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Answered on 13 Apr Learn Unit 11-Some p -Block Elements

Nazia Khanum

Boric acid is technically a weak Lewis acid rather than a protonic acid. Let me break it down: Protonic Acid: Protonic acids, also known as Bronsted acids, are substances that can donate a proton (H⁺ ion) to another substance. In simpler terms, they are acids that readily release hydrogen ions in... read more

Boric acid is technically a weak Lewis acid rather than a protonic acid. Let me break it down:

  1. Protonic Acid: Protonic acids, also known as Bronsted acids, are substances that can donate a proton (H⁺ ion) to another substance. In simpler terms, they are acids that readily release hydrogen ions in solution. Examples include hydrochloric acid (HCl) and sulfuric acid (H₂SO₄).

  2. Lewis Acid: In contrast, Lewis acids are substances that can accept a pair of electrons. They don't necessarily need to donate protons; instead, they can form a coordinate covalent bond by accepting an electron pair from another substance. Boric acid falls into this category.

Boric acid, chemically represented as H₃BO₃ or B(OH)₃, can act as a Lewis acid by accepting a pair of electrons from a Lewis base. Its behavior as an acid is due to the ability of the boron atom to accept an electron pair from a base, forming a coordinate covalent bond. This property allows it to react with substances like alcohols or water to form borate ions.

While boric acid can behave as an acid in certain reactions, it's not as strong as typical protonic acids like hydrochloric acid. Its acidic properties are more subtle and primarily manifest in reactions where it acts as a Lewis acid.

 
 
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Answered on 13 Apr Learn Unit 11-Some p -Block Elements

Nazia Khanum

When boric acid (H3BO3) is heated, it undergoes several chemical changes. Initially, boric acid dehydrates upon heating, losing water molecules to form metaboric acid (HBO2): 2H3BO3 (boric acid) -> H2B4O7 (metaboric acid) + H2O Further heating leads to the conversion of metaboric acid into various... read more

When boric acid (H3BO3) is heated, it undergoes several chemical changes. Initially, boric acid dehydrates upon heating, losing water molecules to form metaboric acid (HBO2):

2H3BO3 (boric acid) -> H2B4O7 (metaboric acid) + H2O

Further heating leads to the conversion of metaboric acid into various polymeric forms of boric anhydride or tetraboric acid (H2B4O7):

4H2B4O7 (metaboric acid) -> 2B2O3 (boric anhydride) + 5H2O

The boron oxide formed can further polymerize to form complex boron oxide networks at higher temperatures.

The exact products and reactions may vary depending on the specific conditions of heating, such as temperature and presence of other substances. Boric acid's thermal decomposition is utilized in various industrial processes, including the production of boron-containing compounds and ceramics.

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Answered on 13 Apr Learn Unit 11-Some p -Block Elements

Nazia Khanum

BF3, or boron trifluoride, has a trigonal planar shape. Each fluorine atom forms a single bond with the central boron atom, resulting in three bonding pairs of electrons around boron. Since there are no lone pairs on boron, the shape is trigonal planar. BH4−, or tetrahydroborate ion, has a tetrahedral... read more

BF3, or boron trifluoride, has a trigonal planar shape. Each fluorine atom forms a single bond with the central boron atom, resulting in three bonding pairs of electrons around boron. Since there are no lone pairs on boron, the shape is trigonal planar.

BH4−, or tetrahydroborate ion, has a tetrahedral shape. Boron in BH4− has four hydrogen atoms bonded to it, resulting in four bonding pairs of electrons around boron. There are no lone pairs on boron, so the shape is tetrahedral.

In both cases, the hybridization of boron can be determined by counting the number of regions of electron density (bonding pairs and lone pairs) around the boron atom.

For BF3:

  • There are 3 regions of electron density (3 bonding pairs), indicating sp2 hybridization.

For BH4−:

  • There are 4 regions of electron density (4 bonding pairs), indicating sp3 hybridization.
 
 
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Answered on 13 Apr Learn Unit 11-Some p -Block Elements

Nazia Khanum

"Aluminum's amphoteric nature is evident when it reacts with both acids and bases. When it encounters an acidic environment, such as hydrochloric acid, it forms aluminum chloride and hydrogen gas. On the other hand, in a basic solution like sodium hydroxide, aluminum reacts to form sodium aluminate... read more
  1. "Aluminum's amphoteric nature is evident when it reacts with both acids and bases. When it encounters an acidic environment, such as hydrochloric acid, it forms aluminum chloride and hydrogen gas. On the other hand, in a basic solution like sodium hydroxide, aluminum reacts to form sodium aluminate and hydrogen gas. This ability to react with both acidic and basic substances showcases its amphoteric behavior."

  2. "One clear demonstration of aluminum's amphoteric nature is its reaction with both strong acids and strong bases. When aluminum reacts with an acid like sulfuric acid, it produces aluminum sulfate and hydrogen gas. Conversely, when it interacts with a strong base like potassium hydroxide, it yields potassium aluminate and hydrogen gas. This dual reactivity illustrates its characteristic amphoteric behavior."

  3. "Aluminum exhibits its amphoteric properties through its reactions with both acidic and basic solutions. For instance, when it reacts with an acid such as nitric acid, it forms aluminum nitrate and releases hydrogen gas. Similarly, when it comes into contact with a base like calcium hydroxide, it produces calcium aluminate and liberates hydrogen gas. These reactions clearly demonstrate aluminum's ability to behave as both an acid and a base."

