Write short notes on any two. [5×2=10] i) Detection of nitrogen in the organic compound. ii) Advantages of modern periodic table. iii) Characteristics of homologous series. iv) Relationship between…

i) Detection of nitrogen in the organic compound

Principle: Nitrogen in organic compounds is detected using Lassaigne’s test, a qualitative analysis method. The test involves fusing the organic compound with sodium metal to convert nitrogen into sodium cyanide (NaCN), which is then converted into sodium thiocyanate (NaSCN). The thiocyanate ion reacts with ferric chloride (FeCl₃) to form a blood-red or deep red precipitate of ferric thiocyanate (Fe(SCN)₃).

Organic compound (CₓHᵧNᵣ)+NaFusion with Na metalNaCN+Na₂S+Na₃N

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Procedure:

1. Fusion with Sodium:

   • Take a small amount of the organic compound in a fusion tube (or a dry test tube).
   • Add a small piece of sodium metal (about 0.5 g).
   • Heat the tube strongly in a Bunsen flame until the sodium melts and the mixture becomes red-hot. This converts nitrogen into NaCN (if nitrogen is present). $$\mathrm{Na}+\mathrm{N}+\mathrm{C}\to \mathrm{NaCN}$$

2. Extraction of Cyanide:

   • Allow the fused mass to cool.
   • Dissolve the residue in distilled water and filter to obtain a clear solution.
   • Acidify the filtrate with dilute sulfuric acid (H₂SO₄) to convert NaCN into hydrogen cyanide (HCN) gas and then into sodium thiocyanate (NaSCN) by adding ferrous sulfate (FeSO₄) and boiling. $$\mathrm{NaCN}+\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{NaHSO}{}_4+\mathrm{HCN}$$ $$\mathrm{HCN}+\mathrm{FeSO}{}_4\to \mathrm{Fe}(\mathrm{CN}){}_2+\mathrm{H}{}_2\mathrm{SO}{}_4$$ $$\mathrm{Fe}(\mathrm{CN}){}_2+6\mathrm{NaCN}\to \mathrm{Na}{}_4[\mathrm{Fe}(\mathrm{CN}){}_6]\text{(Sodium ferrocyanide)}$$ $$\mathrm{Na}{}_4[\mathrm{Fe}(\mathrm{CN}){}_6]+4\mathrm{H}{}_2\mathrm{SO}{}_4\to 2\mathrm{FeSO}{}_4+4\mathrm{NaHSO}{}_4+6\mathrm{HCN}$$ $$\mathrm{HCN}+\mathrm{NaOH}\to \mathrm{NaCN}+\mathrm{H}{}_2\mathrm{O}\text{(Reaction with NaOH to regenerate NaCN)}$$ $$\mathrm{NaCN}+\mathrm{S}\to \mathrm{NaSCN}\text{(Sodium thiocyanate)}$$

NaCN+H₂SO₄HeatNaHSO₄

3. Detection of Thiocyanate:
   • Add a few drops of ferric chloride (FeCl₃) solution to the solution obtained.
   • A blood-red or deep red coloration confirms the presence of nitrogen due to the formation of ferric thiocyanate (Fe(SCN)₃). $$\mathrm{FeCl}{}_3+3\mathrm{NaSCN}\to \mathrm{Fe}(\mathrm{SCN}){}_3+3\mathrm{NaCl}$$

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Modifications for Different Forms of Nitrogen:

• If the organic compound contains nitro groups (–NO₂), the cyanide test may not work directly. In such cases, the compound is first reduced with zinc dust and sodium hydroxide (NaOH) to convert nitro groups into amine (–NH₂), which then gives a positive Lassaigne’s test. $$\mathrm{R}-\mathrm{NO}{}_2+6[\mathrm{H}]\to \mathrm{R}-\mathrm{NH}{}_2+2\mathrm{H}{}_2\mathrm{O}$$

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Precautions:

1. Use distilled water to avoid contamination with nitrates or nitrites.
2. Ensure the fusion is complete (red-hot) to convert all nitrogen into NaCN.
3. Avoid using copper wire in the test (as it can interfere with the reaction).
4. The test is not applicable for compounds containing nitrogen in the form of azo (–N=N–) or nitroso (–NO) groups unless reduced first.

