Tag: study of diborane

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

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

  1. $CH _4$
  2. $NH _3$
  3. $H _2O$
  4. $B _2H _6$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  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 $HI$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\text{B}$
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 molecular hydrides acts as a Lewis acid?

  1. NH$ _{3}$
  2. H$ _{2}$O
  3. B$ _{2}$H$ _{6}$
  4. CH$ _{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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. 
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

Why do boron and aluminium halides behave as Lewis acids?

  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.
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

Which of the following is not true regarding the nature of halides 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.
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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:

  1. $BCl _3, B _2H _6$
  2. $B _2H _6,BCl _3$
  3. $BF _3,Al _2O _6$
  4. $B _2H _6,BF _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $Br$)
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

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

  1. Hydrogen peroxide

  2. Diborane

  3. Ice

  4. Lithium hydride
Reveal answer Fill a bubble to check yourself
B Correct answer
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.