  4. "The amphoteric nature of aluminum becomes apparent in its reactions with acids and bases. When treated with an acid like hydrochloric acid, aluminum undergoes a reaction to form aluminum chloride and hydrogen gas. In contrast, when exposed to a base such as sodium hydroxide, aluminum reacts to produce sodium aluminate and hydrogen gas. These reactions underscore aluminum's dual propensity to interact with both acidic and basic environments, thus exhibiting its amphoteric character."

 
 
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Answered on 13 Apr Learn Unit 11-Some p -Block Elements

Nazia Khanum

Electron deficient compounds are molecules that possess fewer electrons than what is typically expected based on the octet rule or the duet rule in the case of hydrogen. These compounds often exhibit incomplete valence electron shells and tend to be highly reactive, seeking to gain electrons to achieve... read more

Electron deficient compounds are molecules that possess fewer electrons than what is typically expected based on the octet rule or the duet rule in the case of hydrogen. These compounds often exhibit incomplete valence electron shells and tend to be highly reactive, seeking to gain electrons to achieve a more stable electronic configuration.

BCl3 (boron trichloride) and SiCl4 (silicon tetrachloride) are indeed examples of electron deficient species. Let's examine each:

  1. BCl3 (boron trichloride):

    • Boron has three valence electrons, and in BCl3, it forms three covalent bonds with chlorine atoms.
    • However, boron itself only has six electrons around it, leaving it short of the octet rule. Thus, BCl3 is electron deficient.
    • Due to its electron deficiency, BCl3 is highly reactive and acts as a Lewis acid, readily accepting a pair of electrons from a Lewis base.
  2. SiCl4 (silicon tetrachloride):

    • Silicon has four valence electrons, and in SiCl4, it forms four covalent bonds with chlorine atoms.
    • Similar to boron, silicon ends up with only eight electrons around it, which is still short of a complete octet.
    • Therefore, SiCl4 is also electron deficient.
    • Silicon tetrachloride is also reactive due to its electron deficiency, though it's not as reactive as boron trichloride.

In both cases, the central atom (boron or silicon) lacks a full complement of valence electrons, making these molecules electron deficient. This electron deficiency makes them susceptible to reacting with species that can donate electron pairs, making them important reagents in various chemical processes.

 
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Answered on 13 Apr Learn Unit 11-Some p -Block Elements

Nazia Khanum

Sure, I'd be happy to help you with that! To draw resonance structures for CO₃²⁻ (carbonate ion) and HCO₃⁻ (bicarbonate ion), let's start with CO₃²⁻: Step 1: Draw the skeletal structure of CO₃²⁻ with single bonds between each oxygen atom and the central carbon atom, and a double bond... read more

Sure, I'd be happy to help you with that!

To draw resonance structures for CO₃²⁻ (carbonate ion) and HCO₃⁻ (bicarbonate ion), let's start with CO₃²⁻:

  1. Step 1: Draw the skeletal structure of CO₃²⁻ with single bonds between each oxygen atom and the central carbon atom, and a double bond between one oxygen and the carbon:
mathematica
O | O = C = O | O
  1. Step 2: Identify the lone pair on one of the terminal oxygen atoms. Move this lone pair to form a double bond between the terminal oxygen and the central carbon atom:
mathematica
O || O = C = O | O
  1. Step 3: Now, the central carbon atom has a formal charge of +1. To neutralize this charge, move one of the lone pairs from the carbon-oxygen double bond to form a new bond between the carbon and oxygen:
mathematica
O || O = C = O⁻ | O
  1. Step 4: Finally, the adjacent oxygen now carries a formal charge of -1. To stabilize this, move one lone pair from this oxygen to form a double bond with the carbon:
mathematica
O || O = C = O⁻ || O

This completes the resonance structure for the carbonate ion (CO₃²⁻).

Now, let's move on to HCO₃⁻ (bicarbonate ion):

  1. Step 1: Draw the skeletal structure of HCO₃⁻ with a single bond between each oxygen atom and the central carbon atom, and a double bond between one oxygen and the carbon:
mathematica
H | O = C = O | O
  1. Step 2: Similar to before, identify the lone pair on one of the terminal oxygen atoms. Move this lone pair to form a double bond between the terminal oxygen and the central carbon atom:
mathematica
H || O = C = O | O
  1. Step 3: Now, the central carbon atom has a formal charge of +1. To neutralize this charge, move one of the lone pairs from the carbon-oxygen double bond to form a new bond between the carbon and oxygen:
mathematica
H || O = C - O⁻ | O
  1. Step 4: In the final step, the adjacent oxygen now carries a formal charge of -1. To stabilize this, move one lone pair from this oxygen to form a double bond with the carbon:
mathematica
H || O = C = O⁻ || O

This completes the resonance structure for the bicarbonate ion (HCO₃⁻).

Keep in mind that in both cases, the actual structure of the ion is a hybrid of these resonance structures, with the true structure being an average of the contributing resonance forms.

 
 
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Answered on 11/12/2018 Learn Unit 11-Some p -Block Elements +1 Chemistry

Dhananjay Kumar Sharma

Chemistry Tutor

Ga is present in 3rd period after the d-block in the periodic table, In this due to poor shielding effect of d electron effective nuclear charge increases that pull outershell electron more strongly hence has smaller size.
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Answered on 02/01/2019 Learn Unit 11-Some p -Block Elements +1 Chemistry

Bansal Classes Private Limited

Protonic acid is an acid which ionizes in solution and produces protons. But in the case of boric acid, it takes electrons which were produced due to the ionisation of water.
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Answered on 23/10/2018 Learn Unit 11-Some p -Block Elements +1 Chemistry

Bijendra Kumar Sharma

Teacher

C....solid co2
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