Draw charge minimized Lewis structures for the following compounds: (2 marks each) a. HCN b. IOF5
The Correct Answer and Explanation is :
a. Hydrogen Cyanide (HCN):
Hydrogen cyanide (HCN) is a linear molecule consisting of hydrogen (H), carbon (C), and nitrogen (N). To draw its Lewis structure:
- **Count Valence Electrons:**
- Hydrogen (H): 1 valence electron
- Carbon (C): 4 valence electrons
- Nitrogen (N): 5 valence electrons
- Total: 1 + 4 + 5 = 10 valence electrons
- **Determine the Central Atom:**
- Carbon (C) is the least electronegative atom and serves as the central atom.
- **Connect Atoms with Single Bonds:**
- Place a single bond between H and C, and between C and N.
- **Distribute Remaining Electrons:**
- After forming single bonds, 8 electrons remain.
- Complete the octet for nitrogen by placing 6 electrons (3 lone pairs) on nitrogen.
- Place the remaining 2 electrons as a lone pair on carbon.
- **Form Multiple Bonds to Satisfy Octet Rule:**
- Carbon has 6 electrons (2 from the lone pair and 4 from bonds) and needs 2 more to complete its octet.
- Form a triple bond between carbon and nitrogen by converting the lone pair on carbon into a bonding pair.
The resulting structure is H–C≡N, with a triple bond between C and N and a single bond between H and C. This configuration minimizes formal charges, as all atoms have formal charges of zero. The molecule is linear with a bond angle of 180°. citeturn0search4
b. Iodine Oxide Pentafluoride (IOF₅):
Iodine oxide pentafluoride (IOF₅) consists of iodine (I), oxygen (O), and fluorine (F) atoms. To draw its Lewis structure:
- **Count Valence Electrons:**
- Iodine (I): 7 valence electrons
- Oxygen (O): 6 valence electrons
- Fluorine (F): 7 valence electrons × 5 = 35 valence electrons
- Total: 7 + 6 + 35 = 48 valence electrons
- **Determine the Central Atom:**
- Iodine (I) is the central atom due to its larger size and ability to accommodate more than 8 electrons.
- **Connect Atoms with Single Bonds:**
- Place a single bond between iodine and oxygen, and between iodine and each fluorine atom.
- **Distribute Remaining Electrons:**
- After forming single bonds, 42 electrons remain.
- Complete the octet for each fluorine by placing 3 lone pairs on each.
- Place the remaining electrons as lone pairs on oxygen.
- **Form Multiple Bonds if Necessary:**
- Iodine can accommodate more than 8 electrons.
- To minimize formal charges, form a double bond between iodine and oxygen by converting one lone pair on iodine into a bonding pair.
The resulting structure has iodine at the center with a double bond to oxygen and single bonds to five fluorine atoms. This configuration minimizes formal charges, as iodine has a formal charge of zero, oxygen has a formal charge of zero, and each fluorine has a formal charge of zero. The molecular geometry is octahedral, with bond angles of 90° and 180°. citeturn0search1
For a visual representation of the HCN Lewis structure, you can refer to the following video:
videoHow to Draw the Lewis Structure for HCNturn0search0