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Lesson Posted on 08 May Learn Unit XII: Aldehydes, Ketones and Carboxylic Acids

Chemistry, oxidation of ketones

Monika Katoch

Experienced chemistry teacher with 6+ years of experience teaching at the high school and Senior Secondary...

Oxidation of ketones is carried by strong oxidizing agent like concentrated HNO3, KmNO4/H2SO4, dil K2Cr 2O7/H2SO4 For eg CH3COCH2CH3 +Conc. HNO3= CH3COOH + CH3COOH CH3COCH2CH2CH3 + conc.HNO3 = CH3COOH + CH3CH2COOH. Breaking of C ---CO bond takes place according to —poppof's rule Which states... read more

Oxidation of ketones is carried by strong oxidizing agent like concentrated HNO3, KmNO4/H2SO4, dil K2Cr 2O7/H2SO4

For eg

CH3COCH2CH3 +Conc. HNO3= CH3COOH + CH3COOH

CH3COCH2CH2CH3 + conc.HNO3 = CH3COOH + CH3CH2COOH. 

Breaking of C ---CO bond takes place according to —poppof's rule

Which states that carbonyl group stays together with smaller alkyl group

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Answered on 13 Apr Learn Unit VII: p - Block Elements

Nisarg

Make Fitness Your Lifestyle With Nisarg Khatri

Sure! The molecule you're referring to is ozone (O3). It consists of three oxygen atoms bonded together. Here's a simple representation of its structure: mathematica O / \ O = O read more

Sure! The molecule you're referring to is ozone (O3). It consists of three oxygen atoms bonded together. Here's a simple representation of its structure:

mathematica
O / \ O = O
 
 
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Answered on 13 Apr Learn Unit VII: p - Block Elements

Nazia Khanum

Fluorine, being the most electronegative element on the periodic table, does not exhibit any positive oxidation state because it strongly attracts electrons. In chemical reactions, fluorine tends to gain electrons rather than lose them, leading to negative oxidation states. Its high electronegativity... read more

Fluorine, being the most electronegative element on the periodic table, does not exhibit any positive oxidation state because it strongly attracts electrons. In chemical reactions, fluorine tends to gain electrons rather than lose them, leading to negative oxidation states. Its high electronegativity makes it very difficult for fluorine to lose electrons and form positive oxidation states. Consequently, in most compounds, fluorine exists with an oxidation state of -1, where it gains an electron to complete its valence shell.

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Answered on 13 Apr Learn Unit VII: p - Block Elements

Nazia Khanum

Nitrogen is indeed more inert compared to phosphorus, primarily due to differences in their atomic structures and the stability of their compounds. Bond Strength: Nitrogen forms a very strong triple bond (N≡N) in molecular nitrogen (N2), which is difficult to break. This makes nitrogen gas... read more

Nitrogen is indeed more inert compared to phosphorus, primarily due to differences in their atomic structures and the stability of their compounds.

  1. Bond Strength: Nitrogen forms a very strong triple bond (N≡N) in molecular nitrogen (N2), which is difficult to break. This makes nitrogen gas quite unreactive under normal conditions. Phosphorus, on the other hand, tends to form weaker single bonds (P-P) in its elemental form (P4), making it more reactive.

  2. Electronegativity: Nitrogen has a higher electronegativity compared to phosphorus. This means that nitrogen atoms attract electrons more strongly, which stabilizes the molecules they form and makes them less prone to reacting with other substances.

  3. Size of Atom: Nitrogen atoms are smaller than phosphorus atoms, which affects their ability to form stable bonds. Nitrogen's smaller size allows for stronger overlap of atomic orbitals in the formation of multiple bonds, contributing to the stability of nitrogen compounds.

  4. Hybridization: Nitrogen often undergoes sp2 hybridization, leading to planar geometry in many of its compounds. This geometric arrangement can enhance the stability of nitrogen compounds. Phosphorus, however, can exhibit various hybridizations and geometries, which may render its compounds more reactive.

