Write the Lewis formula for each of the following covalent compounds. Which ones contain at least one atom with a share in less than an octet of valence electrons?(a) ${\mathrm{BeBr}}_2$, (b) ${\mathrm{BBr}}_3$, (c) ${\mathrm{NCl}}_3$,(d) ${\mathrm{AlCl}}_3$

Step by Step Solution: Step 1. Write the Lewis symbols of the atoms to show valence electrons Step 2. Determine the total number of valence electrons for the compound Step 3. Draw a single bond between atoms and distribute electrons to satisfy octet rule Step 4. If any atoms lack an octet, form multiple bonds by converting lone pairs into bonding pairs on adjacent atoms Step 5. Determine if any atoms have fewer than 8 electrons around it Final Answer: (a) BeBr2: Br:Be:Br with Beryllium as the central atom. Beryllium has less than an octet. (b) BBr3: Br:B:B:Br with Boron as the central atom. Boron has less than an octet. (c) NCl3: Cl:N:Cl with Nitrogen as the central atom. Nitrogen has a complete octet. (d) AlCl3: Cl:Al:Cl with Aluminum as the central atom. Aluminum has less than an octet.