Questions: Draw a Lewis structure for SeBr2O.

Draw a Lewis structure for SeBr2O.
Transcript text: Draw a Lewis structure for $\mathrm{SeBr}_{2} \mathrm{O}$.
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Solution

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Solution Steps

Step 1: Count the Total Number of Valence Electrons

To draw the Lewis structure, we first need to determine the total number of valence electrons available. Selenium (Se) is in group 16 and has 6 valence electrons. Bromine (Br) is in group 17 and has 7 valence electrons, and there are two bromine atoms. Oxygen (O) is also in group 16 and has 6 valence electrons.

\[ \text{Total valence electrons} = 6 (\text{Se}) + 2 \times 7 (\text{Br}) + 6 (\text{O}) = 26 \]

Step 2: Determine the Central Atom

The central atom is usually the least electronegative element that is not hydrogen. In this case, selenium (Se) is the central atom.

Step 3: Arrange the Atoms and Distribute Electrons

Place the selenium atom in the center, with the two bromine atoms and one oxygen atom surrounding it. Connect each of these atoms to selenium with a single bond initially.

Step 4: Complete the Octets of the Surrounding Atoms

Each bromine and oxygen atom should have a complete octet. After forming single bonds, distribute the remaining electrons to complete the octets of the surrounding atoms.

Step 5: Check the Octet of the Central Atom

After completing the octets of the surrounding atoms, check if the central atom (Se) has a complete octet. If not, adjust by forming double bonds if necessary.

Step 6: Calculate Formal Charges

Calculate the formal charges to ensure the most stable structure. The formal charge is calculated as:

\[ \text{Formal charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{\text{Bonding electrons}}{2} \]

Adjust the structure to minimize formal charges, ideally achieving zero formal charges for all atoms.

Final Answer

The Lewis structure for \(\mathrm{SeBr}_{2} \mathrm{O}\) is:

\[ \begin{array}{c} \text{Br} \\ \vert \\ \text{Se} = \text{O} \\ \vert \\ \text{Br} \end{array} \]

Each bromine atom is connected to selenium with a single bond, and the oxygen is connected with a double bond. All atoms have complete octets, and the formal charges are minimized.

\(\boxed{\text{Lewis structure is drawn as described above.}}\)

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