Beryllium DichlorideBeCl2Valence e = 2 + (2x7) = 16RHED = 2 (2 individual Be-Cl bonds)Electronic geometry is LINEAR.No lone pairs on Be so molecular geometry is LINEARBeClCl180ºδ+δ-δ-POLAR covalent bonds; NON POLAR molecule3p[He]2sBeCl[Ne]3sNeed 2 unpaired e for Be..How about this?NO- Use this.. See over..Cl already has one unpaired e in a p orbital, ready to bond2p2p2p[He]2sBe[He]spBe
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Using a 2s and a 2p orbital for Be and the 3p orbital for each Cl, we would get:3p3pClClBe1s2s2pNot very efficient, and very uneven.Instead, we HYBRIDIZE the single 2s orbital and one of the 2p orbitals, (this is basically mixing two equations to get two new ones). The resulting sp hybridized orbitals are degenerate (of the same energy) and look like this:The two sp orbitals are arranged in space 180º apart, and so we can now build our symmetrical BeCl2 molecule:3pCl3pClspspBe1s