Step 1: Understanding the Concept:

This is a redox reaction where iodate (\( IO_3^- \)) and iodide (\( I^- \)) react in an acidic medium to form iodine (\( I_2 \)). This is a comproportionation reaction.


Step 2: Key Formula or Approach:

To find \( x \), we can balance the iodine atoms on both sides of the chemical equation.


Step 3: Detailed Explanation:

1. Look at the right-hand side (RHS) of the equation:

There are 6 molecules of \( I_2 \).

Total number of iodine atoms on RHS = \( 6 \times 2 = 12 \).

2. Look at the left-hand side (LHS) of the equation:

Iodine atoms come from two sources: \( 2IO_3^- \) and \( xI^- \).

Total iodine atoms on LHS = \( 2 + x \).

3. Equating LHS and RHS for iodine balance:
\[ 2 + x = 12 \]
\[ x = 12 - 2 = 10 \]

4. Verification using charge balance:

LHS total charge = \( 2(-1) + x(-1) + 12(+1) \).

Substituting \( x = 10 \):

Charge = \( -2 - 10 + 12 = 0 \).

RHS total charge = 0.

The equation is balanced with \( x = 10 \).


Step 4: Final Answer:

The value of \( x \) is 10.
Quick Tip: In balanced chemical equations involving ions, you can often find the coefficient by balancing the specific atom that appears in only one product or by checking the net charge on both sides.

Step 1: Understanding the Concept:

A galvanic cell is represented by showing the anode on the left and the cathode on the right. We need to identify the oxidation and reduction half-reactions from the given net cell reaction.


Step 2: Key Formula or Approach:

Net reaction: \( \frac{1}{2} H_2(g) + AgCl(s) \rightarrow H^+(aq) + Cl^-(aq) + Ag(s) \)

Identify oxidation and reduction:

Oxidation (Anode): \( \frac{1}{2} H_2(g) \rightarrow H^+(aq) + e^- \)

Reduction (Cathode): \( AgCl(s) + e^- \rightarrow Ag(s) + Cl^-(aq) \)


Step 3: Detailed Explanation:

1. Anode side: The oxidation involves hydrogen gas turning into hydrogen ions. This requires an inert electrode like Platinum. The electrolyte contains \( H^+ \) ions (provided by \( HCl \)). So, the anode representation is \( Pt | H_2(g) | HCl(sol^n) \).

2. Cathode side: The reduction involves Silver Chloride (solid) and Silver metal. This is a metal-insoluble salt electrode. The electrolyte must contain \( Cl^- \) ions (also provided by \( HCl \)). The representation is \( AgCl(s) | Ag \).

3. Combining them, we get: \( Pt | H_2(g) | HCl(sol^n) | AgCl(s) | Ag \).

Note: Both electrodes share the same electrolyte (\( HCl \)), making it a cell without a salt bridge.


Step 4: Final Answer:

The correct cell representation is \( Pt | H_2(g) | HCl(sol^n) | AgCl(s) | Ag \).
Quick Tip: Remember the notation: Anode | Electrolyte || Electrolyte | Cathode. For gas electrodes, always include the inert metal (Pt) at the extreme end.

Step 1: Understanding the Concept:

The Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed per unit mass of solid adsorbent (\( x/m \)) and pressure (\( p \)) at a constant temperature.


Step 2: Key Formula or Approach:

The mathematical expression is:
\[ \frac{x}{m} = k \cdot p^{1/n} \]

Taking logarithm on both sides:
\[ \log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log p \]


Step 3: Detailed Explanation:

1. Graph A: Shows a plot of \( x/m \) vs \( p \). It is a curve that starts at the origin and levels off at higher pressures. This correctly represents the basic equation \( x/m = k \cdot p^{1/n} \).

2. Graph B: Shows a plot of \( \log(x/m) \) vs \( \log p \). Based on the log equation, this should be a straight line with a positive slope (\( 1/n \)) and a positive intercept (\( \log k \)). This is a standard representation.

3. Graph C: Shows \( x/m \) vs \( c \) (concentration) as a straight line with an intercept. This does not follow the power-law dependency of the Freundlich isotherm for solutions (\( x/m = k \cdot c^{1/n} \)).

4. Graph D: Shows \( x/m \) vs \( p^{1/n} \). According to \( y = mx \), a plot of \( x/m \) vs \( p^{1/n} \) should be a straight line passing through the origin. This is a mathematically valid representation.


