Tag: study of diborane

Questions Related to study of diborane

The reaction which gives Borazole as a major product is ?

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

  1. $LiH+B _{2}H _{6}\overset{2moles LiH.1moleB _{2}H _{6}}{\rightarrow}$

  2. $B _{2}H _{6}+NH _{3}\xrightarrow[low temp]{2molesB _{2}H _{6}1moleNH _{3}}$

  3. $B _{2}H _{6}+NH _{3}\xrightarrow[low temp]{1moB _{2}H _{6}.2molesNH _{3}}$

  4. $B _{2}H _{6}+NH _{3}\xrightarrow[high\, temp]{1moB _{2}H _{6}.2molesNH _{3}}$

Show answer
Correct Option: D
Explanation:

$B _{2} H _{6} + NH _{3} \xrightarrow [high\, temp]{1moB _{2} H _{6} .2molesNH _{3}} B _{3} N _{3} H _{6} $

When an inorganic compound (X) having $3c-2e$ as well as $2c-2e$ bonds reacts with ammonia gas at a certain temperature, gives a compound (Y) iso-structural with benzene. Compound (X) with ammonia at a high temperature, produces a hard substance (Z). Then :

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

  1. $(X)$ is $B _{2}H _{6}$

  2. (Z) is known as inorganic graphite

  3. (Z) having structure similar to graphite

  4. (Z) having structure similar to (X)

Show answer
Correct Option: A,B,C
Explanation:

$B _{2}H _{6}+2NH _{3}\rightarrow B _{2}H _{6}.2NH _{3}$
When the addition product is heated at $200^{\circ}C$ a volatile compound borazole or inorganic benzene is formed.
$3B _{2}H _{6}.2NH _{3}\rightarrow 2B _{3}N _{3}H _{6}+12H _{2}$
$(X)$ is $B _{2}H _{6}$
Hence option A is correct.

The type of hybridisation of boron in diborane is :

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

  1. $sp$

  2. $sp^2$

  3. $sp^3$

  4. $dsp^2$

Show answer
Correct Option: C
Explanation:

Boron has three valence electrons, so it is supposed to make 3 bonds in a molecule with hybridization, $sp^{2}$ as only s and two p orbitals are used in hybridization and last p orbitalis vacant.

But diborane, $B _{2}H _{6}$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry. 

Hence, each Boron atom is $sp^{3}$ - hybridized.

The correct option is C.

$BCl _3$ does not exist as dimer but $BH _3$ exists as dimer $(B_2H_6) because.

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

  1. Chlorine is more electronegative than hydrogen

  2. There is pn-pn back bonding in $BCl _3$ but $BH _3$ does not contain such multiple bonding

  3. large sized chlorine atoms do not fit in between the small boron atoms where as small sized hydrogen atoms get fitted in between boron atoms

  4. None of the above

Show answer
Correct Option: C
Explanation:

BCl3 does not exist as dimer but BH3 exists as dimer as dimer B2H6 because,
large sized chlorine atoms do not fit in between the small boron atoms where as small sized hydrogen atoms get fitted in between boron atoms
Because of its instability it exists as a dimer, B2H6, where elextrons are shared between each of the monomers. It is a gas, but is used in chemistry as solutions in THF, pyridine or dimethyl sulfide. In these solutions the lone pair of electrons from the oxyge,, nitrogen or sulfur, are donated to the empty p orbital of boron, stabilising the BH3:solvent complex.

Which among the following is not a borane?

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

  1. $B _2H _6$

  2. $B _3H _6$

  3. $B _4H _10$

  4. None of these

Show answer
Correct Option: B
Explanation:

Boranes comprise a large group of the group 13 hydride compounds with the generic formula of $B _xH _y$.
Following are the general formulae of boranes. So $B _3H _6$ is not a borane as it does not hold the formulae of boranes.
Hence option $B$ is correct.

