Draw the Lewis structure for ICl3

Draw the Lewis structure for ICl3

The correct answer and explanation is :

Lewis Structure of ICl₃

The Lewis structure of ICl₃ (Iodine trichloride) consists of an iodine (I) atom at the center bonded to three chlorine (Cl) atoms and one lone pair of electrons on iodine.

Step-by-Step Process to Draw the Lewis Structure:

1. Count the Valence Electrons:

• Iodine (I) is in Group 17 and has 7 valence electrons.
• Each Chlorine (Cl) atom is also in Group 17 and has 7 valence electrons.
• Total valence electrons:
  [
  7 + (3 \times 7) = 28 \text{ valence electrons}
  ]

1. Determine the Central Atom:

• Iodine (I) is less electronegative than chlorine, so it is the central atom.

1. Distribute Electrons to Form Bonds:

• Form single bonds between iodine and each chlorine. This uses 6 electrons (3 bonds × 2 electrons per bond).
• Remaining electrons: ( 28 – 6 = 22 ).

1. Distribute Remaining Electrons:

• Place 6 electrons (3 lone pairs) around each chlorine to satisfy the octet rule.
• This uses 18 electrons (6 per Cl × 3 Cl atoms).
• Remaining electrons: ( 22 – 18 = 4 ).

1. Place Remaining Electrons on the Central Atom (Iodine):

• The remaining 4 electrons go to iodine as 2 lone pairs.

1. Check Formal Charges and Octet Rule:

• Each chlorine has 8 electrons (octet complete).
• Iodine has more than 8 electrons (10 total: 3 bonds + 2 lone pairs), which is allowed since iodine is in Period 5 and can have an expanded octet.

Final Lewis Structure:

      Cl
      |
Cl—I—Cl
      |
      ..

Iodine (I) has one lone pair above it.

Explanation (300 Words)

Iodine trichloride (ICl₃) is a polar molecule due to its T-shaped molecular geometry, which arises from the presence of a lone pair on the central iodine atom. The valence electron count shows that iodine follows the expanded octet rule, meaning it accommodates more than 8 electrons due to its position in Period 5 of the periodic table.

When drawing the Lewis structure, we begin by identifying the total valence electrons (28). We assign iodine as the central atom since it is the least electronegative. Three single bonds are formed between iodine and chlorine, consuming 6 electrons. Each chlorine gets 6 more electrons to complete their octets, using 18 more electrons, leaving 4 electrons for iodine. These 4 electrons remain as two lone pairs on iodine.

The molecular geometry is determined using VSEPR (Valence Shell Electron Pair Repulsion) theory. The AX₃E₂ notation (3 bonded atoms + 2 lone pairs) suggests a T-shaped structure. The lone pairs on iodine cause a bent geometry instead of a perfect trigonal planar shape.

Additionally, ICl₃ exhibits dipole moments due to an asymmetrical charge distribution, making it polar. The presence of lone pairs affects its shape and polarity, causing repulsion among bonded and non-bonded electron pairs. This leads to deviations from an ideal trigonal planar shape.

In summary, the Lewis structure of ICl₃ follows the expanded octet rule, displays a T-shaped geometry, and is polar due to lone pair-bond pair repulsions.