Step 4: Final Answer:

Graphs A, B, and D correctly represent the Freundlich isotherm.
Quick Tip: The Freundlich isotherm is characterized by the power \( 1/n \), where \( 1/n \) typically ranges from 0.1 to 0.5. If the plot of \( \log(x/m) \) vs \( \log p \) is a straight line, it confirms Freundlich behavior.

Step 1: Understanding the Concept:

Electronegativity is the tendency of an atom to attract a shared pair of electrons. It generally increases across a period (left to right) and decreases down a group (top to bottom).


Step 2: Key Formula or Approach:

We compare the values on the Pauling scale for Carbon (Group 14), Phosphorus (Group 15), Bromine (Group 17), and Astatine (Group 17).


Step 3: Detailed Explanation:

1. Bromine (Br): Being a halogen in the 4th period, it is very electronegative. Its value is approximately 2.96.

2. Carbon (C): Located in Group 14, Period 2. Its electronegativity is 2.55.

3. Astatine (At): Located in Group 17, but in the 6th period. Electronegativity decreases significantly down the group. Its value is approximately 2.2.

4. Phosphorus (P): Located in Group 15, Period 3. Its value is approximately 2.19.

Comparing these values: \( 2.96 (Br) > 2.55 (C) > 2.2 (At) > 2.19 (P) \).

Thus, the order is Br \( > \) C \( > \) At \( > \) P.


Step 4: Final Answer:

The correct order is Br \( > \) C \( > \) At \( > \) P.
Quick Tip: Remember that Halogens are always more electronegative than other elements in their respective periods. Even a lower halogen (At) can have electronegativity comparable to non-metals of earlier groups (P, C).

Step 1: Understanding the Concept:

Leaching is a chemical method used for the concentration of ores where the ore is treated with a suitable reagent (acid or base) such that the metal or its mineral dissolves, while the impurities remain insoluble.


Step 2: Detailed Explanation:

1. Au (Gold): Extracted by leaching with a dilute solution of \( NaCN \) or \( KCN \) in the presence of air (Cyanide process). This is alkaline but specifically "Cyanide leaching".

2. Cu (Copper): Usually concentrated by froth floatation and extracted by smelting.

3. Sn (Tin): Tin is an amphoteric metal. Its oxide (\( SnO_2 \)) or the metal itself can react with concentrated alkali like \( NaOH \) to form soluble stannates:
\[ Sn + 2NaOH + H_2O \rightarrow Na_2SnO_3 + 2H_2 \]

This property allows Tin to be separated from non-amphoteric impurities through alkali leaching.

4. Note: The most common example of alkali leaching in metallurgy is the extraction of Aluminum (\( Al \)) from Bauxite using \( NaOH \) (Bayer's process).


Step 3: Final Answer:

Tin (Sn) can be extracted/concentrated using alkali leaching.
Quick Tip: Only amphoteric metals (like Al, Zn, Pb, Sn) can dissolve in strong alkalis like NaOH. This unique property is exploited in leaching processes to separate them from basic or neutral impurities.

Step 1: Understanding the Concept:

Water gas is a mixture of Carbon Monoxide (\( CO \)) and Hydrogen (\( H_2 \)). It can be used to synthesize various organic compounds depending on the catalyst and conditions used.


Step 2: Detailed Explanation:

1. Water gas (\( CO + H_2 \)) reacts with additional hydrogen in the presence of specific catalysts to produce methanol (\( CH_3OH \)).

2. The reaction is:
\[ CO(g) + 2H_2(g) \xrightarrow{Co catalyst} CH_3OH(l) \]

3. While various catalysts like \( ZnO/Cr_2O_3 \) are commonly used for industrial methanol synthesis, Cobalt-based catalysts are also effective and were historically significant in these transformations (including Fischer-Tropsch synthesis which produces hydrocarbons and alcohols).


Step 3: Final Answer:

The product formed is Methanol.
Quick Tip: Syn-gas (\( CO + H_2 \)) is a versatile precursor. With different catalysts, it can form methane, methanol, or higher hydrocarbons. Methanol is the primary product when stoichiometric hydrogen is added.

Step 1: Understanding the Concept:

Cement setting involves complex chemical reactions between its components and water. Gypsum (\( CaSO_4 \cdot 2H_2O \)) is added in small quantities (2-3%) during the final stages of cement manufacturing.


Step 2: Detailed Explanation:

1. One of the main components of cement is Tri-calcium Aluminate (\( C_3A \)), which reacts very rapidly with water, causing "flash set" or premature hardening.

2. Gypsum acts as a retarder. It reacts with \( C_3A \) to form a protective layer of ettringite on the surface of cement particles.

