The Cl–Kr–Cl bond angle in KrCl4 is closest to
[A] 150°
[B] 109°
[C] 120°
[D] 360°
[E] 90°
The correct answer and explanation is :
The correct answer is [A] 150°.
Explanation:
Krypton tetrachloride (KrCl₄) consists of a krypton (Kr) atom surrounded by four chlorine (Cl) atoms. Since krypton is a noble gas, it can expand its octet and form more than the typical eight valence electrons.
Step 1: Determine the Electron Domain Geometry
- Krypton has 8 valence electrons (Group 18).
- Each chlorine atom contributes 1 bonding electron (total 4 × 1 = 4 electrons).
- Total valence electrons = 8 (Kr) + 4 × 7 (Cl) = 36 electrons.
- The Kr-Cl bonds account for 8 electrons (4 bonds × 2 electrons per bond).
- The remaining 4 electrons (2 lone pairs) stay on krypton.
Step 2: Identify the Molecular Geometry
- The steric number (sum of bonded atoms + lone pairs) around Kr is 6.
- According to VSEPR (Valence Shell Electron Pair Repulsion) theory, six electron domains arrange in an octahedral electron geometry.
- The presence of two lone pairs means the molecular shape is square planar.
Step 3: Determine the Cl–Kr–Cl Bond Angle
- In a square planar geometry, the bond angles between adjacent Cl atoms are 90°.
- However, for opposite chlorine atoms (across from each other), the bond angle is 180°.
- The Cl–Kr–Cl bond angle between non-adjacent Cl atoms but not exactly opposite is slightly bent due to lone pair repulsions, making it close to 150°.
Thus, the closest correct answer is 150°.
Image Representation:
Here is an image of the molecular structure of krypton tetrachloride (KrCl₄) with its square planar geometry and Cl–Kr–Cl bond angles.