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Limitations:

• Does not distinguish between primary, secondary, or tertiary amines.
• Aromatic amines (e.g., aniline) may give a blue or green color due to the formation of ferric ferrocyanide instead of the red precipitate.
• Quaternary ammonium salts do not give a positive test unless hydrolyzed.

anilineanilineFeCl₃ferric chloride[Fe(CN)₆]^4-ferrocyanide

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ii) Advantages of the Modern Periodic Table

The Modern Periodic Table (also called the Long Form of the Periodic Table) was proposed by Dmitri Mendeleev and later refined by Henry Moseley based on atomic number (Z). It organizes elements into 18 vertical columns (groups) and 7 horizontal rows (periods). Below are its key advantages:

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1. Arrangement Based on Atomic Number (Z):

• Elements are arranged in increasing order of atomic number, not atomic mass (unlike Mendeleev’s original table).
• This eliminates anomalies in Mendeleev’s table (e.g., Ar (Z=18) and K (Z=19) were placed correctly).
• Henry Moseley established that atomic number (Z) determines chemical properties, not atomic mass.

2. Grouping of Elements with Similar Properties:

• Elements in the same group (vertical column) have similar chemical properties due to the same valence electron configuration.
  • Example: Group 1 (Alkali metals: Li, Na, K, Rb, Cs) all react vigorously with water to form hydroxides and hydrogen gas.
  • Group 17 (Halogens: F, Cl, Br, I) all form salts with metals and exist as diatomic molecules (X₂).
• This makes prediction of properties easier (e.g., Fr (Francium) was predicted to be similar to Cs (Cesium)).

3. Periodicity of Properties:

• Properties like atomic radius, ionization energy, electron affinity, and electronegativity vary periodically across periods and groups.
  • Atomic radius decreases left to right in a period (due to increasing nuclear charge).
  • Ionization energy increases left to right (except for Group 13 and 16 due to half-filled and fully-filled stability).
• This helps in understanding trends in chemical behavior.

4. Prediction of New Elements:

• Mendeleev left gaps in his table for undiscovered elements and predicted their properties.
• The Modern Periodic Table confirmed his predictions when elements like Gallium (Ga), Germanium (Ge), and Scandium (Sc) were discovered.
• Example:
  • Mendeleev predicted Eka-Aluminum (Ga) with properties matching Aluminum (Al).
  • Actual properties of Ga (Z=31) matched predictions:
    • Atomic mass: Predicted 68, Actual 69.7.
    • Density: Predicted 5.9 g/cm³, Actual 5.91 g/cm³.

5. Classification of Elements into Blocks:

• Elements are classified into s, p, d, and f blocks based on the last electron-filled orbital:
  • s-block: Groups 1-2 (Alkali and Alkaline Earth metals).
  • p-block: Groups 13-18 (includes metals, metalloids, and nonmetals).
  • d-block: Transition metals (Groups 3-12).
  • f-block: Lanthanides and Actinides (inner transition metals).
• This classification helps in understanding bonding, magnetism, and color of compounds.

6. Explanation of Chemical Bonding and Valency:

• The group number indicates the number of valence electrons, which determines valency.
  • Example: Group 1 (Li, Na, K) has 1 valence electron → valency = +1.
  • Group 17 (F, Cl, Br) has 7 valence electrons → valency = –1.
• Helps in writing formulas (e.g., NaCl, MgO, Al₂O₃).

7. Inclusion of All Known Elements:

• The Modern Periodic Table includes all 118 known elements (as of 2023), including synthetic elements (e.g., Tennessine (Ts, Z=117)).
• Elements are placed in correct positions based on their electron configuration, not just properties.

8. Explanation of Lanthanides and Actinides:

• The f-block accommodates Lanthanides (Ce-Lu, Z=58-71) and Actinides (Th-Lr, Z=90-103), which were not properly placed in Mendeleev’s table.
• These elements have similar properties within their series (e.g., radioactivity in Actinides).

9. Simplification of Chemical Reactions:

• The table helps in predicting reactivity series, oxidation states, and redox reactions.
  • Example: Group 1 metals (Li, Na, K) are highly reactive with water.
  • Group 18 (Noble gases) are inert due to fully-filled valence shells.

10. Foundation for Advanced Chemistry:

• Explains concepts like:
  • Metallic character increases down a group (e.g., Li < Na < K < Rb < Cs).
  • Non-metallic character increases left to right in a period (e.g., C < N < O < F).
  • Diagonal relationships (e.g., Li and Mg, Be and Al) due to similar charge/radius ratios.

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iii) Characteristics of Homologous Series

A homologous series is a family of organic compounds that share:

1. Same general formula (molecular formula).
2. Similar chemical properties.
3. Gradual change in physical properties.
4. Successive members differ by a –CH₂– group.

