Tag: study of boron

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

Diborane is formed from the elements as shown in equation (1) $ 2B(s) +3H _2(g) \rightarrow B _2H _6(g) ......(1) $
The $ H^o$ for the reaction (1) is 
Given that 

$ { H } _{ 2 }O(\ell )\rightarrow H _{ 2 }O(g) $ $ \Delta { H } _{ 1 }^{ 0 }=44kJ $
$ 2B(s)+\frac { 3 }{ 2 } { O } _{ 2 }(g)\rightarrow B _{ 2 }O _{ 3 }(s) $ $ \Delta { H } _{ 2 }^{ 0 }=-1273kJ $
$ B _{ 2 }H _{ 6 }(g)+3O _{ 2 }(g)\rightarrow B _{ 2 }O _{ 3 }+3H _{ 2 }O(g) $ $ \Delta { H } _{ 3 }^{ 0 }=-2035kJ $
$ H _{ 2 }(g)+\frac { 1 }{ 2 } O _{ 2 }(g)\rightarrow H _{ 2 }O(\ell ) $ $ \Delta { H } _{ 4 }^{ 0 }=-286kJ $

  1. 36 kJ

  2. 509 kJ

  3. 520kJ

  4. -3550kJ
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Hess's Law, we combine the given equations to find the enthalpy of formation of B2H6. The calculation results in -3550 kJ based on the provided thermochemical data.

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

Which is/are correct for $B _2H _6$ structure ?

  1. It has $1\:B-H$ terminal bonds and two 3C-2e bonds
  2. It has $5\:B-H$ terminal bonds and two 3C-2e bonds
  3. It has $4\:B-H$ terminal bonds and two 3C-2e bonds and one $B-B$ bond
  4. It has ionic interaction between $[BH _2]^+$ and $[BH _4]^-$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the structure of diborane, $4$ terminal $B-H$ bonds are present. These are regular 2C-2e bonds.
In addition, two $B-H-B$ bridges are present. These are 3C-2e bonds.

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

Which of the following statement about boron carbide is wrong?

  1. Its molecular formula is $B _4C$.

  2. It is also called Norbia.

  3. It is the hardest substance.

  4. It is used for cutting glasses.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Boron carbide $B _4C$ is an extremely hard boroncarbon ceramic material used in tank armor, bulletproof vests, engine sabotage powders and  is called Norbia.

In the arts, silicon carbide is a popular abrasive in modern lapidary due to the durability and low cost of the material. In manufacturing, it is used for its hardness in abrasive machining processes such as grinding, honing, water-jet cutting and sandblasting. 

Particles of silicon carbide are laminated to paper to create sandpapers and the grip tape on skateboards.

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

Find the incorrect match.

  1. $Al _2Cl _6$ : $3C - 4e$ bond is present
  2. $Al _2(CH _3) _6$ : All carbon atoms are $sp^{3}$ - hybridized
  3. $I _2Cl _6$ : Nonplanar
  4. $Al _2Br _6$ : Nonpolar
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A) $Al _2Cl _6$ contains, $3C-4e$ bonds. In each such bond, two $Al$ atoms and one $Cl$ atom participate. Each $Al$ atom contributes $1$ electron and $Cl$ atom contributes $2$ electrons. Thus, the option A is correct.
B) In $Al _2(CH _3) _6$, all $C$ atoms are $sp^3$ hybridized and result in tetrahedral geometry. Thus, the option B is correct.
C) In $I _2Cl _6$, each iodine atom is $sp^3d^2$ hybridized with $4$ bond pairs and two lone pairs. Each iodine atom has square planar geometry. The repulsion between the lone pair of electrons on two iodine atoms will be minimized when the molecule is planar. Thus, the option C is incorrect.
D) $Al _2Br _6$ molecule is non planar as each $Al$ atom is $sp^3$ hybridized. Thus, 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

Find out the similarities between $I _2Cl _6$ and $Al _2Cl _6$.

  1. Both have $3C\, -\, 4e^-$ bond
  2. Both have $sp^3$- hybridization for the central atom

  3. Both are nonplanar

  4. All are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A)  Both $ I _2Cl _6$ and $ Al _2Cl _6$ have $ 3C\, -\, 4e^-$ bonds. Hence, option A is correct.
B) In $ I _2Cl _6$, the central $I$ atom undergoes $sp^3d^2$ hybridization and in $ Al _2Cl _6$, the central $Al$ atom undergoes $sp^3$ hybridization. Thus, the option B is incorrect.
C) $ I _2Cl _6$ is planar and $ Al _2Cl _6$ is non planar. Thus, the option C is incorrect.

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

Which of the following molecule has $3C\, -\, 4e^-$ bond?

  1. $Al _2Cl _6$
  2. $Be _2Cl _4$
  3. $I _2Cl _6$
  4. All are having $3C\, -\, 4e^-$ bond
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 $3C-4e^-$ bond is a model used to explain bonding in certain hypervalent molecules such as tetratomic & hexatomic interhalogen compounds, sulphur tetrafluoride, xenon fluoride and the fluoride ions.

$Al _2Cl _6$, $Be _2Cl _4$ and $I _2Cl _6$ contain $3C\, -\, 4e^-$ bond. The chlorine atoms acts as bridges and forms one covalent bond and other coordinate bond.

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

Which one of the following methyl diboranes does not exist?

  1. $B _2H _4(CH _3) _2$
  2. $B _2H _3(CH _3) _3$
  3. $B _2H _2(CH _3) _4$
  4. $B _2H(CH _3) _5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Di borane forms methyl diborane($B _2H _5(CH _3))$,dimethyl diborane($B _2H _4(CH _3) _2$),trimethyl diborane($B _2H _3(CH _3) _3$) ,tetramethyl diborane($B _2H _2(CH _3) _4$).

So $B _2H(CH _3) _5$ can not form,becuase in this compound we have to use bridge hydrogen for the formation of this compound,that never happens.

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

Which of the following overlapping is used for the formation of $3C\, -\, 2e^-$ bond in chain polymer of $BeMe _2$ ?

  1. $sp\, -\, sp\, -\, sp$
  2. $sp^2\, -\, sp^2\,-\, sp^2$
  3. $sp^2\, -\, sp^3\, -\, sp^2$
  4. $sp^3\, -\, sp^3\, -\, sp^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$sp^3\, -\, sp^3\, -\, sp^3$ overlapping is used for the formation of $3C\, -\, 2e^-$ bond in chain polymer of $BeMe _2$. Thus, the $Be$ atom and $C$ atom of each methyl group is $sp^3$ hybridized.