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Q.(i) Use molecular orbital theory to explain why Be2 molecule does not exist? [1]

(ii) Which out of NH3 and NF3 has higher dipole moment and why? [2] OR Define Octet rule and write its limitations.
Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2026Subjective· 3mImportance★★★★★
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Be2Be_2's MO configuration gives equal electrons in bonding and antibonding orbitals (bond order zero), so no stable molecule forms; in NH3NH_3 the lone pair's dipole adds to the N–H bond dipoles, while in NF3NF_3 it opposes the N–F bond dipoles, so NH3NH_3's net dipole moment is larger.

(i) Be2Be_2 via MOT: Be has electronic configuration 1s22s21s^2 2s^2, so Be2Be_2 has 8 electrons total, filling the MOs as:

σ1s2 σ∗1s2 σ2s2 σ∗2s2\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2

Bond order =Nb−Na2=4−42=0= \dfrac{N_b - N_a}{2} = \dfrac{4-4}{2} = 0. A bond order of zero means there is no net bonding interaction, so Be2Be_2 is not a stable molecule and does not exist under ordinary conditions.

(ii) NH3NH_3 vs NF3NF_3 dipole moment: Both molecules are pyramidal with one lone pair on N.

  • In NH3NH_3: N is more electronegative than H, so each N–H bond dipole points towards N — in the same direction as the lone pair's dipole. These reinforce each other, giving a larger net dipole moment (≈1.47 D\approx 1.47\ D). …

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