One type of exception to the octet rule are compounds with central atoms having fewer than eight electrons around them. \(\mathrm{BeH}_{2}\) and \(\mathrm{BH}_{3}\) are examples of this type of exception. Draw the Lewis structures for \(\mathrm{BeH}_{2}\) and \(\mathrm{BH}_{3}\).

For both BeH2 and BH3, we need to count the total number of valence electrons. Be has 2 valence electrons, H has 1 valence electron, and B has 3 valence electrons. For BeH2: \(1 \times (\text{number of valence electrons for Be}) + 2 \times (\text{number of valence electrons for H}) = 1 \times 2 + 2 \times 1 = 4\) For BH3: \(1 \times (\text{number of valence electrons for B}) + 3 \times (\text{number of valence electrons for H}) = 1 \times 3 + 3 \times 1 = 6\)

For both structures, the central atom will be the least electronegative element in the compound other than hydrogen, which is Be in BeH2 and B in BH3.

For BeH2: 1. Place the central Be atom. 2. Distribute the 4 valence electrons around the Be atom by forming single bonds with 2 H atoms (2 electrons per bond). 3. The resulting Lewis structure is BeH2, with Be at the center, forming single bonds with 2 H atoms and having a total of 4 electrons around it. For BH3: 1. Place the central B atom. 2. Distribute the 6 valence electrons around the B atom by forming single bonds with 3 H atoms (2 electrons per bond). 3. The resulting Lewis structure is BH3, with B at the center, forming single bonds with 3 H atoms and having a total of 6 electrons around it. Both BeH2 and BH3 have central atoms (Be and B) with fewer than eight electrons around them, making them exceptions to the octet rule.