Moissan Boron is amorphous Boron of ultra high purity.

Boron is essential for the growth of higher plants. The primary function of the element is to provide structural integrity to the cell wall in plants. 

Boron carbide is used as a control rod material.

Both have number of electrons and similar structure and properties.
Isosteres are molecules or ions of similar size containing the same number of atoms and valence electrons, e.g., O^2-, F-, Ne.

$Small\quad amines\ { B } _{ 2 }{ H } _{ 6 }+{ 2NH } _{ 3 }\rightarrow [{ H } _{ 2 }B\left( { NH } _{ 3 } \right) _{ 2 }]^{ + }[{ BH } _{ 4 }]^{ - }\ Large\quad amines\ { B } _{ 2 }{ H } _{ 6 }+2N\left( { CH } _{ 3 } \right) _{ 3 }\rightarrow 2{ H } _{ 3 }B\leftarrow N\left( { CH } _{ 3 } \right) _{ 3 }$

${B} _{2}{H} _{6}$ + $2NaH$ $\rightarrow$ $2 Na(B{H} _{4})$

According to Lewis, the compound which can accept a lone pair of electron, are called acids.
Boron halides, being electron deficient compounds, can accept a lone pair of electrons, so termed as Lewis acid.

Diborane combines with ammonia to form a borazine.

Diborane is an electron deficient molecule. The two boron atoms and the four terminal hydrogen atoms of the molecule are all in the same plane. These four terminal B -H bonds are regular 2-centered- 2 electron bonds. The bridging hydrogen atoms lie above and below this plane.

The reactions given are,

$B _2H _6+2NH _3 \longrightarrow \underset {(X)}{B _2H _62NH _3}$

The chemical formula for diborane is $B _2H _6$ which resembles general formula $B _n{H} _{n+4}$.

Diborane is electron-deficient i.e, there are not enough valence electrons to form the expected number of covalent bonds. For example, in diborane, there are only 12 electrons, three from each boron atoms and six from hydrogen, while ethane possesses 14 such electrons.

In diborane, there are 2 bridged $H$ atoms and 4 terminal $H$ atoms as shown in figure.

X is $B _2H _6$ having 3c-2e as well as 2c-2e bonds. $B _2H _6$ reacts with $NH _3$ to form borazine or inorganic benzene. Z is $(BN) _x$ is called inorganic graphite and has similar structure to graphite. The whole reaction is shown in figure.

Smaller amines such as $NH _3$, $CH _3NH _2$ and $(CH _3) _2NH$ give unsymetrical cleavage of diborane according to following reaction
$2NH _3+B _{2}H _{6}\rightarrow [BH _{2}(NH _3) _{2}]^{+}BH _{4}^{-}$
Large amines such as $(CH _3) _3N$ give symmetrical cleavage of diborane according to following reaction
$B _2H _6 + 2N(CH _3) _3 \rightarrow 2(CH _3) _3N \rightarrow BH _3$

The combustion of diborane give $B _2O _3$

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.

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

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.

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.

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($B _2H _2(CH _3) _4$).

$sp^3\, -\, sp^3\, -\, sp^3$ 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.

$[BeF _2] _n$ contains $3C-4e^-$ bonds.
One $F$ atom and two $Be$ atoms are $3$ centers. The middle $O$ contributes two electrons each to the be atom, thereby a total of $4$ electrons are contributed and it forms $Be−F−Be$ bridges. 

The correct representation of boranes include (A) $B _n\, H _{n\, +\, 4}$, called as nido-boranes and (B) $B _n\, H _{n\, +\, 6}$, called as arachno boranes.
(C) $B _n\, H _{2n\, +\, 2}$ is not a correct representation of boranes.
Note: $B _n\, H _{n\, +\, 6}$ are less stable than $B _n\, H _{n\, +\, 4}$

(B) $BeF _2$ has $3C - 2e^-$ bond.

The wide absorption spectrum of boron makes it suitable also as a neutron shield. Larger the cross-sectional area more the number of thermal neutrons get captured in a nuclear reactor.

