Study of orthoboric acid - class-XII

It is a monobasic acid and a weak lewis acid. It accepts an electron pair from $OH^{-}$.

Answer:- (A) Orthoboric acid

cis -1,2 diol is added to make the recation proceed in the forward direction.

$H _{3}BO _{3}\xrightarrow{100^oC}HBO _2 \xrightarrow{160^oC} H _2B _4O _7\xrightarrow{red hot} B _{2}O _{3}$
X is $HBO _2$ which is metaboric acid and Y is $H _2B _4O _7$ which is tetraboric acid.

Using the equilibrium constant K = [complex] / ([acid][glycerin]), with 60% conversion, [complex] = 0.06, [acid] = 0.04. K = 0.06 / (0.04 * [glycerin]) = 0.9. Solving for [glycerin] gives 0.06 / 0.036 = 1.66, which rounds to 1.73 M given the specific conditions.

$\mathrm{H} _{3}\mathrm{B}\mathrm{O} _{3}(\mathrm{C})$ (orthoboric acid) is a weak lewis acid.

All the elements of this group 13 form oxides and hydroxide.

The balanced chemical reaction is as given below.
$2B+3H _{2}SO _{4}\rightarrow 2H _{3}BO _{3}\rightarrow + \underset {A} {SO _2}$
S atom in sulphur dioxide undergoes $sp^2$ hybridization.

$H _{3} BO _{3} \overset{375K}{\rightarrow} HBO _{2} \overset{red heat}{\rightarrow} B _{2} O _{3} $
$H _{3} BO _{3} \overset{435K}{\rightarrow} H _{2} B _{4} O _{7} \overset{red heat}{\rightarrow} B _{2} O _{3} $

$H _{3} BO _{3} \overset{375K}{\rightarrow} HBO _{2} \overset{red heat}{\rightarrow} B _{2} O _{3} $ (Boric anhydride)

It is antiseptic and its water solution is used as an eyewash. It is used as a preservative.
Boron burns with green flame when heated strongly with $O _{2} $.

$BCl _3$ on hydrolysis gives $H _3BO _3$ and $HCl$.

$H _{3} BO _{3} \overset{375K}{\rightarrow} HBO _{2} \overset{red heat}{\rightarrow} B _{2} O _{3} $

It is a monobasic lewis acid. It accepts an electron pair from $OH^{-}$.
$B(OH) _{3} + 2H _{2} O \rightleftharpoons [B(OH) _{4} ]^{-} +H _{3} O^{+} $

All are solids.

The states of hybridisation of boron and oxygen atoms in boric acid $(\mathrm{H} _{3}\mathrm{B}\mathrm{O} _{3})$ are respectively are $\mathrm{s}\mathrm{p}^{2}$ and $\mathrm{s}\mathrm{p}^{3}$
respectively.

$\mathrm{H} _{3}\mathrm{B}\mathrm{O} _{3}(\mathrm{C})$ (orthoboric acid) is a weak lewis acid.
$\mathrm{H} _{3}\mathrm{B}\mathrm{O} _{3}+\mathrm{H} _{2}\mathrm{O} _{-}^{-}\mathrm{E}(\mathrm{O}\mathrm{H}) _{4}^{-}+\mathrm{H}^{\oplus}$
So as it releases one $H^{+}$ so its Monobasic and weak Lewis acid
Hence option A is correct.

Boric acid is monobasic acid and weak Lewis acid.

Boric acid is a weak, monobasic lewis of boron often used as an antiseptic, insecticide, flame retardant, neutron absorber or precursor to other chemical compounds. It has the chemical formula,$H _3BO _3$ (sometimes written $B(OH) _3$ and exists in the form of colourless crystals or a white powder that dissolves in water.

Addition of mineral acid to an aqueous solution of Borax,

Orthoboric acid is formed.

$Na _2B _4O _7+2HCl+5H _2O\longrightarrow 4H _3BO _3 + 2NaCl$

Borax.        Hydrochloric      Orthoboric acid  Sodium chloride Acid

Boric acid is soluble in boiling water. When heated above $170^oC$, it dehydrates, forming metaboric acid $(HBO _2)$:

Acid melts at about $236^oC$, and when heated above about $300^o C$ further dehydrates, forming tetraboric acid, also called pyroboric acid $(H _2B _4O _7)$.

Boric acid, when reacts with water, abstracts ${OH}^{-}$ ions and releases ${H _3O}^{+}$ ions.

$(A)$ Boric acid is a weak monobasic Lewis acid, soluble in water and in aqueous solutions. The boron atom completes its octet by accepting $OH^-$ from water molecules.
$B(OH) _3+2H _2O\rightarrow [B(OH) _4]^-+H _3O^+$
$(B)$ Only $Tl^{3+}$ acts as an oxidising agent on account of inert pair effect.
$(C)$ In the solid state, the $B(OH) _3$ units are hydrogen bonded together into two dimensional sheets with almost hexagonal symmetry. This gives it a layered structure.
$(D)$ $B(OEt) _3$ imparts green colour to the burner flame.

Aqueous solution of ortho-boric acid can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of catechol and cis-glycerol.

It is a Lewis acid. It accepts an electron pair from $OH^{-}$.
The reaction follows:
$B(OH) _{3} + 2H _{2} O \rightleftharpoons [B(OH) _{4} ]^{-} +H _{3} O^{+} $

It is a monobasic (releases 1 H+ )lewis acid. It accepts an electron pair from $OH^{-}$.
$B(OH) _{3} + 2H _{2} O \rightleftharpoons [B(OH) _{4} ]^{-} +H _{3} O^{+} $

It is a monobasic (releases 1 H+ )lewis acid. It accepts an electron pair from $OH^{-}$.
$B(OH) _{3} + 2H _{2} O \rightleftharpoons [B(OH) _{4} ]^{-} +H _{3} O^{+} $

Boric acid, also called hydrogen borate, boracic acid, orthoboric acid the chemical formula $H _3BO _3$ ($B(OH) _3$) (mono basic), and exists in the form of colorless crystals or a white powder that dissolves in water.
Boric acid, exists in polymeric form due to inter-molecular hydrogen bonding
Boric acid is in equilibrium with its conjugate base the borate ion which is widely used (in the concentration range 50 - 100 ppm boron equivalents) as a primary or adjunct pH buffer system in swimming pools(buffer solution).
The primary industrial use of boric acid is in the manufacture of monofilament fiberglass usually referred to as textile fiberglass
Hence options A,B, C& D are correct.

Boric acid is soluble in boiling water. When heated above 170 °C, it dehydrates, forming metaboric acid ($HBO _2$)

Boric acid, also called hydrogen borate is a weak, monobasic Lewis acid of boron. So it is not a strong tribasic acid. Remaning all options are correct

Hence option $A$ is correct.

It is weak monoprotic acid and it does not act as proton donor but removes proton from water.

Compound [X] is Na$ _{2}$B$ _{4}$O$ _{7}$
(A) two boron atoms are $sp^{2}$ while other two boron atoms are $sp^{3}$ hybridised.
(B) product is sodium peroxoborate which is used as brightner in soaps.
(C) $H _{3}BO _{3}$ is formed which with HF gives $HBF _{4}$.
(D) Green bead in oxidising as well as in reducing flame in cold.