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B.E.: F-F Cl-Cl 158.8 242.6. Along the period, I.P. generally increases but not regularly. Be and B are exceptions. First I.P. increases in moving from left to right in a period, but I.P. of B is lower than Be.

1. Be = 1s²2s²; B = 1s²2s²2p¹; C = 1s²2s²2p²; N = 1s²2s²2p³; O = 1s²2s²2p⁴. I.P. increases along the period. But I.P. of Be > B. Further I.P. of O < N because atoms with fully or partly filled orbitals are most stable and hence have high ionisation energy.
2. Mg = 1s²2s²2p⁶3s². After the removal of 2 electrons, the magnesium ion will acquire noble gas configuration hence removal of 3rd electron will require large amount of energy.
3. First ionisation potential of Be is greater than boron due to following configuration: Be = 1s²2s²; B = 1s²2s²2p¹.
4. Order of attraction of electrons towards nucleus: 2s > 2p, so more amount of energy is required to remove electrons from 2s orbital.

Correct answer: First ionisation potential of Be is greater than boron due to following configuration: Be = 1s²2s²; B = 1s²2s²2p¹.

Solution

The first ionization potential of Be is greater than B because the 2s orbital in Be is fully filled, making it more stable than the partially filled 2p orbital in B. This stability requires more energy to remove an electron from Be.

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