Tag: study of diborane

Quizzes (0) Questions (93)

Questions Related to study of diborane

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

Boron can undergo the following reactions with the given enthaly changes:
Assume no other reactions are occurring. If in a container (operating at constant pressure) which is isolated from the surrounding, mixture of are passed over excess of B(s), then calculate the molar ratio so that temperature of the container do not change :
${\text{2B}}\left( {\text{s}} \right){\text{ + }}\dfrac{{\text{3}}}{{\text{2}}}{{\text{O}} _{\text{2}}}\left( {\text{g}} \right) \to {{\text{B}} _{\text{2}}}{{\text{O}} _{\text{3}}}\left( {\text{s}} \right);\;\Delta H =  - 1260\;KJ$
${\text{2B}}\left( {\text{s}} \right){\text{ + 3}}{{\text{H}} _{\text{2}}}\left( {\text{g}} \right) \to {{\text{B}} _2}{H _6}\left( g \right);\;\Delta H = 30KJ$

  1. 15 : 3

  2. 42 : 1

  3. 1 : 42

  4. 1 : 84
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the temperature of the isolated container to remain constant, the total enthalpy change for the reactions must be zero. Let x moles of B react to form B2O3 and y moles of B react to form B2H6. Using the given stoichiometry and enthalpy values, we solve for the ratio x:y to find the correct molar ratio.

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

For healthy growth of plants, boron is an essential:

  1. minor nutrient

  2. major nutrient

  3. both

  4. major metal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Boron is essential for the growth of higher plants. The primary function of the element is to provide structural integrity to the cell wall in plants. 


Other functions likely include the maintenance of the plasma membrane and other metabolic pathways.

So it is essential micronutrient for plants.

Option A is correct.

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

Diborane undergoes cleavage and gives aduct with
(i) CO
(ii) $ H _2O $
(iii) $ N(Me) _3 $
(iv) $ NH _3 $

  1. i, ii

  2. i, iii

  3. ii, iii

  4. i, iv
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Diborane (B2H6) undergoes symmetrical or unsymmetrical cleavage with various Lewis bases. It forms adducts with CO and NH3, among others. Option A is the most standard representation of these reactions.

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.