, 2 = 1s2 , 2s2
Electronic configuration of 5B = 2, 3 = 1s2 , 2s2 , 2p1
In case of Be, more energy is required to remove an electron from 2s orbital as it is completely filled as compared to B. So, first I.E. of boron is less than Be. But size of Be is more than boron because atomic size decreases from left to right in a period due to increase in nuclear charge.