3. This layer slows down the initial hydration process, allowing workers enough time for mixing, transporting, and placing the cement before it hardens.

4. Therefore, the primary role is to increase the setting time of the cement.


Step 3: Final Answer:

The purpose of adding gypsum is to slow down the process of setting.
Quick Tip: Gypsum is a classic example of a chemical retarder. Without it, cement would harden almost instantly upon mixing with water, making it useless for construction.

Step 1: Understanding the Concept:

This question explores the properties of Group 1 and Group 2 elements and the diagonal relationship between Lithium and Magnesium.


Step 2: Detailed Explanation:

1. Statement I: Superoxides (\( O_2^- \)) are typically formed by larger alkali metals like \( K, Rb, \) and \( Cs \). Lithium (\( Li \)) is small and forms a stable normal oxide (\( Li_2O \)). Magnesium (\( Mg \)), being a Group 2 metal, also forms a normal oxide (\( MgO \)) or peroxide (\( MgO_2 \)) under specific conditions, but not a superoxide. Thus, Statement I is correct.

2. Statement II: Although they are in different groups and periods, \( Li \) and \( Mg \) exhibit a diagonal relationship. The ionic radius of \( Li^+ \) is approximately 76 pm, while that of \( Mg^{2+} \) is approximately 72 pm. Because of its higher nuclear charge (\( +2 \) vs \( +1 \)), \( Mg^{2+} \) pulls its electron cloud more tightly than \( Li^+ \). Thus, \( Li^+ \) is slightly larger than \( Mg^{2+} \). Statement II is correct.


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: The diagonal relationship exists because these pairs (\( Li/Mg, Be/Al, B/Si \)) have similar ionic sizes and charge/radius ratios, leading to many similar chemical properties.

Step 1: Understanding the Concept:

This reaction represents the reduction of Copper(II) to Copper(I) by a halide ion, while the halide ion is oxidized to the halogen. This occurs when the halide is a strong enough reducing agent.


Step 2: Detailed Explanation:

1. Standard reduction potential of \( Cu^{2+}/Cu^+ \) is \( +0.15 V \).

2. Standard reduction potentials for halogens are: \( F_2 (+2.87 V), Cl_2 (+1.36 V), Br_2 (+1.09 V), I_2 (+0.54 V) \).

3. For the reaction to be spontaneous, the halogen must be easily reduced, or the halide must be easily oxidized.

4. Actually, \( Cu^{2+} \) is not strong enough to oxidize \( Cl^- \) or \( Br^- \). However, it can oxidize \( I^- \) because \( I^- \) is a very strong reducing agent.

5. In aqueous solution, \( CuI_2 \) is unstable and immediately decomposes to form white precipitate of \( Cu_2I_2 \) and violet vapors of \( I_2 \):
\[ 2Cu^{2+} + 4I^- \rightarrow Cu_2I_2(s) + I_2 \]

6. \( CuCl_2 \) and \( CuBr_2 \) are stable in aqueous solutions.


Step 3: Final Answer:

Only Iodine causes this reaction.
Quick Tip: This reaction is the basis for iodometric estimation of copper. Iodide ions reduce Cu(II) quantitatively, and the liberated \( I_2 \) is then titrated with sodium thiosulfate.

Step 1: Understanding the Concept:

We need to evaluate the coordination number, magnetic property (presence of unpaired electrons), and thermodynamic stability of the given complexes.


Step 2: Detailed Explanation:

1. Option A: \( K_3[Co(CN)_6] \)

- Coordination Number = 6 (Octahedral).

- Metal ion is \( Co^{3+} \) (\( d^6 \) system).

- \( CN^- \) is a very strong field ligand (SFL). It causes pairing of electrons in \( t_{2g} \) orbitals: \( t_{2g}^6 e_g^0 \).

- Since all electrons are paired, it is diamagnetic.

- Complexes of \( Co^{3+} \) with SFL are exceptionally stable due to high Crystal Field Stabilization Energy (CFSE).

2. Option B: \( [Co(H_2O)_6]Cl_2 \)

- Metal is \( Co^{2+} \) (\( d^7 \)). Regardless of ligand, \( d^7 \) octahedral is paramagnetic.

3. Option C: \( [Ni(NH_3)_6]Cl_2 \)

- Metal is \( Ni^{2+} \) (\( d^8 \)). In an octahedral field, \( d^8 \) always has 2 unpaired electrons (\( t_{2g}^6 e_g^2 \)). It is paramagnetic.

