Tag: chemistry

Quizzes (490) Questions (9892)

Questions Related to chemistry

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

Which of the following statements(s) is/are correct?

  1. Dipole moment of diborane is zero

  2. Diborane is lewis acid

  3. Diborane has incomplete octet
  4. Diborane has four $2C-2e^-$ bond
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Diborane is the chemical compound consisting of boron and hydrogen with the formula $B _2H _6$.In diborane, boron is $sp^3$ hybridized and each boron is attached to two hydrogens through two centres two-electron bond ($2C-2e^-$ bond) and to another two hydrogens (bridging hydrogen) through three centres two electron bonds ($3C-2e^-$ bond). $BH _3$ (monomer of $B _2H _6$), is an electron-deficient molecule (i.e. the boron is surrounded by only six electrons; not eight). It means $B _2H _6$ is surrounded by $12$ electrons not with $16$ electrons(electron deficient molecule is Lewis acid). And the dipole moment is zero. 

Hence options $A,\ B,\ C$ and $D$ are correct.

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

$Mg _3B _2\xrightarrow {HCl _{(aq)}}[X]+MgCl _2$ 

For [X] and [Y], the incorrect choice is :

  1. $[X]+H _2O\xrightarrow {HCl _{(aq)}}[Y]+H _2$
  2. [X] is $BCl _3$ and [Y] is $H _3BO _3$

  3. [X] with air and [Y] on strong heating (red heat) give same compound
  4. in [Y], Boron completes its octet by removing $H^+$ from water molecule
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$Mg _3B _2 + HCl \rightarrow MgCl _2 +B _2H _6$
X is $B _2H _6$
Y is $H _3BO _3$
boric acid octet by removing hydrogrn from water molecule 
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

In the reaction $2X+B _2H _6\rightarrow [BH _2X _2]^+[BH _4]^-$ the amine(s) X is (are) :

  1. $NH _3$
  2. $CH _2NH _2$
  3. $(CH _3) _2NH$
  4. $(CH _3) _3N$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation
Diborane forms addition compounds with lewis bases as it is electron deficient compound act as lewis acid. If the lewis base is small without steric strain unsymmetric cleavage takes place. If the base is large with steric strain symmetric cleavage takes place. 
$B _2H _6 + NH _3 \rightarrow [BH _2(NH _3) _2]^+[BH _4]^- \rightarrow B _3N _3H _6 (Borazol)$
$B _2H _6 + CH _3NH _2 \rightarrow [BH _2(CH _3NH _2) _2]^+[BH _4]^-$
$B _2H _6 + (CH _3) _2NH \rightarrow [B^-H _3-N^+H(CH _3) _2] \rightarrow [BH _2N(CH _3) _2]$
$B _2H _6 + (CH _3) _3N \rightarrow [(CH _3) _3N^+-B^-H _3]$

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

$BCl _{3}$ does not exist as a dimer but $BH _{3}$ exists as a dimer because:

  1. chlorine is more electropositive than hydrogen
  2. there is $p\pi -p\pi $ back-bonding in $BCl _{3}$ but $BH _{3}$ does not contain such multiple bonding

  3. large sized chlorine atoms cannot fit in between small sized boron atoms, whereas small sized hydrogen atoms can easily fit in between boron atoms

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The dimerization of BH3 is favored because the small size of hydrogen atoms allows them to bridge the two boron atoms effectively. In contrast, the larger chlorine atoms in BCl3 cause steric hindrance, preventing the formation of a stable dimer.

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

The reaction which gives Borazole as a major product is ?

