Draw the Lewis structure for BrCl3 and answer the following question

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Draw the Lewis structure for BrCl3 and answer the following question. What are the approximate bond angles about the central atom?

Select all that apply.

no v 60°
yes 90°
no 1090
yes 120°
no v 180

The Correct Answer and Explanation is:

To answer this question, let’s first draw the Lewis structure for BrCl₃ (bromine trichloride) and analyze its geometry to determine the approximate bond angles around the central atom.

Step 1: Count Valence Electrons

  • Br (Bromine) has 7 valence electrons.
  • Cl (Chlorine) has 7 valence electrons.
  • There are 3 Cl atoms, so:
    Total = 7 (from Br) + 3 × 7 (from 3 Cl) = 28 valence electrons.

Step 2: Choose Central Atom and Arrange Electrons

Bromine is less electronegative than chlorine, so Br will be the central atom.

  • Connect each Cl atom to Br with a single bond. This uses up 6 electrons (3 bonds × 2 electrons each).
  • Distribute the remaining 22 electrons to satisfy the octets of the Cl atoms first (6 electrons on each of the 3 Cl atoms = 18 electrons).
  • Leftover: 22 – 18 = 4 electrons, which go on the central atom Br as 2 lone pairs.

So, the central atom Br is surrounded by:

  • 3 bonding pairs (Br–Cl bonds)
  • 2 lone pairs

This gives a total of 5 electron domains (regions of electron density).

Step 3: Determine Electron Geometry

With 5 electron domains, the electron geometry is trigonal bipyramidal.

  • In trigonal bipyramidal geometry:
    • Axial bonds are 90° from equatorial positions.
    • Equatorial bonds are 120° apart from each other.

Step 4: Determine Molecular Geometry

Since 2 of the 5 electron domains are lone pairs, the molecular geometry is T-shaped (not trigonal planar or linear).

In a T-shaped molecule:

  • The lone pairs occupy equatorial positions to minimize repulsion.
  • The bonding pairs occupy 2 axial and 1 equatorial position.

Step 5: Approximate Bond Angles

  • The bond angle between axial and equatorial positions is about 90°.
  • The bond angle between equatorial atoms (if bonded) would be 120°, but here the lone pairs occupy equatorial positions.

So, the actual bond angles in BrCl₃ are approximately:

✅ 90°
✅ 120°

✅ Final Answer:

  • Yes: 90°
  • Yes: 120°
  • No: 60°
  • No: 109°
  • No: 180°

These values come from the electron domain geometry (trigonal bipyramidal) and the resulting molecular shape (T-shaped) of BrCl₃.

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