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If diborane each boron has three electro...

If diborane each boron has three electrons bonds around each B atom and form

A

Two bonds

B

Three bonds

C

Four bonds

D

Five bonds

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To solve the question regarding the bonding in diborane (B2H6) and how many bonds each boron atom forms, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Structure of Diborane**: - Diborane has the chemical formula B2H6. It consists of two boron (B) atoms and six hydrogen (H) atoms. 2. **Drawing the Structure**: - In diborane, the two boron atoms are connected to each other and also bonded to hydrogen atoms. The structure can be visualized as: ``` H H \ / B - B / \ H H \ H ``` - In this structure, two hydrogen atoms are terminally bonded to each boron atom, and there are two additional hydrogen atoms that create bridge bonds between the two boron atoms. 3. **Identifying the Bonds**: - Each boron atom forms: - Two terminal B-H bonds (one with each of the two terminal hydrogen atoms). - One bridge bond with the other boron atom, which can be considered as a unique type of bond known as a "banana bond" due to its shape. 4. **Counting the Total Bonds**: - For each boron atom: - 2 bonds with terminal hydrogen atoms. - 1 bridge bond with the other boron atom. - Therefore, each boron atom forms a total of 3 bonds. 5. **Conclusion**: - The question states that if diborane has three electrons around each boron atom, we can conclude that each boron atom forms a total of 4 bonds (2 terminal B-H bonds and 2 bridge bonds). - Thus, the correct answer is that each boron atom forms **4 bonds**. ### Final Answer: - Each boron atom in diborane forms **4 bonds**.
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Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)) . The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product. CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH=CH_(2) underset((ii)H_(2)O_(2)//OH^(bar(..)))overset((i)BH_(3)//THF)rarrZ . Z is :

Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)) . The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product. Y is :

Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)) . The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product. CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH=CH_(2) underset((ii)H_(2)O_(2)//OH^(bar(..)))overset((i)BH_(3)//THF)rarrZ . Z is :

Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)) . The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product.