4. Option D: \( Na_3[CoCl_6] \)

- Metal is \( Co^{3+} \) (\( d^6 \)). \( Cl^- \) is a weak field ligand (WFL). It forms a high-spin complex with 4 unpaired electrons. It is paramagnetic.


Step 3: Final Answer:

The complex \( K_3[Co(CN)_6] \) satisfies all three conditions.
Quick Tip: For \( d^6 \) metal ions (like \( Co^{3+}, Fe^{2+} \)) in octahedral fields, strong field ligands always lead to diamagnetic low-spin complexes with very high stability.

Step 1: Understanding the Concept:

Safe drinking water standards are established by regulatory bodies like the WHO or EPA.

Different ions and elements have specific maximum permissible limits (in ppm or mg/L) to ensure health and prevent toxicity or aesthetic issues.


Step 2: Detailed Explanation:

Based on environmental chemistry standards:

1. Fluoride (\(F^-\)): Excess fluoride causes fluorosis. The permissible limit is usually around 1.5-2.0 ppm.

Therefore, A matches with III (\(< 2\) ppm).

2. Sulphate (\(SO_4^{2-}\)): High sulphate levels can have a laxative effect. The limit is generally \(< 500\) ppm.

Therefore, B matches with IV (\(SO_4^{2-} \rightarrow 500\) ppm is standard, but looking at options, B corresponds to a value). Wait, let's re-examine.

Looking at standard JEE/competitive exam values:
\(F^- \rightarrow 2\) ppm (III)
\(Zn \rightarrow 5\) ppm (II)
\(NO_3^- \rightarrow 50\) ppm (I)
\(SO_4^{2-} \rightarrow 500\) ppm (IV)

So, the match is: A-III, B-IV, C-I, D-II.

Checking the option provided in the template: Option (D) lists A-III, B-IV, C-I, D-II. (Wait, the screenshot text for D is A-III, B-II, C-I, D-IV? Let me re-read).

Actually, looking at the labels:

A matches with III (\(< 2\) ppm).

C matches with I (\(< 50\) ppm).

D matches with II (\(< 5\) ppm).

B matches with IV (\(< 500\) ppm).

The correct match sequence is A-III, B-IV, C-I, D-II.

Looking at the options in screenshot 1:

Option 3666949384: A-III, B-IV, C-I, D-II? No, it says A-III, B-II, C-I, D-IV. Let's re-verify standards.

Permissible limit of \(Zn\) is 5 ppm. \(SO_4^{2-}\) is 500 ppm.

If the paper says B-II (\(SO_4^{2-}\) < 5) and D-IV (\(Zn\) < 500), that would be incorrect chemistry.

Let's re-read the prompt's image 1 carefully.

Option 3666949384: A-III, B-IV, C-I, D-II. (Matches standard chemistry).


Step 3: Final Answer:

The matching is Fluoride (\(< 2\) ppm), Sulphate (\(< 500\) ppm), Nitrate (\(< 50\) ppm), and Zinc (\(< 5\) ppm).
Quick Tip: Mnemonic: Nitrates (\(NO_3^-\)) limit is 50 ppm (N for Fifty), Fluoride is 2 ppm, Zinc is 5 ppm, and Sulphate is 500 ppm.

Step 1: Understanding the Concept:

The spin-only magnetic moment (\(\mu\)) is determined by the number of unpaired electrons (\(n\)) using the formula \(\mu = \sqrt{n(n+2)}\) Bohr Magnetons (BM).

The number of unpaired electrons depends on the oxidation state of the metal and the strength of the ligand (Strong Field Ligands cause pairing).


Step 2: Key Formula or Approach:

1. Find oxidation state of the metal.

2. Determine d-electron configuration.

3. Determine if the complex is high spin (WFL) or low spin (SFL).

4. Count unpaired electrons (\(n\)).


Step 3: Detailed Explanation:

1. \([Fe(CN)_6]^{3-}\): \(Fe^{3+}\) is \(d^5\). \(CN^-\) is SFL \(\rightarrow\) Low Spin.

Config: \(t_{2g}^5 e_g^0\). Unpaired \(n = 1\).

2. \([Mn(CN)_6]^{3-}\): \(Mn^{3+}\) is \(d^4\). \(CN^-\) is SFL \(\rightarrow\) Low Spin.

Config: \(t_{2g}^4 e_g^0\). Unpaired \(n = 2\).

3. \([CoF_6]^{3-}\): \(Co^{3+}\) is \(d^6\). \(F^-\) is WFL \(\rightarrow\) High Spin.

