Tag: study of diborane

Questions Related to study of diborane

With which of the following diborane does not produce hydrogen?

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. NaOH

  2. $Cl _{2}$

  3. $H _{2}O$

  4. HCl

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Correct Option: B
Explanation:

$B _{2} H _{6} + 6Cl _{2} \rightarrow 2BCl _{3} + 6HCl $ 

Thus, with chlorine diborane does not form hydrogen.

The non planar molecule among the following is:

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. $B _{2}H _{6}$

  2. $C _{2}H _{4}$

  3. $C _{6}H _{6}$

  4. $BCl _{3}$

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Correct Option: A
Explanation:

diborane has non-planar structure. It has 2 hydrogen atoms which are present above and below the plane of the boron and hydrogen bond. 

In the case of small cuts, bleeding is stopped by applying potash alum, hence alum acts as:

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. germicide

  2. disinfectant

  3. fungicide

  4. coagulating Agent

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Correct Option: D
Explanation:

Clotting of blood is called coagulation and a substance promoting this is called coagulating agent.

All the $Al-Cl$ bonds in $Al _2Cl _6$ are equivalent.
chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. True

  2. False

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Correct Option: B
Explanation:

The given statement is False.
All the  $\displaystyle Al-Cl$ bonds in $\displaystyle Al _2Cl _6$ are not equivalent.
Two types of bonds are present in $\displaystyle Al _2Cl _6$, terminal $\displaystyle Al-Cl$ bonds that are normal $\displaystyle 2c-2e$ bonds and $\displaystyle Al-Cl-Al$ bridges that are $\displaystyle 3c-4e$ bonds.

Which one of the following does not exist in the free form ?

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. $BF _{3}$

  2. $BCl _{3}$

  3. $BBr _{3}$

  4. $BH _{3}$

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Correct Option: D

From $B _2H _6$, all the following can be prepared except :

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. $B _2O _3$

  2. $H _3BO _3$

  3. $B _2(CH _3) _6$

  4. $NaBH _4$

Show answer
Correct Option: C
Explanation:
$B _2H _6+3O _2\rightarrow B _2O _3+3H _2O+Heat$

$B _2H _6+6H _2O\rightarrow H _3BO _3+6H _2$

$2NaH+B _2H _6\rightarrow etherNaBH _4$

From $B _2H _6$, all can be prepared except $B _2(CH _3) _6$

Hence option $C$ is correct.

The hydrides of boron are called :

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. Boron hydrogen compounds

  2. Hydrogen borides

  3. Boranes

  4. Hydroboric acids

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Correct Option: C

The reaction between diborane and chlorine yields :

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. $ B _2H _5Cl $

  2. $ H _2 $

  3. $ BCl _3 $

  4. (B) and (C) both

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Correct Option: C

Which of the following statement is/are correct regarding $B _{2}H _{6}$?

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. Banana bonds are longer but stronger than normal $B-H$ bonds

  2. $B _{2}H _{6}$ is also known $3c-2e$ compound

  3. The hybrid state of B in $B _{2}H _{6}$ is $sp^{3}$ while that of $sp^{2}$ in $BH _{3}$

  4. It cannot be prepared by reacting $BF _{3}$ with $LiBH _{3}$ in the presence of dry ether

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Correct Option: B,C

In which of the following molecules is hydrogen bond absent?

chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

  1. Water

  2. Inorganic benzene

  3. Diborane

  4. Methanol

Show answer
Correct Option: C
Explanation:

As there is no lone pair of electrons present on the boron in diborane it do not form a hydrogen bond.But in water inorganic benzene and methanol there is a lone pair of electrons on oxygen, nitrogen and oxygen respectively so they form hydrogen bonding between most electronegative atom like nitrogen oxygen and hydrogen.
Hence option $D$ is correct.