Questions: Draw the Lewis structure of IF5 and then determine its electron domain and molecular geometries. A tetrahedral / t-shaped B octahedral / t-shaped C linear / see-saw D pentagonal / bent (109.5) E octahedral / square pyramidal

Solution Steps

Iodine (I) has 7 valence electrons, and each fluorine (F) has 7 valence electrons. Since there are 5 fluorine atoms, the total number of valence electrons is: \[ 7 + 5 \times 7 = 7 + 35 = 42 \]

Place iodine in the center and arrange the five fluorine atoms around it. Connect each fluorine atom to iodine with a single bond. Each bond represents 2 electrons, so: \[ 5 \times 2 = 10 \text{ electrons used in bonds} \] Subtract these from the total valence electrons: \[ 42 - 10 = 32 \text{ electrons remaining} \] Distribute the remaining electrons to complete the octets of the fluorine atoms. Each fluorine needs 6 more electrons to complete its octet: \[ 5 \times 6 = 30 \text{ electrons used for fluorine octets} \] Subtract these from the remaining electrons: \[ 32 - 30 = 2 \text{ electrons remaining} \] Place the remaining 2 electrons as a lone pair on the iodine atom.

The electron domains around iodine include 5 bonding pairs and 1 lone pair, making a total of 6 electron domains. The electron domain geometry for 6 electron domains is octahedral.

With 5 bonding pairs and 1 lone pair, the molecular geometry is square pyramidal.

Final Answer