Which hybrid orbitals would NOT be possible in electrically neutral molecules containing the elements shown?
1. sp3/Al and H
2. sp2/B and F
3. sp3/O and F
4. sp2/N and O
5. sp/C and H
sp3/Al and H
To determine which hybrid orbitals would not be possible in electrically neutral molecules containing the given elements, we need to consider the valence electron configuration of each element and their ability to undergo hybridization.
1. sp3/Al and H:
The valence electron configuration of aluminum (Al) is 3s2 3p1, which means it has three valence electrons. Hydrogen (H) has only one valence electron. Both aluminum and hydrogen can undergo hybridization.
2. sp2/B and F:
The valence electron configuration of boron (B) is 2s2 2p1, and fluorine (F) has one electron less than a stable noble gas configuration. In this case, boron can undergo hybridization, but fluorine cannot as it already has a complete octet.
3. sp3/O and F:
The valence electron configuration of oxygen (O) is 2s2 2p4, and fluorine (F) has one electron less than a stable noble gas configuration. Both oxygen and fluorine can undergo hybridization.
4. sp2/N and O:
The valence electron configuration of nitrogen (N) is 2s2 2p3, and oxygen (O) has six valence electrons. Both nitrogen and oxygen can undergo hybridization.
5. sp/C and H:
The valence electron configuration of carbon (C) is 2s2 2p2, and hydrogen (H) has only one valence electron. Both carbon and hydrogen can undergo hybridization.
Based on the analysis, the hybrid orbitals that would NOT be possible in electrically neutral molecules containing the given elements are sp2/B and F.
To determine which hybrid orbitals would not be possible in electrically neutral molecules containing the given elements, we need to consider the valence electron configuration of each element and the number of bonds it can form.
1. sp3/Al and H: Al is in group 13 and has a valence electron configuration of 3s2 3p1. It can form three bonds, which means it can hybridize its orbitals to form sp2 hybrid orbitals. Therefore, sp3 hybrid orbitals are not possible for Al in this case. H has only one valence electron and can form only one bond, so it doesn't involve hybridization.
2. sp2/B and F: B is in group 13 and has a valence electron configuration of 2s2 2p1. It can form three bonds, making sp2 hybridization possible. F has seven valence electrons and can form one bond, so it doesn't involve hybridization.
3. sp3/O and F: O is in group 16 and has a valence electron configuration of 2s2 2p4. It can form two bonds, making sp2 hybridization possible. F has seven valence electrons and can form one bond, so it doesn't involve hybridization.
4. sp2/N and O: N is in group 15 and has a valence electron configuration of 2s2 2p3. It can form three bonds, making sp2 hybridization possible. O is in group 16 and can form two bonds, so it can also undergo sp2 hybridization.
5. sp/C and H: C is in group 14 and has a valence electron configuration of 2s2 2p2. It can form four bonds, which allows it to undergo sp3 hybridization. H has only one valence electron and can form only one bond, so it doesn't involve hybridization.
Therefore, the hybrid orbitals that would NOT be possible are:
1. sp3/Al and H
3. sp3/O and F
The correct answer is option 1.