Which of the following best represents the Lewis structure for dichlorine monoxide (${\mathrm{Cl}}_{2}O$)?

• A. A central $\mathrm{Cl}$ atom double bonded to an $O$ atom and single bonded to another $\mathrm{Cl}$ atom, with three lone pairs on each $\mathrm{Cl}$.
• B. An $O$ atom single bonded to two $\mathrm{Cl}$ atoms, with two lone pairs on each $\mathrm{Cl}$ and two lone pairs on the $O$.
• C. A linear structure with a $\mathrm{Cl}$ atom double bonded to an $O$ atom, and the $O$ atom single bonded to another $\mathrm{Cl}$ atom, with one lone pair on each atom.
• D. An $O$ atom triple bonded to one $\mathrm{Cl}$ atom and single bonded to another $\mathrm{Cl}$ atom, with no lone pairs on $O$.

Answer: (B)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons for dichlorine monoxide (Cl\_2O). Chlorine (Cl) has 7 valence electrons each, and oxygen (O) has 6 valence electrons. So, total valence electrons = 2 \times 7 + 6 = 20 electrons.
2. Step 2: Identify the central atom. Oxygen is less electronegative than chlorine and can form two bonds, so oxygen will be the central atom bonded to two chlorine atoms.
3. Step 3: Draw single bonds between the oxygen atom and each chlorine atom. Each single bond accounts for 2 electrons, so 4 electrons are used, leaving 16 electrons to be distributed as lone pairs.
4. Step 4: Distribute the remaining electrons as lone pairs to satisfy the octet rule. Place three lone pairs (6 electrons) on each chlorine atom to complete their octets, and place the remaining electrons as lone pairs on the oxygen atom to complete its octet.
5. Step 5: Verify that all atoms have a complete octet and that the total number of electrons used equals the total valence electrons calculated in Step 1. This confirms the correct Lewis structure is oxygen single bonded to two chlorines, with two lone pairs on each chlorine and two lone pairs on oxygen.