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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Answer:
The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.Explanation:
There are 66 neutrons in a single atom of indium-115. The atomic number of indium-115 is 49, meaning there are 49 protons. Then the atomic mass is 115, so 115-49 = 66.
When it has sunlight and water
Answer:
Option B. 3.0 M
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 3.0 L
Mole of MgCl₂ = 9 moles
Molarity =?
Molarity can simply be defined as the mole of solute per unit litre of the solution. Mathematically, it can be expressed as:
Molarity = mole of solute /Volume of solution
With the above formula, we can obtain the molarity of the solution as follow:
Volume of solution = 3.0 L
Mole of MgCl₂ = 9 moles
Molarity =?
Molarity = mole of solute /Volume of solution
Molarity = 9 / 3
Molarity = 3 mol/L = 3.0 M
Thus, the molarity of the solution is 3 M
