Tag: some important compounds of boron

Questions Related to some important compounds of boron

Which of the following is not true regarding the nature of halides of boron?

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

Show answer
Correct Option: D
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.

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:

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

  1. $BCl _3, B _2H _6$

  2. $B _2H _6,BCl _3$

  3. $BF _3,Al _2O _6$

  4. $B _2H _6,BF _3$

Show answer
Correct Option: A
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 $Br$)

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

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

  1. Hydrogen peroxide

  2. Diborane

  3. Ice

  4. Lithium hydride

Show answer
Correct Option: B
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.

On hydrolysis diborane produces ?

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

  1. $H _3BO _2+H _2O _2$

  2. $H _3BO _3+H _2$

  3. $B _2O _3+O _2$

  4. $H _3BO _3+H _2O _2$

Show answer
Correct Option: B
Explanation:

Answer:- (B) $H _3BO _3 + H _2$

Diborane react with water to produce boric acid and hydrogen.
$B _2H _6 + 6H _2O \; \longrightarrow \; 2H _3BO _3 + 6H _2$

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

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

  1. $NH _3$

  2. $CH _3NH _2$

  3. $(CH) _3) _2NH$

  4. $(CH _3) _3N$

Show answer
Correct Option: D
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:-
${B} _{2}{H} _{6} + 2N{(C{H} _{3})} _{3} \; \longrightarrow \; 2B{H} _{3}.N{(C{H} _{3})} _{3}$

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

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

  1. $H _3BO _3$

  2. $B _2O _3$

  3. $B _2H _6$

  4. $B _3N _3H _6$

Show answer
Correct Option: D
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

Select the correct statement(s):

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

  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 $H _3O^+$.

Show answer
Correct Option: A,B,C,D
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.

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

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

  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

Show answer
Correct Option: A,C,D
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.

Select the correct statement(s).

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

  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.

Show answer
Correct Option: A,B,D
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".

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

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

  1. Dipole moment of diborane is zero

  2. Diborane is lewis acid

  3. Diborane has incomplete octet

  4. Diborane has four $2C-2e^-$ bond

Show answer
Correct Option: A,B,C,D
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.