  1. $LiH+B _{2}H _{6}\overset{2moles LiH.1moleB _{2}H _{6}}{\rightarrow}$
  2. $B _{2}H _{6}+NH _{3}\xrightarrow[low temp]{2molesB _{2}H _{6}1moleNH _{3}}$
  3. $B _{2}H _{6}+NH _{3}\xrightarrow[low temp]{1moB _{2}H _{6}.2molesNH _{3}}$
  4. $B _{2}H _{6}+NH _{3}\xrightarrow[high\, temp]{1moB _{2}H _{6}.2molesNH _{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$B _{2} H _{6} + NH _{3} \xrightarrow [high\, temp]{1moB _{2} H _{6} .2molesNH _{3}} B _{3} N _{3} H _{6} $

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

When an inorganic compound (X) having $3c-2e$ as well as $2c-2e$ bonds reacts with ammonia gas at a certain temperature, gives a compound (Y) iso-structural with benzene. Compound (X) with ammonia at a high temperature, produces a hard substance (Z). Then :

  1. $(X)$ is $B _{2}H _{6}$

  2. (Z) is known as inorganic graphite

  3. (Z) having structure similar to graphite

  4. (Z) having structure similar to (X)
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$B _{2}H _{6}+2NH _{3}\rightarrow B _{2}H _{6}.2NH _{3}$
When the addition product is heated at $200^{\circ}C$ a volatile compound borazole or inorganic benzene is formed.
$3B _{2}H _{6}.2NH _{3}\rightarrow 2B _{3}N _{3}H _{6}+12H _{2}$
$(X)$ is $B _{2}H _{6}$
Hence option A is correct.

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

The type of hybridisation of boron in diborane is :

  1. $sp$
  2. $sp^2$
  3. $sp^3$
  4. $dsp^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Boron has three valence electrons, so it is supposed to make 3 bonds in a molecule with hybridization, $sp^{2}$ as only s and two p orbitals are used in hybridization and last p orbitalis vacant.

But diborane, $B _{2}H _{6}$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry. 

Hence, each Boron atom is $sp^{3}$ - hybridized.

The correct option is C.

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

$BCl _3$ does not exist as dimer but $BH _3$ exists as dimer $(B_2H_6) because.

  1. Chlorine is more electronegative than hydrogen
  2. There is pn-pn back bonding in $BCl _3$ but $BH _3$ does not contain such multiple bonding

  3. large sized chlorine atoms do not fit in between the small boron atoms where as small sized hydrogen atoms get fitted in between boron atoms

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

BCl3 does not exist as dimer but BH3 exists as dimer as dimer B2H6 because,
large sized chlorine atoms do not fit in between the small boron atoms where as small sized hydrogen atoms get fitted in between boron atoms
Because of its instability it exists as a dimer, B2H6, where elextrons are shared between each of the monomers. It is a gas, but is used in chemistry as solutions in THF, pyridine or dimethyl sulfide. In these solutions the lone pair of electrons from the oxyge,, nitrogen or sulfur, are donated to the empty p orbital of boron, stabilising the BH3:solvent complex.

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

Which among the following is not a borane?

  1. $B _2H _6$
  2. $B _3H _6$
  3. $B _4H _10$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Boranes comprise a large group of the group 13 hydride compounds with the generic formula of $B _xH _y$.
Following are the general formulae of boranes. So $B _3H _6$ is not a borane as it does not hold the formulae of boranes.
Hence option $B$ is correct.

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

A compound $A$ of boron reacts with $NMe _3$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$. Compound $C$ is an acid.


Gas $D$ is :

  1. hydrogen

  2. oxygen

  3. water

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Since compound $'A'$  of Boron reacts with ${ NMe } _{ 3 }$  to form an adduct $'B'$.  Thus compound $'A'$ is lewis acid.  Hydrolysis of $'B'$ gives  an acid $'C'$ and hydrogen gas.

 Diborane  $(A)$ ${ B } _{ 2 }{ H } _{ 6 }+{ 2NMe } _{ 3 }\longrightarrow { 2BH } _{ 3 }\cdot { NMe } _{ 3 }$  Adduct  $(B)$

${ BH } _{ 3 }\cdot { NMe } _{ 3 }+3{ H } _{ 2 }O\longrightarrow { H } _{ 3 }{ BO } _{ 3 }+{ NMe } _{ 3 }+6{ H } _{ 2 }$  Boric Acid $(C)$