Show a Lewis structure for urea

Show a Lewis structure for urea. CH4N2O. Both N’s and the O are bonded to the C. The H’s are bonded to the N’s. None of the atoms has a formal charge.

The correct answer and explanation is :

The Lewis structure for urea (CH₄N₂O) can be drawn by considering the bonding and electron distribution for each atom. Here’s a step-by-step explanation of how to draw the structure:

Step 1: Count the total valence electrons.

Carbon (C) is in group 14 and has 4 valence electrons.
Nitrogen (N) is in group 15 and has 5 valence electrons. Since there are two nitrogen atoms, the total is ( 5 \times 2 = 10 ) valence electrons.
Hydrogen (H) is in group 1 and has 1 valence electron. Since there are four hydrogen atoms, the total is ( 1 \times 4 = 4 ) valence electrons.
Oxygen (O) is in group 16 and has 6 valence electrons.

The total number of valence electrons is:
[
4 \text{ (from C)} + 10 \text{ (from N)} + 4 \text{ (from H)} + 6 \text{ (from O)} = 24 \text{ valence electrons.}
]

Step 2: Determine the skeletal structure.

From the molecular formula (CH₄N₂O), we know:

The carbon (C) is the central atom.
Each nitrogen (N) is bonded to carbon (C).
The hydrogens (H) are bonded to the nitrogen atoms.
Oxygen (O) is also bonded to carbon.

The skeletal structure is:

C is the central atom, bonded to both N atoms and O.
Each N atom is bonded to two H atoms.

Step 3: Distribute the electrons to form bonds.

Each single bond consists of two electrons. We place single bonds between C and N, and between N and H. We also place a single bond between C and O.

C-N bond: 2 electrons.
N-H bonds: Each N has two H atoms, so 4 electrons per N are used for N-H bonds.
C=O: A double bond between C and O uses 4 electrons.

Step 4: Fill the octet rule.

Now, distribute the remaining electrons to satisfy the octet rule for each atom:

C: The carbon has 4 bonds (two N-H bonds, one C-N bond, and one C=O bond), satisfying its octet.
N: Each nitrogen is involved in three bonds (C-N and N-H), which satisfies the octet.
O: Oxygen has two bonds (C=O) and two lone pairs, completing its octet.
H: Each hydrogen is bonded to nitrogen and satisfies its duet rule.

Step 5: Verify the formal charges.

To ensure that the atoms have no formal charge:

The formal charge on C is 0 because it has 4 bonds and 0 lone pairs.
The formal charge on N is 0 because each nitrogen has 3 bonds and 1 lone pair.
The formal charge on O is 0 because it has 2 bonds and 2 lone pairs.
The formal charge on each H is 0 because each hydrogen has 1 bond and no lone pairs.

Final Lewis Structure for Urea:

   H2N - C - NH2
        ||
        O

Explanation:

The structure has no formal charges on any of the atoms.
Carbon (C) is bonded to two nitrogen atoms (N) and one oxygen atom (O).
Nitrogen atoms (N) are each bonded to two hydrogen atoms (H).
Oxygen (O) is double-bonded to the central carbon.

Thus, the Lewis structure for urea is complete, satisfying the octet rule for all atoms involved, with no formal charges.