These factors collectively contribute to the relative inertness of nitrogen compared to phosphorus. However, despite nitrogen's inertness in its diatomic form, it can react vigorously under certain conditions to form a wide variety of compounds, especially when it reacts with highly reactive elements or under specific catalytic conditions.

 
 
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Answered on 13 Apr Learn Unit VIII: d and f Block Elements

Nazia Khanum

The increase in density from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements can be explained by several factors: Atomic Mass: As you move from titanium to copper in the periodic table, the atomic mass generally increases due to the addition of more protons, neutrons,... read more

The increase in density from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements can be explained by several factors:

  1. Atomic Mass: As you move from titanium to copper in the periodic table, the atomic mass generally increases due to the addition of more protons, neutrons, and electrons. Since density is mass per unit volume, an increase in atomic mass tends to increase density.

  2. Atomic Radius: While the atomic radius generally decreases across a period in the periodic table due to increasing effective nuclear charge, the increase in atomic mass across the transition metals offsets this effect to some extent. As you move from titanium to copper, the increase in atomic mass generally outweighs the decrease in atomic radius, contributing to the increase in density.

  3. Crystal Structure: Transition metals typically have a close-packed crystal structure, which means that their atoms are densely packed together in a regular pattern. Changes in atomic size and mass can influence how tightly packed these atoms are, affecting the density of the material.

  4. Electron Configuration: Transition metals have complex electron configurations, with electrons occupying different sublevels within the d-block. Changes in electron configuration can influence the interactions between atoms and hence affect the density of the material.

  5. Transition Metals' Special Properties: Transition metals often exhibit unique properties such as high melting points, hardness, and metallic bonding characteristics, all of which can influence the density of the elements in this series.

Overall, the increase in density from titanium to copper in the first series of transition elements is a result of various interplaying factors including atomic mass, atomic radius, crystal structure, electron configuration, and special properties of transition metals.

 
 
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Answered on 13 Apr Learn Unit VIII: d and f Block Elements

Nazia Khanum

(i) Transition elements generally form colored compounds: The color exhibited by transition metal compounds arises from the d-d transition, which involves the movement of electrons between the d orbitals of the metal ions. Transition metals have partially filled d orbitals, which allow for the absorption... read more

(i) Transition elements generally form colored compounds:

The color exhibited by transition metal compounds arises from the d-d transition, which involves the movement of electrons between the d orbitals of the metal ions. Transition metals have partially filled d orbitals, which allow for the absorption of visible light. When light strikes a transition metal complex, it can promote an electron from a lower-energy d orbital to a higher-energy d orbital, resulting in the absorption of certain wavelengths of light and the reflection or transmission of others. The color observed depends on the energy difference between the d orbitals involved in the transition.

The intensity and nature of the color can be influenced by various factors such as the oxidation state of the metal ion, the ligands surrounding the metal ion, and the coordination geometry of the complex. Ligands with different electron-donating abilities can lead to different splitting patterns of the d orbitals, resulting in different absorption spectra and hence different colors.

(ii) Zinc is not regarded as a transition element:

Zinc is often not considered a transition element because it lacks partially filled d orbitals in its common oxidation states. In its most common oxidation state, +2, the 3d orbitals are completely filled, which means there are no available d electrons for d-d transitions to occur. Therefore, zinc typically forms colorless compounds.

Transition metals, by definition, have incompletely filled d orbitals in at least one oxidation state, which allows them to exhibit characteristic transition metal properties such as forming colored compounds and acting as catalysts. Since zinc does not fulfill this criterion, it is often excluded from the list of transition elements despite being located in the d-block of the periodic table.

