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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Answer:
Copper(II) sulphate – sodium hydroxide reaction
The reaction between copper(Il) sulphate and sodium hydroxide solutions is a good place to start. If you slowly add one to the other while stirring, you will get a precipitate of copper(II) hydroxide, Cu(OH)2.
<em>The statement that gives the relationship between energy needed in breaking a bond and the one that is released after breakin</em>g is
The amount of energy it takes to break a bond is always less than the amount of energy released when the bond is formed.
- Bond energy can be regarded as amount of energy that is required in breaking a particular bond.
- For a bond to be broken Energy will be added and when a bond is broken there will be release of energy
- Bond breaking can be regarded as endothermic process, it is regarded as endothermic because there is a lot of energy required to be absorbed.
- Where ever a bond is broken, there must be formation of another bond
- Bond forming on the other hand can be regarded as exothermic process, since there is a release of releases energy.
Therefore, more energy is required in breaking of bond compare to energy released after breaking of bond.
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your answer will be :
B. <u>Na has a lower</u> <u>electronegativity than H</u>
because Na belongs to alkali metals which are least electronegative (most electro positive) but hydrogen is a non metal, it has higher electronegativity as compared to metals like Sodium (Na).