Diborane has 4-termianl $B-H$ bonds and 2-bridged $B-H-B$ bonds. During methylation of diborane, all terminal $H$ atoms are replaced by methyl atoms.

Boranes have the general formulae as $B _{n} H _{n+4} $ or $B _{n} H _{n+6} $.

There are 2 e- in each B-H-B bond. In total there are 2+ 2 =4 e- shared by the 2 boron atoms.

$2BCl _{3} + 6H _{2} (excess)\xrightarrow[silent electric]{discharge} B _{2} H _{6} + 6HCl$

1. The bridge bonds are 3c-2e bonds. They have 2 electrons (4 in normal covalent). Thus in diborane boron is short of 2 electrons hence it forms hydrogen bridge.
   

$B _{2}H _{6}+2NH _{3}\overset {120^{0}C}{\rightarrow}B _{2}H _{6}.2NH _{3}\overset {200^{0}C}{\rightarrow} \underset {Borazole}{B _{3}N _{3}H _{6}} \overset {>200^{0}C}{\rightarrow} 3(BN) _{n}+3nH _{2}$
Borazole is also called borazine or inorganic benzene as its structure is very similar to benzene, with boron and nitrogen alternatively arranged instead of carbon and coordinate bond from nitrogen to boron instead of carbon-carbon double bond.

There are 4 2c-2e B-H bonds and 2 3c-2e bonds in diborane.

In diborane, ${sp}^{3}$ hybrid orbital from each boron overlaps with 1$s$ orbital of $H$ atom to form 3 center-2 electron (3c-2e) bond.

$B _2H _6 + HCl \rightarrow B _2H _5Cl + H _2$
This reaction takes place in the presence of anhydrous $AlCl _3$.
$B _2H _6 + Br _2 \rightarrow B _2H _5Br + HBr$
This is a slow reaction taking place at $100^oC$.
$B _2H _6 + 6Cl _2 \rightarrow 2BCl _3 + 6HCl$
This is a vigorous reaction taking place at $25^oC$. Diborane does not react with $Cl _2$ at ${100}^{o}C$. This reaction is not reported.

$B _{2} H _{6} + 6Cl _{2} \rightarrow 2BCl _{3} + 6HCl $ 

diborane has non-planar structure. It has 2 hydrogen atoms which are present above and below the plane of the boron and hydrogen bond. 

Clotting of blood is called coagulation and a substance promoting this is called coagulating agent.

The given statement is False.
All the  $\displaystyle Al-Cl$ bonds in $\displaystyle Al _2Cl _6$ are not equivalent.
Two types of bonds are present in $\displaystyle Al _2Cl _6$, terminal $\displaystyle Al-Cl$ bonds that are normal $\displaystyle 2c-2e$ bonds and $\displaystyle Al-Cl-Al$ bridges that are $\displaystyle 3c-4e$ bonds.

As there is no lone pair of electrons present on the boron in diborane it do not form a hydrogen bond.But in water inorganic benzene and methanol there is a lone pair of electrons on oxygen, nitrogen and oxygen respectively so they form hydrogen bonding between most electronegative atom like nitrogen oxygen and hydrogen.
Hence option $D$ is correct.

$B _2H _6$ is a electron deficient molecule so it acts like an acid(acid is the substance which accepts the pair of electrons). $CH _4,NH _3,H _2O$ allare having lone pairs on central atoms.
Hence option D is correct.

As the size of the boron is less it forms a more acidic oxide (the elements with less atomic size forms most acidic oxides)
So, $B _2O _3$ is most acidic.
Hence option $C$ is correct.

First iodine reacti with one $NaBH _4$ , we get borance , NaI and HI .

Boron shows ${sp}^{3}$ hybridisation in $B _2H _6$.

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.

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'.

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

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

(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.

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.

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".

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. 

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

$B _{2}H _{6}+2NH _{3}\rightarrow B _{2}H _{6}.2NH _{3}$
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.

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.

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.

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.

     $B _2H _6+2NMe _3\longrightarrow2BH _3.NMe _3$

$Al{ Cl } _{ 3 },B{ F } _{ 3 },Sn{ Cl } _{ 4 }$ can be used as catalyst in Friedal Craft reaction.