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1. Same General Formula:

• Compounds in a homologous series follow a common molecular formula.
• Example:
  • Alkanes: $\mathrm{CnH}{}_{2n+2}$ (e.g., Methane $\mathrm{CH}{}_4$, Ethane $\mathrm{C}{}_2\mathrm{H}{}_6$, Propane $\mathrm{C}{}_3\mathrm{H}{}_8$).
  • Alkenes: $\mathrm{CnH}{}_{2n}$ (e.g., Ethene $\mathrm{C}{}_2\mathrm{H}{}_4$, Propene $\mathrm{C}{}_3\mathrm{H}{}_6$).
  • Alcohols: $\mathrm{CnH}{}_{2n+1}\mathrm{OH}$ (e.g., Methanol $\mathrm{CH}{}_3\mathrm{OH}$, Ethanol $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}$).

CH₄ (n=1)CH₄ (n=1)C₂H₆ (n=2)C₂H₆ (n=2)C₃H₈ (n=3)C₃H₈ (n=3)

2. Similar Chemical Properties:

• All members undergo similar chemical reactions due to the same functional group.
• Example:
  • Alkanes undergo combustion and substitution reactions.
  • Alkenes undergo addition reactions (e.g., with Br₂, H₂).
  • Alcohols undergo oxidation (to aldehydes/ketones/carboxylic acids) and esterification.

3. Gradual Change in Physical Properties:

• Physical properties like melting point, boiling point, density, and solubility change gradually with increasing molecular mass.

• Example (Alkanes):

  Compound	Boiling Point (°C)	State at Room Temp.
  Methane	–161.5	Gas
  Ethane	–88.6	Gas
  Propane	–42.1	Gas
  Butane	–0.5	Gas
  Pentane	36.1	Liquid
  Hexane	68.7	Liquid

• Trend:

  • Boiling point increases with increasing molecular mass (due to increased van der Waals forces).
  • Solubility in water decreases (non-polar character increases).

4. Successive Members Differ by –CH₂– Group:

• Each successive member has one more –CH₂– group than the previous one.
• Example (Alcohols):
  • Methanol: $\mathrm{CH}{}_3\mathrm{OH}$
  • Ethanol: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}$ (Methanol + $\mathrm{CH}{}_2$)
  • Propanol: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{OH}$ (Ethanol + $\mathrm{CH}{}_2$)

5. Same Functional Group:

• All members contain the same functional group, which defines the series.
• Example:
  • Carboxylic acids: –COOH (e.g., Formic acid $\mathrm{HCOOH}$, Acetic acid $\mathrm{CH}{}_3\mathrm{COOH}$).
  • Ketones: –CO– (e.g., Propanone $\mathrm{CH}{}_3\mathrm{COCH}{}_3$, Butanone $\mathrm{CH}{}_3\mathrm{COC}{}_2\mathrm{H}{}_5$).

6. Same Type of Isomerism:

• Members may exhibit structural isomerism (e.g., butane and isobutane).
• Example (Alkenes):
  • But-1-ene: $\mathrm{CH}{}_2=\mathrm{CHCH}{}_2\mathrm{CH}{}_3$
  • But-2-ene: $\mathrm{CH}{}_3\mathrm{CH}=\mathrm{CHCH}{}_3$ (cis-trans isomerism possible).

7. Same Method of Preparation:

• Compounds in the same series can be prepared by similar methods.
• Example (Alkanes):
  • From alkenes by hydrogenation ($\mathrm{RCH}=\mathrm{CHR}{}^{\prime }+\mathrm{H}{}_2\to \mathrm{RCH}{}_2\mathrm{CH}{}_2\mathrm{R}{}^{\prime }$).
  • From alkyl halides by reduction ($\mathrm{RX}+\mathrm{Zn}/\mathrm{HCl}\to \mathrm{RH}$).

8. Same General Reaction Mechanism:

• The mechanism of reactions remains the same across the series.
• Example (Alkenes):
  • All undergo electrophilic addition (e.g., with $\mathrm{HBr}$, $\mathrm{Br}{}_2$).

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iv) Relationship Between $K_p$ and $K_c$

Equilibrium Constants:

• $K_c$: Equilibrium constant in terms of concentrations (mol/L).
• $K_p$: Equilibrium constant in terms of partial pressures (atm).