 
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Answered on 13 Apr Learn Unit VIII: d and f Block Elements

Nazia Khanum

(i) Copper (I) ion is not known in aqueous solution primarily because copper tends to exist in the +2 oxidation state in aqueous solutions. This is due to the relative stability of the Cu(II) oxidation state compared to Cu(I) in aqueous environments. The standard reduction potential for the Cu(II)/Cu(I)... read more

(i) Copper (I) ion is not known in aqueous solution primarily because copper tends to exist in the +2 oxidation state in aqueous solutions. This is due to the relative stability of the Cu(II) oxidation state compared to Cu(I) in aqueous environments. The standard reduction potential for the Cu(II)/Cu(I) couple is higher than that for many other metal ions, making the Cu(II) state more stable in water. Additionally, Cu(II) ions readily hydrolyze in water, forming insoluble Cu(OH)₂, further reducing the concentration of Cu(I) ions in solution.

(ii) Actinoids exhibit a greater range of oxidation states than lanthanoids due to the presence of f-orbitals in their electron configurations. Actinoid elements have more extended series of f-orbitals available for electron configuration, leading to a greater variety of possible oxidation states. The lanthanoid series, on the other hand, have electrons filling 4f orbitals, which are relatively shielded from the outer environment by the 5s and 5p orbitals. As a result, lanthanoid elements generally exhibit fewer accessible oxidation states compared to actinoids. Additionally, the actinoid series is longer than the lanthanoid series, providing more elements with a greater variety of electron configurations and oxidation states.

 
 
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Answered on 13 Apr Learn Unit IX: Coordination Compounds

Nazia Khanum

When undecomposed silver bromide (AgBr) is washed with hypo solution (sodium thiosulfate) in photography, it forms a complex ion known as the tetrathionate complex, 3-. This complex ion helps in removing the unexposed silver bromide from the photographic film during the fixing process, leaving behind... read more

When undecomposed silver bromide (AgBr) is washed with hypo solution (sodium thiosulfate) in photography, it forms a complex ion known as the tetrathionate complex, [Ag(S2O3)2]3-. This complex ion helps in removing the unexposed silver bromide from the photographic film during the fixing process, leaving behind the developed silver image.

 
 
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Answered on 13 Apr Learn Unit IX: Coordination Compounds

Nazia Khanum

The ionization isomer of Cl is formed when one of the ligands is replaced by the counterion. So, in this case, one of the NH3 ligands will be replaced by Cl. The IUPAC name of the original compound is tris(ammine)nitronickel(II) chloride. Now, replacing one NH3 ligand with Cl, the IUPAC name would... read more

The ionization isomer of [Ni(NH3)3NO3]Cl is formed when one of the ligands is replaced by the counterion. So, in this case, one of the NH3 ligands will be replaced by Cl.

The IUPAC name of the original compound is tris(ammine)nitronickel(II) chloride.

Now, replacing one NH3 ligand with Cl, the IUPAC name would be: tris(ammine)chloronitronickel(II) nitrate.

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Answered on 13 Apr Learn Unit IX: Coordination Compounds

Nazia Khanum

Sure, here are two examples of ligands commonly used in coordination compounds for analytical chemistry: Ethylenediamine (en): Ethylenediamine is a bidentate ligand, meaning it can coordinate to a central metal ion through two of its nitrogen atoms. This ligand forms stable complexes with many metal... read more

Sure, here are two examples of ligands commonly used in coordination compounds for analytical chemistry:

  1. Ethylenediamine (en): Ethylenediamine is a bidentate ligand, meaning it can coordinate to a central metal ion through two of its nitrogen atoms. This ligand forms stable complexes with many metal ions, such as copper, cobalt, and nickel. In analytical chemistry, ethylenediamine complexes are often used in qualitative and quantitative analysis of metal ions in solution, including complexometric titrations.

  2. 1,10-Phenanthroline: 1,10-Phenanthroline is a heterocyclic aromatic compound that acts as a chelating ligand. It forms stable complexes with various metal ions, including iron, copper, and zinc. These complexes are often intensely colored, making them useful for colorimetric determination of metal ions in solution. 1,10-Phenanthroline complexes are widely used in analytical chemistry for applications such as spectrophotometric analysis and metal ion detection.

 
 
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