Electron group arrangement

2-D Lewis structure 3-D Lewis structure IF4+ Atom # of valence e per atom Total # of alence e v Te FITO ion Total (0.25 pt for electron accounting) F: 5 (0.75 pt) (0.25 pt) Electron group arrangement (0.25 pt): Molecular shape & class (0.25 pt ea): Shape: Class (AXmE.): Ideal Bond Angle(s) (list all angles) (0.25 pt): Polar or Non-Polar (0.25 pt): (if polar, show dipole moment in the 3D structure) 

The Correct Answer and Explanation is :

Lewis Structure of IF₄⁺ (Iodine Tetrafluoride Ion)

Step 1: Determining Valence Electrons

• Iodine (I) is in group 17 of the periodic table, so it has 7 valence electrons.
• Fluorine (F) is in group 17 as well, so each fluorine atom has 7 valence electrons.
• Since we have 4 fluorine atoms, the total valence electrons from fluorine atoms is 4 × 7 = 28.
• The positive charge of the ion (IF₄⁺) means it has lost 1 electron, so we subtract 1 from the total count.

Total Valence Electrons:

• 7 (I) + 28 (4 F) − 1 (charge) = 34 valence electrons.

Step 2: Drawing the 2D Lewis Structure

• The iodine atom will be the central atom because it is less electronegative than fluorine.
• Each fluorine atom will form a single bond with iodine, using 2 electrons per bond.
• After placing bonds, distribute the remaining electrons to fulfill the octet rule for fluorine atoms.
• There will be no lone pairs on iodine in this structure because iodine has expanded its valence shell to accommodate more than 8 electrons due to its position in period 5 of the periodic table.

The structure looks like this:

    F
    |
F - I - F
    |
    F

Step 3: Electron Group Arrangement

• Iodine has 4 bonding pairs of electrons (one for each F atom) and no lone pairs. So, the electron group arrangement around iodine is tetrahedral.

Step 4: Molecular Shape & Class

• The molecular shape of IF₄⁺ is tetrahedral because there are 4 bonding pairs and no lone pairs on the central iodine atom.
• The class is AX₄, where A is the central atom (iodine), and X represents the fluorine atoms.

Step 5: Ideal Bond Angles

• The ideal bond angles for a tetrahedral geometry are approximately 109.5° between all the bonds.

Step 6: Polarity

• The molecule IF₄⁺ is non-polar. This is because the 4 fluorine atoms are symmetrically arranged around the central iodine atom in a tetrahedral shape, and the dipoles cancel out.

Explanation:

• Electron Accounting: Each bond involves two electrons. The iodine has 4 bonds (8 electrons around it) plus any lone pairs if there were any. The fluorine atoms each have 3 lone pairs of electrons and one bonding electron shared with iodine. The net charge affects the electron count, leading to a subtraction of one electron for the positive charge.
• Geometry: The electron groups around iodine cause it to adopt a tetrahedral arrangement for minimal repulsion between the electron pairs.