For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
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Chemical energy is the answer to your question
Answer:
Carbon-13
Explanation:
Carbon have three isotopes. Isotopes are the atoms of the same element which has a different number of neutrons. Carbon has 3 isotopes.
Carbon-12 : 6 electrons I 6 protons I 6 neutrons
Carbon-13 : 6 electrons I 6 protons I 7 neutrons
Carbon-14 : 6 electrons I 6 protons I 8 neutrons
Answer:
A. 1:3
Explanation:
If we look at the ions shown in the image attached to the question, we will notice that we have aluminum (Al^3+), a trivalent ion combining with the iodide ion (I^-).
Aluminum can easily give out its three outermost electrons to three atoms of iodine. If aluminum gives out its three electrons, it achieves the stable octet structure. Iodine atoms have seven electrons in their outermost shell. They only need one more electrons to complete their octet. This one electron can be gotten by the combination of three iodine atoms with one atom of aluminum. One electron each is transferred from the aluminum atom to each iodine atom to form AlI3 with a ratio of 1:3.
