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swat32
2 years ago
4

Use VSEPR theory to draw structures, with ideal bond angles, for boric acid and the anion it forms when it reacts with water.

Chemistry
1 answer:
Mrrafil [7]2 years ago
7 0

According to VSEPR theory, the molecular geometry of boric acid is found to be trigonal planar.

<h3>What is Boric acid? </h3>

Boric Acid, which is also known as orthoboric acid, hydrogen borate, boracic acid, is basically a weak acid. It is an acidic hydrate of boric oxide having antifungal, mild, antiseptic and antiviral properties.

<h3>Structure of Boric acid</h3>

It has two-dimensional structure, that lies on a plane. According to VSEPR theory, the molecular geometry of boric acid is found to be trigonal planar.

The oxygen atom in both the compounds are sp3 hybridised and ideal bond angle of boric acid is 109°28′.

<h3>Prepartion of boric acid</h3>
  • Boric acid is prepared by the reaction borax with the mineral acid,

Na2B4O7·10H2O + 2HCl → 4 B(OH)3 + 2 NaCl + 5 H2O

  • It can be formed as a by-product of the hydrolysis of boron trihalides and diborane. The reaction is as below:

B2H6 + 6H2O→ 2B(OH)3 + 6H2

BX3 + 3H2O → B(OH)3 + 3HX (X = Cl, Br, I)

<h3>Uses of boric acid</h3>
  • It is mainly used in the manufacturing of monofilament fibre glass.
  • It is also used in the combination along with denatured alcohol in the jewellery industries.
  • It is typically used in electroplating.
  • It is also used in manufacturing of flat LCD panel displays.

Thus, we concluded that the molecular geometry of boric acid is found to be trigonal planar.

learn more about Boric acid:

brainly.com/question/11626749

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To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

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For oxygen gas:

\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol

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By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

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Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of Al_2O_3

So, 0.092 moles of aluminium will produce = \frac{2}{4}\times 0.092=0.046moles of Al_2O_3

Now, calculating the mass of aluminium oxide:

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\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield  = 3.5 g

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Putting values in above equation, we get:

\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%

Hence, the percent yield of the reaction is 74 %

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