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

In the reaction, 
$2X+B _2H _6 \rightarrow [BH _2(X _2)]^+[BH _4]^-$
'X' cannot be ?

  1. $NH _3$
  2. $CH _3NH _2$
  3. $(CH) _3) _2NH$
  4. $(CH _3) _3N$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Answer:- (D) ${(C{H} _{3})} _{3}N$
${B} _{2}{H} _{6}$ reacts with primary(1º)  and secondary(2º)  amine and form an ionic compound and gives unsymmetrical cleavage of diborane. However with tertiary(3º) amine, ${B} _{2}{H} _{6}$ shows it's symmetrical cleavage and forms an adduct as:-
${B} _{2}{H} _{6} + 2N{(C{H} _{3})} _{3} \; \longrightarrow \; 2B{H} _{3}.N{(C{H} _{3})} _{3}$
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

A mixture of boron trichloride and hydrogen is subjected to silent electric discharge to form $'A'$ and $HCl. 'A'$ is miced with $NH _3$ and heated to $200^oC$ to form $'B'$. The formula of $'B'$ is

  1. $H _3BO _3$
  2. $B _2O _3$
  3. $B _2H _6$
  4. $B _3N _3H _6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$2BCl _3+3H _2\underset {discharge}{\xrightarrow {electric}}\underset {(A)}{B _2H _6}+6HCl$.


$A$  is $B _2H _6$

$3B _2H _6+6NH _3\rightarrow 2B _3H _6N _3 + 12H _2$

So $B$  is $B _3N _3H _6$.

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

Select the correct statement(s):

  1. The crystal structure of $NaHCO _3$ and $KHCO _3$ both show hydrogen bonding, but are different. In $NaHCO _3$ the $HCO^{-} _{3}$ ions are linked into an infinite chain, while in $KHCO _3$ a dimeric anion is formed.
  2. The $BeX _2$ molecules polymerize to form chains containing bridging halogen groups; for example, in $(BeF _2) _n$ and $(BeCl _2) _n$ each halogen from one normal covalent bond and use a lone pair to form a co-ordinate bond.
  3. $[Be(Me _2) _n]$ has essentially the same structure as $(BeCl _2) _n$ but the bonding in the methyl compound is best regarded as three center two electron bonds covering one $Me$ and $Be$ atoms.
  4. Beryllium salts are acidic when dissolved in pure water because the hydrated ion hydrolyzed producing $H _3O^+$.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

(A) Even though both sodium bicarbonate and potassium bicarbonate shows hydrogen bonding in the  crystal structure, sodium bicarbonate forms a polymer and potassium bicarbonate forms a dimer.
Hence, the option A is correct.
(B)  Beryllium dihalide is a polymer in which halogen atoms acts as bridges between two Be atoms.
Thus, each halogen atom forms two bonds, one is coordinate and the other is covalent.
Thus, the option B is correct.
(C) The polymeric dimethyl beryllium and the polymeric beryllium dichloride have similar structures involving bridging but in polymeric dimethyl beryllium, each methyl group form bridges with two Be atoms. These bridges are 3 C - 2 e bonds.
Thus, the option C is correct.
(D) Beryllium ion on hydration forms protonium ion. Hence, beryllium salts are acidic.
$Be^{2+} + 2H-OH \rightarrow Be(OH)^+ +H _3O^+$.
Hence, the 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 statements(s) is/are correct?

  1. $B _2H _6$ is non-planar
  2. $B _2H _6$ is polar
  3. $B _2H _6$ is $e^-$ deficient
  4. $B _2H _6$ has two $3C-2e^-$ bond
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

Diborane is non-planar molecule as each $B$ atom has tetrahedral geometry.It is electron deficient molecule as each $B$ atom has only $6$ valence electrons. It contains two $3C-2e$ bonds.
Thus, all the options are correct.

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

Select the correct statement(s).

  1. In diborane 12 valence $e^-$ are involved in bonding

  2. In diborane, maximum six atoms, two boron and four terminal hydrogen, lie in the same plane.

  3. Diborane has ethane-like structure

  4. In diborane, bridging bonds are stronger and longer than the terminal bonds.
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Diborane is an electron deficient molecule. The two boron atoms and the four terminal hydrogen atoms of the molecule are in the same plane. The bridging hydrogen atoms lie above and below this plane. The four B-H bonds are regular 2-centered 2-electron bonds. The bridging B-H bonds are unusual 3-centered 2-electron bonds. There are 12 valence electrons, out of them 8 are used in four non bridging hydrogen bonds and other 4 are shared in forming 2-centered 2-electron bond. The bridging hydrogen bonds are stronger that means the bridging H-atoms can not be replaced in chemical reactions. The lengths of these bonds are $1.33 A°$ which is longer than that of terminal $H-$bond $1.19A°$. This is because of the electrostatic repulsion felt by the positively charged nuclei of the two hydrogen atoms that form the hydrogen bridge will cause the bond to be bent- referred to as a "banana bond".

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.