Tag: some important compounds of boron

Questions Related to some important compounds of boron

The reaction between diborane and chlorine yields :

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

  1. $ B _2H _5Cl $

  2. $ H _2 $

  3. $ BCl _3 $

  4. (B) and (C) both

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Correct Option: C

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

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

  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

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Correct Option: B,C

In which of the following molecules is hydrogen bond absent?

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

  1. Water

  2. Inorganic benzene

  3. Diborane

  4. Methanol

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Correct Option: C
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.

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. $B _2H _6$

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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. $As _2O _3$

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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

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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

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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.