Answer
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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.
Step 3:: Observe the formation of AlBr₃ and BeF₂
- AlBr₃ is formed by Al sharing its 3 valence electrons with 3 Br atoms, each of which contributes 7 valence electrons. - Al: 3s³ 3p³, forming 3 bonds with Br - Br: 4s² 4p⁵, accepting an electron from Al to complete the octet - BeF₂ is formed by Be sharing its 2 valence electrons with 2 F atoms, each of which contributes 7 valence electrons. - Be: 2s², forming 2 bonds with F - F: 2s² 2p⁵, accepting an electron from Be to complete the octet
Step 4:: Analyze the electron configurations of AlBr₃ and BeF₂
- In AlBr₃, Al has 5 electrons in its 3p orbital, not satisfying the octet rule. - In BeF₂, Be has 4 electrons in its 2s orbital, not satisfying the octet rule.
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.
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