We have just seen the MO diagram of BeH2. Predict the molecular geometry of BH2 based on the MO diagram. Linear, like BeH2, because it has the same MO diagram, just with one more e- (B) Bent, because the overlap of the 1s(H) orbital with the 2py(B) will stabilize a bent configuration. (C) Linear, because the “extra” e- will not go into the 2py orbital, but in the 2px orbital instead. No stabilization, no bend. (D) BH2 will be linear because the two protons will repel each other, and the best geometry is therefore linear. The overlap of the 1s(H) orbital with the 2py(B) is net non-bonding anyway. (E) BH2 will be bent, because the extra electron will go predominantly to one of the two lobes of the 2py(B) orbital, destroy the symmetry and force the bend.

We have just seen the MO diagram of BeH2. Predict the molecular geometry of BH2 based on the MO diagram (Fig. 10.9 in McQuarrie & Simon). (A) BH2 will be linear, like BeH2, because it has the same MO diagram, just with one more electron. The extra electron changes the interactions! (B) BH2 will be bent, because the overlap of the 1s(H) orbital with the 2py(B) will stabilize a bent configuration. (C) BH2 will be linear, because the “extra” electron will not go into the 2py orbital, but in the 2px orbital instead. No stabilization, no bend. The 2py and 2px orbitals are equivalent! (D) BH2 will be linear because the two protons will repel each other, and the best geometry is therefore linear. The overlap of the 1s(H) orbital with the 2py(B) is net non-bonding anyway. Not the dominant effect. (E) BH2 will be bent, because the extra electron will go predominantly to one of the two lobes of the 2py(B) orbital, destroy the symmetry and force the bend. None of the 2 lobes is a priori better than the other!