Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1: Recall the Octet Rule and its conditions
- The Octet Rule states that atoms tend to gain, lose, or share electrons to achieve a stable electron configuration with eight valence electrons. - This rule is most applicable to main-group elements and less so for transition metals.
Step 2: Examine the electron configurations of Al, Br, Be, and F
- Aluminum (Al) has 3 valence electrons, and its electron configuration is 1s² 2s² 2p⁶ 3s² 3p¹. - Bromine (Br) has 7 valence electrons, and its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵. - Beryllium (Be) has 2 valence electrons, and its electron configuration is 1s² 2s². - Fluorine (F) has 7 valence electrons, and its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
Final Answer
AlBr₃ and BeF₂ do not obey the Octet Rule because Aluminum in AlBr₃ has 5 electrons in its 3p orbital, and Beryllium in BeF₂ has 4 electrons in its 2s orbital. This violates the Octet Rule, which states that atoms tend to gain, lose, or share electrons to achieve a stable electron configuration with eight valence electrons.
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students