A compound $A$ of boron reacts with $NMe _3$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$. Compound $C$ is an acid.


Gas $D$ is :

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

  1. hydrogen

  2. oxygen

  3. water

  4. none of these

Show answer
Correct Option: A
Explanation:
Since compound $'A'$  of Boron reacts with ${ NMe } _{ 3 }$  to form an adduct $'B'$.  Thus compound $'A'$ is lewis acid.  Hydrolysis of $'B'$ gives  an acid $'C'$ and hydrogen gas.

 Diborane  $(A)$ ${ B } _{ 2 }{ H } _{ 6 }+{ 2NMe } _{ 3 }\longrightarrow { 2BH } _{ 3 }\cdot { NMe } _{ 3 }$  Adduct  $(B)$

${ BH } _{ 3 }\cdot { NMe } _{ 3 }+3{ H } _{ 2 }O\longrightarrow { H } _{ 3 }{ BO } _{ 3 }+{ NMe } _{ 3 }+6{ H } _{ 2 }$  Boric Acid $(C)$

A compound $A$ of boron reacts with $NMe _3$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$. Compound $C$ is an acid.

Compound $C$ is :

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

  1. diborane

  2. boric acid

  3. borate salt

  4. none of these

Show answer
Correct Option: B
Explanation:

     $B _2H _6+2NMe _3\longrightarrow 2BH _3.NMe _3$

Diborane (A)                        Adduct (B)

$BH _3.NMe _3+3H _2O\longrightarrow H _3BO _3+NMe _3+3H _2$
                                                Boric acid                     $\downarrow$
                                                      (C)                          (D) gas

Compound "C" is boric acid $-H _3BO _3$

A compound $A$ of boron reacts with $NMe _3$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$. Compound $C$ is an acid.


Compound $A$ is :

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

  1. diborane

  2. boric acid

  3. borate salt

  4. none of these

Show answer
Correct Option: A
Explanation:

     $B _2H _6+2NMe _3\longrightarrow2BH _3.NMe _3$

Diborane (A)                        Adduct (B)

$BH _3.NMe _3+3H _2O\longrightarrow H _3BO _3+NMe _3+3H _2$
                                               Boric acid                      $\downarrow$
                                                    (C)                            (D) gas
Compound $A$ is diborane.

Which of the following are used as catalyst in Friedal-craft reaction?

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

  1. $AlCl _3$

  2. $SiCl _4$

  3. $BF _3$

  4. $SnCl _4$

Show answer
Correct Option: A,C,D
Explanation:

$Al{ Cl } _{ 3 },B{ F } _{ 3 },Sn{ Cl } _{ 4 }$ can be used as catalyst in Friedal Craft reaction.

An alkali metal hydride (NaH) react with diborane in 'A' to give a tetrahedral compound 'B' which is extensively used as reducing agent in organic synthesis. The compounds 'A' and 'B' respectively are :

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

  1. $CH _3COCH _3\,\,\, and \,\,\,B _3N _3H _6$

  2. $(C _2H _5) _2O \,\,\,and\,\,\, NaBH _4$

  3. $C _2H _6\,\,\,and\,\,C _2H _5Na$

  4. $C _6H _6\,\,\,and\,\,\,NaBH _4$

Show answer
Correct Option: B
Explanation:
Answer:-
When an alkali metal hydride $(NaH)$ react with diborane $({B} _{2}{H} _{6})$ in the presence of ether $({({C} _{2}{H} _{5})} _{2}O)$, a tetrahedral compound (Metal borohydride) is formed which act as a reducing agent in organic synthesis.
$2NaH + {B} _{2}{H} _{6} \; \xrightarrow{{({C} _{2}{H} _{5})} _{2}O} \; \underset{\text{sodium borohydride}}{2NaB{H} _{4}}$
Thus, A is  ${({C} _{2}{H} _{5})} _{2}O$ and B is $NaB{H} _{4}$.