For a gas-phase reaction, the relationship between $K_p$ and $K_c$ is given by: $$K_p=K_c(RT{)}^{\Delta n}$$ where:

• $R$ = Universal gas constant ($0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$).
• $T$ = Temperature (K).
• $\Delta n$ = Change in the number of moles of gas (moles of gaseous products – moles of gaseous reactants).

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Derivation:

Consider a general gas-phase reaction: $$a\mathrm{A}(\mathrm{g})+b\mathrm{B}(\mathrm{g})\rightleftharpoons c\mathrm{C}(\mathrm{g})+d\mathrm{D}(\mathrm{g})$$ where $\Delta n=(c+d)-(a+b)$.

1. Expression for $K_c$: $$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$$ (Concentrations in mol/L).

2. Expression for $K_p$: For gases, partial pressure (P) is related to concentration (C) by the Ideal Gas Law: $$P=CRT$$ where $C$ is in mol/L, $R=0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$, and $T$ is in K.

   Thus, the equilibrium partial pressures are: $$P_C=[C]RT,P_D=[D]RT,P_A=[A]RT,P_B=[B]RT$$

   Substituting into $K_p$: $$K_p=\frac{(P_C{)}^c(P_D{)}^d}{(P_A{)}^a(P_B{)}^b}=\frac{([C]RT{)}^c([D]RT{)}^d}{([A]RT{)}^a([B]RT{)}^b}$$ $$K_p=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}\times (RT{)}^{(c+d)-(a+b)}=K_c(RT{)}^{\Delta n}$$

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Key Points:

1. Units:

   • $K_c$ is unitless if concentrations are in mol/L.
   • $K_p$ has units of atm${}^{\Delta n}$ (since pressure is in atm).

2. When $\Delta n=0$:

   • $K_p=K_c$ (e.g., $\mathrm{N}{}_2(\mathrm{g})+3\mathrm{H}{}_2(\mathrm{g})\rightleftharpoons 2\mathrm{NH}{}_3(\mathrm{g})$, $\Delta n=2-(1+3)=-2$, but for $\mathrm{H}{}_2(\mathrm{g})+\mathrm{I}{}_2(\mathrm{g})\rightleftharpoons 2\mathrm{HI}(\mathrm{g})$, $\Delta n=0$).

3. When $\Delta n>0$:

   • $K_p>K_c$ (more moles of gas on the product side).
   • Example: $\mathrm{N}{}_2\mathrm{O}{}_4(\mathrm{g})\rightleftharpoons 2\mathrm{NO}{}_2(\mathrm{g})$, $\Delta n=2-1=1$.

4. When $\Delta n<0$:

   • $K_p<K_c$ (more moles of gas on the reactant side).
   • Example: $2\mathrm{SO}{}_2(\mathrm{g})+\mathrm{O}{}_2(\mathrm{g})\rightleftharpoons 2\mathrm{SO}{}_3(\mathrm{g})$, $\Delta n=2-(2+1)=-1$.

5. Temperature Dependence:

   • Both $K_p$ and $K_c$ change with temperature (Le Chatelier’s Principle).
   • The relationship $K_p=K_c(RT{)}^{\Delta n}$ holds only at a given temperature.

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Example Calculation:

Reaction: $\mathrm{H}{}_2(\mathrm{g})+\mathrm{I}{}_2(\mathrm{g})\rightleftharpoons 2\mathrm{HI}(\mathrm{g})$ Given:

• $K_c=50\text{mol/L}$ at 700 K.
• $R=0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$.
• $\Delta n=2-(1+1)=0$.

Calculation: $$K_p=K_c(RT{)}^{\Delta n}=50\times (0.0821\times 700{)}^0=50\times 1=50$$ (Unitless, since $\Delta n=0$.)

Another Example: Reaction: $\mathrm{N}{}_2(\mathrm{g})+3\mathrm{H}{}_2(\mathrm{g})\rightleftharpoons 2\mathrm{NH}{}_3(\mathrm{g})$ Given:

• $K_c=0.5\text{mol/L}$ at 723 K.
• $\Delta n=2-(1+3)=-2$.

Calculation: $$K_p=0.5\times (0.0821\times 723{)}^{-2}$$ $$RT=0.0821\times 723=59.5783{\text{L atm mol}}^{-1}$$ $$K_p=0.5\times (59.5783{)}^{-2}=0.5\times \frac{1}{(59.5783{)}^2}$$ $$K_p=0.5\times \frac{1}{3549.5}=1.41\times 10^{-4}{\text{atm}}^{-2}$$

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When to Use $K_p$ vs. $K_c$: