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Answered 5 days ago Learn Unit 11-Some p -Block Elements

Nazia Khanum

The variation in oxidation states across the elements of a group or period often reflects underlying trends in electronic configurations and chemical reactivity. Let's explore the pattern of oxidation states for the elements: (i) Boron (B) to Thallium (Tl): Boron (B): Boron typically exhibits an... read more

The variation in oxidation states across the elements of a group or period often reflects underlying trends in electronic configurations and chemical reactivity. Let's explore the pattern of oxidation states for the elements:

(i) Boron (B) to Thallium (Tl):

  • Boron (B): Boron typically exhibits an oxidation state of +3 in its compounds, as it prefers to lose its three valence electrons to achieve a stable electronic configuration.
  • Aluminum (Al): Aluminum predominantly shows an oxidation state of +3 in its compounds, similar to boron, due to the loss of its three valence electrons.
  • Gallium (Ga): Gallium mainly displays an oxidation state of +3, following the trend set by boron and aluminum.
  • Indium (In): Indium is more versatile in its oxidation states compared to the preceding elements. It can exhibit both +1 and +3 oxidation states, with the +3 oxidation state being more common.
  • Thallium (Tl): Thallium, like indium, shows a wider range of oxidation states. It can exhibit +1, +3, and even +5 oxidation states. However, the +1 oxidation state is more stable and common than the +3 or +5 states.

The trend across this group shows a general progression from +3 oxidation state (B, Al, Ga) to a more varied set of oxidation states as we move down the group, with elements like indium and thallium displaying a wider range of oxidation states due to the increasing ease of losing electrons as we move down the group.

(ii) Carbon (C) to Lead (Pb):

  • Carbon (C): Carbon typically exhibits oxidation states of -4 (in compounds like methane) to +4 (in compounds like carbon dioxide). However, its most common oxidation states are +4 in carbon dioxide and -4 in methane.
  • Silicon (Si): Silicon primarily shows oxidation states of +4 (in compounds like silicon dioxide) and sometimes +2 (in compounds like silane, although less common).
  • Germanium (Ge): Germanium mainly exhibits an oxidation state of +4, following the trend set by carbon and silicon.
  • Tin (Sn): Tin is more versatile in its oxidation states compared to the preceding elements. It can exhibit oxidation states of +2 and +4, with the +2 oxidation state being more common.
  • Lead (Pb): Lead, like tin, can exhibit multiple oxidation states. It commonly shows oxidation states of +2 and +4, with the +2 oxidation state being more stable and common.

The trend across this group shows a general progression from a wider range of oxidation states (-4 to +4) for carbon to a narrower range of oxidation states (+2 to +4) for lead. This narrowing occurs due to the increasing size and decreasing electronegativity of the elements as we move down the group, making it less favorable for higher oxidation states to be stabilized.

 
 
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Answered 5 days ago 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 5 days ago 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 5 days ago 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 13/12/2018 Learn Unit 11-Some p -Block Elements +1 Chemistry

Snehal Apturkar

Teacher

Anhydrides of inorganic acids are mostly oxides of organometallic elements hence carbon dioxide is the anhydride of carbonic acid.
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Answered on 14/12/2018 Learn Unit 11-Some p -Block Elements +1 Chemistry

Ramakrishna Dhooli

An Excellent Tutor for Mths and Science with Kannada,Hindi and english

CARBON
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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

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C....solid co2
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