Tag: study of boron
Questions Related to study of boron
Which of the following is not true regarding the nature of halides of boron?
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Boron trihalides are covalent.
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Boron trihalides are planar triangular with $sp^2$ hybridisation.
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Boron trihalides act as Lewis acids.
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Boron trihalides cannot be hydrolysed easily.
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:
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$BCl _3, B _2H _6$
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$B _2H _6,BCl _3$
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$BF _3,Al _2O _6$
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$B _2H _6,BF _3$
Hydrogen form a 'bridge' in the chemical structure of which of the following compound?
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Hydrogen peroxide
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Diborane
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Ice
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Lithium hydride
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'.
On hydrolysis diborane produces ?
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$H _3BO _2+H _2O _2$
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$H _3BO _3+H _2$
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$B _2O _3+O _2$
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$H _3BO _3+H _2O _2$
Answer:- (B) $H _3BO _3 + H _2$
In the reaction,
$2X+B _2H _6 \rightarrow [BH _2(X _2)]^+[BH _4]^-$
'X' cannot be ?
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$NH _3$
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$CH _3NH _2$
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$(CH) _3) _2NH$
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$(CH _3) _3N$
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
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$H _3BO _3$
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$B _2O _3$
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$B _2H _6$
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$B _3N _3H _6$
$2BCl _3+3H _2\underset {discharge}{\xrightarrow {electric}}\underset {(A)}{B _2H _6}+6HCl$.
Select the correct statement(s):
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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.
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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.
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$[Be(Me _2) _n]$ 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.
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Beryllium salts are acidic when dissolved in pure water because the hydrated ion hydrolyzed producing $H _3O^+$.
(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?
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$B _2H _6$ is non-planar
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$B _2H _6$ is polar
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$B _2H _6$ is $e^-$ deficient
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$B _2H _6$ has two $3C-2e^-$ bond
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).
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In diborane 12 valence $e^-$ are involved in bonding
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In diborane, maximum six atoms, two boron and four terminal hydrogen, lie in the same plane.
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Diborane has ethane-like structure
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In diborane, bridging bonds are stronger and longer than the terminal bonds.
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?
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Dipole moment of diborane is zero
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Diborane is lewis acid
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Diborane has incomplete octet
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Diborane has four $2C-2e^-$ bond
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.