Tag: study of diborane

Questions Related to study of diborane

Which of the following molecular hydride acts as a lewis acid?

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

  1. $CH _4$

  2. $NH _3$

  3. $H _2O$

  4. $B _2H _6$

Show answer
Correct Option: D
Explanation:

$B _2H _6$ is a electron deficient molecule so it acts like an acid(acid is the substance which accepts the pair of electrons). $CH _4,NH _3,H _2O$ allare having lone pairs on central atoms.
Hence option D is correct.

The most acidic compound among the following is:

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

  1. $P _2O _3$

  2. $Sb _2O _3$

  3. $B _2O _3$

  4. $As _2O _3$

Show answer
Correct Option: C
Explanation:

As the size of the boron is less it forms a more acidic oxide (the elements with less atomic size forms most acidic oxides)
So, $B _2O _3$ is most acidic.
Hence option $C$ is correct.

When $NaBH _4$ reacts with $HI$,the product formed is:

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

  1. $B _2H _6$

  2. $B _2H _4$

  3. $H _3BO _3$

  4. none of these

Show answer
Correct Option: A
Explanation:

First iodine reacti with one $NaBH _4$ , we get borance , NaI and HI .

Again, HI is react with another $NaBH _4$ , we get another borane 
so,the answer is $B _2H _6$

All the products formed in the oxidation of $NaBH _{4}$ by $I _{2}$, are:

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

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

  2. $B _{2}H _{6}, H _{2}$ and $NaI$

  3. $BI _{3}$ and $NaH$

  4. $NaBI _{4}$ and $HI$

Show answer
Correct Option: B
Explanation:
The reaction is as follows:

$2NaBH _4(s)+I _2(s)\rightarrow B _2H _6(g)+2NaI(s)+H _2(g)$

Hence, the correct option is $\text{B}$

Which one of the following molecular hydrides acts as a Lewis acid?

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

  1. NH$ _{3}$

  2. H$ _{2}$O

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

  4. CH$ _{4}$

Show answer
Correct Option: C
Explanation:
According to the definition a molecule which can accept a lone pair is called a lewis acid.
$A)$ Ammonia has a lone pair on nitrogen,so it can donate the lone pair rather than accepting a lonepair.So,it is a lewis base.
$B)$Water has $2$ lone pairs on oxygen so it cannot accept any further lonepairs,so water is a lewis base not a lewis acid.
$C)$In diborane the bonds found are banana bonds or tau bonds so it has a tendency to accept a lone pair because it has empty orbitals.So it can be considered as a lewis acid.
$D)$Carbon usually doesn't accept or donate lonepair,so it is neither a lewis base nor lewis acid.We can consider it as a neutral molecule. 

In diborane, boron involves:

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

  1. no hybridization

  2. sp hybridization

  3. $sp^2$ hybridization

  4. $sp^3$ hybridization

Show answer
Correct Option: D
Explanation:

Boron shows ${sp}^{3}$ hybridisation in $B _2H _6$.

Why do boron and aluminium halides behave as Lewis acids?

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

  1. Both halides $(MX _3)$ can accept electrons from a donor to complete their octet.

  2. Both halides $(MX _3)$ can donate a pair of electrons.

  3. Both halides $(MX _3)$ are covalent polymeric structures.

  4. Both halides $(MX _3)$ react with water to give hydroxides and HCI.

Show answer
Correct Option: A
Explanation:
Answer:- (A) Both halides can accept electrons from a donor to complete their octet.
$\because$ Lewis acid is defined as an electron-pair acceptor and both boron and aluminium in their tri-halides $(MX _3)$ possess six electrons in their valence shell. 
Hence, to complete their octet, they can accept a lone pair of electrons.
Thus, both behave as Lewis acid.

Which of the following is not true regarding the nature of halides of boron?

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

  1. Boron trihalides are covalent.

  2. Boron trihalides are planar triangular with $sp^2$ hybridisation.

  3. Boron trihalides act as Lewis acids.

  4. Boron trihalides cannot be hydrolysed easily.

Show answer
Correct Option: D
Explanation:

Boron trihalides like $BF _{3}$ are covalent in nature. These are forming with $sp^{2}$ hybridization in the shape of triangular planar. All trihalides are strong in acidic nature, as Lewis acids. They react with water to form boric acid. 

The sequence for the Lewis acidity is $BF _{3} < BCl _{3} < BBr _{3}$, where $BBr _{3}$ is the strongest Lewis acid. But these trihalides can be easily hydrolyzed except $BF _3$ due to the highly stable nature of $BF _3$.

Thus option D is correct.

Compound $(X)$ on reduction with $LiAlH _4$ gives a hydride $(Y)$ containing 21.72% hydrogen along with other products. The compound $(Y)$ react with air explosively resulting in boron trioxide. Compounds $X$ and $Y$ are respectively:

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

  1. $BCl _3, B _2H _6$

  2. $B _2H _6,BCl _3$

  3. $BF _3,Al _2O _6$

  4. $B _2H _6,BF _3$

Show answer
Correct Option: A
Explanation:
$(A)$ $B{Cl} _{3}, \; {B} _{2}{H} _{6}$
Since ${B} _{2}{O} _{3}$ is formed by reaction of $(Y)$ with air, $(Y)$ therefore should be ${B} _{2}{H} _{6}$ in which % of hydrogen is 21.72. 
${B} _{2}{H} _{6} + 3{O} _{2} \; \longrightarrow \; {B} _{2}{O} _{6} + 3{H} _{2}O + heat$
The compound $(X)$ on reduction with $LiAI{H} _{4}$ gives ${B} _{2}{H} _{6}$. Thus it is boron trihalide.
$4B{X} _{3} + 3LiAl{H} _{4} \; \longrightarrow \; 2{B} _{2}{H} _{6} + 3LiX + 3Al{X} _{3}$ ($X = CI$ or $Br$)

Hydrogen form a 'bridge' in the chemical structure of which of the following compound?

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

  1. Hydrogen peroxide

  2. Diborane

  3. Ice

  4. Lithium hydride

Show answer
Correct Option: B
Explanation:

In $B _2H _6$,the bonding between the boron atoms and the bridging hydrogen atoms is, however, different from that in molecules such as hydrocarbons. Having used two electrons in bonding to the terminal hydrogen atoms, each boron has one valence electron remaining for additional bonding. The bridging hydrogen atoms provide one electron each. Thus the $B _2H _2$ ring is held together by four electrons, an example of 3-center 2-electron bonding. This type of bond is sometimes called a 'banana bond'.

Hence option B is correct answer.