QQuestionChemistry
QuestionChemistry
"Draw the Lewis structure of HCN. Include lone pairs."" is a complete question. It clearly asks for:
The Lewis structure of hydrogen cyanide (HCN).
That lone pairs should be shown."
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Answer
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Step 1:: Identify the number of valence electrons in each atom
1 \ \text{valence electron (H)} + 5 \ \text{valence electrons (N)} + 5 \ \text{valence electrons (C)} = 11 \ \text{valence electrons}
Hydrogen (H) has 1 valence electron, and nitrogen (N) and carbon (C) each have 5 valence electrons. Since there is one H atom, one N atom, and one C atom, the total number of valence electrons is:
Step 2:: Place the atoms in the Lewis structure
Place the H atom and C atom next to each other, followed by the N atom. Connect them with a single bond, which uses 2 valence electrons: H - C - N
Step 3:: Add remaining valence electrons as lone pairs and multiple bonds (if necessary)
There are 11 valence electrons in total, and 2 have been used for the single bond between H and C. This leaves 9 valence electrons to be distributed as lone pairs and multiple bonds: H : C ::: N: Since N is more electronegative than C, it will attract the remaining electrons more strongly. Add a double bond between C and N: H : C ::: N: This uses 2 more valence electrons, leaving 7 valence electrons to be distributed as lone pairs: H : C ::: N: Add a lone pair to the N atom: H : C ::: N: This uses 2 more valence electrons, leaving 5 valence electrons to be distributed as lone pairs: H : C ::: N: Add a lone pair to the H atom: H : C ::: N: This uses 2 more valence electrons, leaving 3 valence electrons to be distributed as a lone pair on the C atom: H : C ::: N: Add a lone pair to the C atom: H : C ::: N: This uses the last 2 valence electrons, resulting in a Lewis structure with the correct number of valence electrons.
Step 4:: Check the formal charges
\text{Formal Charge} = \text{Valence Electrons} - \text{Non-bonding Electrons} - \frac{1}{2} \ \text{Bonding Electrons}
Formal charge is calculated as: Calculate the formal charges for each atom: - Hydrogen (H): 1 valence electron - 0 non-bonding electrons - 1 bonding electron / 2 = 0 formal charge - Carbon (C): 4 valence electrons - 2 non-bonding electrons - 4 bonding electrons / 2 = 0 formal charge - Nitrogen (N): 5 valence electrons - 4 non-bonding electrons - 4 bonding electrons / 2 = 0 formal charge The formal charges are all 0, which is ideal in a Lewis structure.
Final Answer
The Lewis structure of HCN with lone pairs is: H : C ::: N: The formal charges are all 0, and the octet rule is satisfied for all atoms.
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