Tag: study of boron

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

The number of simple covalent bonds (B-H) present in diborane:

  1. 2

  2. 4

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The molecular formula of diborane is $B _{2}H _{6}$
Diborane consists 2 3center-2electrons bonds and 4 2center-2electrons bonds.
In diborane, the two boron groups are bonded to each other by a bridge-like structure. Thus, it contains 4 covalent bonds.
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

Which of the following statement is correct for diborane ?

  1. Small amines like $NH _3,\, CH _3NH _2$ give unsymmetrical cleavage of diborane
  2. Large amines such as $(CH _3) _3N$ and pyridine gives symmetrical cleavage of diborane

  3. Small as well as large amines both gives symmetrical cleavage of diborane

  4. (A) and (B) both
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Small\quad amines\ { B } _{ 2 }{ H } _{ 6 }+{ 2NH } _{ 3 }\rightarrow [{ H } _{ 2 }B\left( { NH } _{ 3 } \right) _{ 2 }]^{ + }[{ BH } _{ 4 }]^{ - }\ Large\quad amines\ { B } _{ 2 }{ H } _{ 6 }+2N\left( { CH } _{ 3 } \right) _{ 3 }\rightarrow 2{ H } _{ 3 }B\leftarrow N\left( { CH } _{ 3 } \right) _{ 3 }$

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

Which one of the following statement is not true regarding $B _2H _6$ ?

  1. It react with $NaH$ to form $NaBH _4$

  2. It is highly inflamable gas
  3. It contain four equal $B-H$ bonds

  4. It react with lewis base to form always adduct
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${B} _{2}{H} _{6}$ + $2NaH$ $\rightarrow$ $2 Na(B{H} _{4})$

It is highly Inflamable Gas and it reacts with Lewis base to form always adduct
It Doesn't contain 4 equal B-H bonds
So Option C is correct.

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

Boron halides behave as Lewis acids because of their ________ nature.

  1. proton donor

  2. covalent

  3. electron deficient

  4. ionising
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Lewis, the compound which can accept a lone pair of electron, are called acids.
Boron halides, being electron deficient compounds, can accept a lone pair of electrons, so termed as Lewis acid.

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

Diborane combines with ammonia at ${120}^{o}C$ to give:

  1. ${B} _{2}{H} _{6}.{NH} _{3}$
  2. ${B} _{2}{H} _{6}.2{NH} _{3}$
  3. ${B} _{2}{H} _{6}.3{NH} _{3}$

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Diborane combines with ammonia to form a borazine.

${ B } _{ 2 }{ H } _{ 6 }+2N{ H } _{ 3 }\rightarrow { B } _{ 2 }{ H } _{ 6 }.2N{ H } _{ 3 }$
The reaction proceeds further to form inorganic benzene.
$2{ B } _{ 2 }{ H } _{ 6 }.2N{ H } _{ 3 }\rightarrow 2{ B } _{ 3 }{ N } _{ 3 }{ H } _{ 6 }+12{ H } _{ 2 }$
Here, ${ B } _{ 3 }N _{ 3 }{ H } _{ 6 }$ is inorganic benzene.

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

Number of terminal hydrogen atoms present in diborane?

  1. 2

  2. 4

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Diborane is an electron deficient molecule. The two boron atoms and the four terminal hydrogen atoms of the molecule are all in the same plane. These four terminal B -H bonds are regular 2-centered- 2 electron bonds. The bridging hydrogen atoms lie above and below this plane.

Hence option B is correct answer.

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

Borazine is sometimes called inorganic benzene. Which of the following reactions is expected to give this compound?

  1. $B _{2}H _{6}+NH _{3} \xrightarrow [ excess\ { NH } _{ 3 } ]{ low\ tem } $
  2. $B _{2}H _{6}+NH _{3} \xrightarrow [ excess\ { NH } _{ 3 } ]{ high\ tem } $
  3. $B _{2}H _{6}+NH _{3} \xrightarrow [ ratio\ 2\ { NH } _{ 3 }:1{ B } _{ 2 }{ H } _{ 6 } ]{ high\ tem } $
  4. $All\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Borazine is an inorganic compound with the chemical formula ${ (BH } _{ 3 }{ )(NH } _{ 3 })$ In this cyclic compound, the three BH units and three NH units alternate. The compound is isoelectronic and isostructural with benzene. Like benzene, borazine is a colourless liquid. For this reason, borazine is sometimes referred to as "inorganic benzene"

Borazine is synthesized from diborane and ammonia in a 1:2 ratio at 250–300 °C with a conversion of 50%.

$3B _2H _6 + 6NH _3 \rightarrow 2B _3H _6N _3 + 12H _2$

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

$BCl _3+LiAlH _4\rightarrow A+LiCl+AlCl _3$
$A+H _2O\rightarrow B+H _2$
$B\rightarrow C$.
in this reaction sequence A, B, and C compounds respectively are?

  1. $B _2H _6,B _2O _3,B$
  2. $B _2H _6,H _3BO _3,B _2O _3$
  3. $B _2H _6,H _3BO _3,B$
  4. $HBF _4,H _3BO _3,B _2O _3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The reactions given are,

$LiAlH _4 + BCl _3 \rightarrow B _2H _6 + LiCl + AlCl _3$
Hence A is $B _2H _6$

Now, second reaction is,
$B _2H _6 +H _2O \rightarrow H _3BO _3 +H _2$
Hence B is $H _3BO _3$

On heating B i.e. $H _3BO _3$ we get $B _2O _3$

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

$B _2H _6 + NH _3 \rightarrow$ Addition compound $(X)$
$(X) \overset{450K}{\rightarrow} Y + Z(g)$
In the above sequence $Y$ and $Z$ are respectively:

  1. borazine, $H _2O$
  2. boron, $H _2$
  3. boron nitride, $H _2$

  4. borazine and hydrogen
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$B _2H _6+2NH _3 \longrightarrow \underset {(X)}{B _2H _62NH _3}$

$3B _2H _62NH _3\xrightarrow[]{450K}\underset {(Y)}{2B _3N _3H _6}+\underset {(Z)}{3H _2}$
$(Y)$ & $(Z)$ are borazine and $H _2$ respectively.