Answer
See the explanation
Work Step by Step
Let's draw the Lewis structures for BeH2 and BH3, which are examples of compounds with central atoms having fewer than eight electrons around them, an exception to the octet rule.
BeH2 (Beryllium Dihydride):
The Lewis structure for BeH2 is:
H - Be - H
Beryllium (Be) has an atomic number of 4, which means it has 4 valence electrons. Each hydrogen (H) atom has 1 valence electron.
In the BeH2 molecule, the beryllium atom forms two covalent bonds with the two hydrogen atoms. This results in the beryllium atom having only 4 electrons around it, which is less than the octet rule of 8 electrons.
BH3 (Boron Trifluoride):
The Lewis structure for BH3 is:
\[
\begin{array}{ccc}
& \mathrm{H} & \\
& \mathrm{|} & \\
\mathrm{H} & -\mathrm{Be}- & \mathrm{H}
\end{array}
\]
Boron (B) has an atomic number of 5, which means it has 3 valence electrons. Each hydrogen (H) atom has 1 valence electron.
In the BH3 molecule, the boron atom forms three covalent bonds with the three hydrogen atoms. This results in the boron atom having only 6 electrons around it, which is less than the octet rule of 8 electrons.
Both BeH2 and BH3 are examples of compounds with central atoms having fewer than eight electrons around them, which is an exception to the octet rule.
