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
To know more about hybrisisation:
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These are formed when, after formation of the first atom, the pi-orbitals of Sulfur and Oxygen align and share another pair of electrons. This occurs between sulfur and both of the oxygen atoms, with each oxygen atom forming one pi-bond.
From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of fluorine = 18.99 grams
molecular mass of chlorine = 35.5 grams
Therefore:
one mole of CF2Cl2 = 12 + 2(18.99) + 2(35.5) = 120.98 grams
Therefore, we can use cross multiplication to find the number of moles in 79.34 grams as follows:
mass = (79.34 x 1) / 120.98 = 0.6558 moles
Now, one mole contains 6.022 x 10^23 molecules, therefore:
number of molecules in 0.65548 moles = 0.6558 x 6.022 x 10^23
= 3.949 x 10^23 molecules
Answer:
2
Explanation:
Pb(NO3)2 + 2NaI 2NaNO3 + PbI2
Lithium fluoride is the compound name
