Tag: some important compounds 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

Which of the following statement is/are correct regarding $B _{2}H _{6}$?

  1. Banana bonds are longer but stronger than normal $B-H$ bonds
  2. $B _{2}H _{6}$ is also known $3c-2e$ compound
  3. The hybrid state of B in $B _{2}H _{6}$ is $sp^{3}$ while that of $sp^{2}$ in $BH _{3}$
  4. It cannot be prepared by reacting $BF _{3}$ with $LiBH _{3}$ in the presence of dry ether
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

In which of the following molecules is hydrogen bond absent?

  1. Water

  2. Inorganic benzene

  3. Diborane

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

As there is no lone pair of electrons present on the boron in diborane it do not form a hydrogen bond.But in water inorganic benzene and methanol there is a lone pair of electrons on oxygen, nitrogen and oxygen respectively so they form hydrogen bonding between most electronegative atom like nitrogen oxygen and hydrogen.
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

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.