IF6+ and IF6- both have an octahedral molecular shape (point group Oh). Draw Lewis structures for both including all valence electrons and formal atom charges. Based on Lewis and VSEPR theories, are the molecular geometries expected or unexpected?
IF6+ and IF6- both have an octahedral molecular shape (point group Oh). Draw Lewis structures for both including all valence electrons and formal atom charges. Based on Lewis and VSEPR theories, are the molecular geometries expected or unexpected?
The Correct Answer and Explanation is:
Let’s start by drawing the Lewis structures for IF6+ and IF6- and understanding how these molecules adopt their octahedral shapes.
IF6+ (Iodine Hexafluoride cation)
- Total Valence Electrons:
- Iodine (I) is in Group 17, so it has 7 valence electrons.
- Each Fluorine (F) is in Group 17, so each fluorine atom has 7 valence electrons.
- There are 6 fluorine atoms, contributing a total of 6×7=426 \times 7 = 426×7=42 electrons.
- The +1 charge means we subtract one electron from the total.
- Structure and Formal Charges:
- Iodine, being a Group 17 element, can expand its octet and accommodate more than 8 electrons around it. It will bond to 6 fluorine atoms.
- Since we have 48 electrons to distribute, each of the 6 iodine-fluorine bonds uses 2 electrons, so 6×2=126 \times 2 = 126×2=12 electrons are used in bonding.
- The remaining 48−12=3648 – 12 = 3648−12=36 electrons are placed as lone pairs on the fluorine atoms. Each fluorine gets 3 lone pairs, utilizing 6×6=366 \times 6 = 366×6=36 electrons.
- Iodine is the central atom with 6 fluorine atoms around it. Each F-I bond is single, and each fluorine has 3 lone pairs.
- Formal charge on iodine = 0, and formal charge on each fluorine = 0.
IF6- (Iodine Hexafluoride anion)
- Total Valence Electrons:
- Iodine (I) has 7 valence electrons.
- Each fluorine (F) has 7 valence electrons.
- There are 6 fluorine atoms, contributing 6×7=426 \times 7 = 426×7=42 electrons.
- The -1 charge means we add one electron to the total.
- Structure and Formal Charges:
- The iodine atom still bonds to 6 fluorine atoms. The remaining electrons are distributed as lone pairs.
- The 50 electrons are distributed similarly: each of the 6 I-F bonds takes 2 electrons, leaving 50−12=3850 – 12 = 3850−12=38 electrons to be placed as lone pairs on fluorine atoms.
- Each fluorine gets 3 lone pairs, using 6×6=366 \times 6 = 366×6=36 electrons.
- The remaining 2 electrons go to iodine as lone pairs.
- Iodine is the central atom with 6 fluorine atoms surrounding it, each having 3 lone pairs.
- Formal charge on iodine = 0, and formal charge on each fluorine = 0.
Geometrical Considerations:
- IF6+ and IF6- both adopt octahedral geometry, where the iodine atom is at the center and the fluorine atoms are arranged symmetrically around it.
- VSEPR Theory predicts that the arrangement of electron pairs (bonding and lone pairs) around a central atom with 6 bonded atoms will result in an octahedral geometry.
- Expected vs. Unexpected:
- For IF6+: The molecular geometry is expected. In this case, the central iodine atom has an expanded octet, and the geometry is driven by the need to minimize electron pair repulsion.
- For IF6-: The molecular geometry is also expected. The addition of an electron (increasing electron-electron repulsion) still leads to an octahedral arrangement.
Thus, both the cation and anion follow the predicted octahedral shape according to Lewis and VSEPR theories. The difference in their electron counts (48 vs. 50) doesn’t change the overall geometry significantly because both configurations minimize electron repulsion in a symmetric arrangement.
