Answer
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Step 1:: Identify the total number of valence electrons in the molecule
The given molecule is CH^2O^2, where C is carbon, H is hydrogen, and O is oxygen. The number of valence electrons for each atom is as follows: - C (carbon) has 4 valence electrons (2s^2 2p^2 configuration) - H (hydrogen) has 1 valence electron (1s^1 configuration) - O (oxygen) has 6 valence electrons (2s^2 2p^4 configuration) There are two hydrogen atoms, one carbon atom, and two oxygen atoms in the molecule. So, the total number of valence electrons in CH^2O^2 is: Total valence electrons = (2 * 1) + 4 + (2 * 6) = 18 valence electrons
Step 2:: Place the atoms in the Lewis structure
Place the atoms in the Lewis structure, starting with the central atom. In this case, carbon (C) is the central atom. Then, add the remaining atoms around it. The structure will look like this: H — C — O — O
Step 3:: Distribute the valence electrons
Distribute the 18 valence electrons around the atoms in the Lewis structure. Since carbon is the central atom, it will share electrons with the oxygen and hydrogen atoms.
Step 4:
Add a single bond (2 electrons) between C and each O atom: H — C — O — O (with 12 electrons now)
Step 5:
Add a single bond (2 electrons) between C and each H atom: H — C — O — O (with 16 electrons now)
Step 6:
Add the remaining 2 electrons to the oxygen atoms to complete their octets: H — C — O:- O: (with 18 electrons now)
Step 7:: Check the formal charges
Calculate the formal charges for each atom in the Lewis structure: - Formal charge of C = (4 valence electrons - 4 non-bonding electrons) = + 1 - Formal charge of O with a double bond = (6 valence electrons - 6 non-bonding electrons) = 0 - Formal charge of O with a single bond = (6 valence electrons - 2 non-bonding electrons) = - 1 The formal charges add up to zero, which is correct for a neutral molecule.
Step 8:: Draw the final Lewis structure
The final Lewis structure for CH^2O^2 is: H — C — O:- O:
Final Answer
The Lewis structure for CH^2O^2 is: H — C — O:- O:
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