Config: \(t_{2g}^4 e_g^2\). Unpaired \(n = 4\).

4. \([MnBr_4]^{2-}\): \(Mn^{2+}\) is \(d^5\). Tetrahedral, \(Br^-\) is WFL \(\rightarrow\) High Spin.

Config: \(e^2 t_2^3\). Unpaired \(n = 5\).

Order of unpaired electrons: \(1 < 2 < 4 < 5\).

Since \(\mu\) increases with \(n\), the magnetic moment order follows the same sequence.


Step 4: Final Answer:

The correct order is \([Fe(CN)_6]^{3-} < [Mn(CN)_6]^{3-} < [CoF_6]^{3-} < [MnBr_4]^{2-}\).
Quick Tip: Strong Field Ligands (like \(CN^-\), \(CO\)) force pairing of electrons, reducing the magnetic moment. Weak Field Ligands (like Halides) usually lead to high-spin complexes.

Step 1: Understanding the Concept:

In an \(S_N1\) mechanism, the rate-determining step is the formation of a carbocation. The reactivity is directly proportional to the stability of the carbocation intermediate.


Step 2: Detailed Explanation:

A: Leaving \(Br_{(b)}\) forms a secondary benzylic carbocation, which is resonance-stabilized. Leaving \(Br_{(a)}\) forms a primary carbocation. Thus, \(Br_{(b)}\) is more reactive.

B: Leaving \(I_{(b)}\) forms an allylic carbocation, which is stabilized by resonance. Leaving \(I_{(a)}\) is vinylic, which is extremely unstable. Thus, \(I_{(b)}\) is more reactive.

C: Leaving \(Br_{(a)}\) forms a secondary carbocation. Leaving \(Br_{(b)}\) would place a carbocation at a bridgehead carbon, which is highly unstable (Bredt's Rule). Thus, \(Br_{(a)}\) is more reactive.

D: Leaving \(Br_{(a)}\) forms a tertiary carbocation (\(3^\circ\)), while \(Br_{(b)}\) would form a primary carbocation (\(1^\circ\)). Stability order: \(3^\circ > 2^\circ > 1^\circ\). Thus, \(Br_{(a)}\) is more reactive.


Step 3: Final Answer:

The most reactive halogens are A - \(Br_{(b)}\), B - \(I_{(b)}\), C - \(Br_{(a)}\), D - \(Br_{(a)}\).
Quick Tip: For \(S_N1\), check for resonance stabilization (benzylic, allylic) and degree of substitution (\(3^\circ > 2^\circ > 1^\circ\)). Avoid forming carbocations at vinylic or bridgehead positions.

Step 1: Understanding the Concept:

Lithium Borohydride (\(LiBH_4\)) is a selective reducing agent. It is more reactive than \(NaBH_4\) but less reactive than \(LiAlH_4\).


Step 2: Detailed Explanation:

The reactant is an o-substituted benzene derivative containing a carboxylic acid group (\(-CO_2H\)) and an ester group (\(-CO_2Et\)).

1. \(LiBH_4\) in Ethanol/THF selectively reduces ester groups (\(-COOR\)) to primary alcohols (\(-CH_2OH\)).

2. Under these standard conditions, \(LiBH_4\) does not reduce simple carboxylic acids.

3. Therefore, the ester group (\(-CO_2Et\)) is reduced to \(-CH_2OH\), while the acid group remains intact.


Step 3: Final Answer:

The product contains a \(-COOH\) group and a \(-CH_2OH\) group in the ortho positions.
Quick Tip: While \(LiAlH_4\) reduces both acids and esters to alcohols, \(LiBH_4\) allows for selective reduction of esters in the presence of carboxylic acids.

Step 1: Understanding the Concept:

Boiling point depends on the strength of intermolecular forces. Stronger forces lead to higher boiling points.


Step 2: Detailed Explanation:

1. Butan-1-ol (\(CH_3CH_2CH_2CH_2OH\)): Contains a polar \(-OH\) group. The oxygen-hydrogen bond allows for the formation of intermolecular hydrogen bonds between alcohol molecules.

2. Ethoxyethane (\(CH_3CH_2OCH_2CH_3\)): Is an ether. While it is polar, it lacks hydrogen atoms directly bonded to oxygen, so it cannot form intermolecular hydrogen bonds with its own molecules. It only exhibits weaker dipole-dipole interactions.

3. Since hydrogen bonding is significantly stronger than dipole-dipole interactions, Butan-1-ol molecules are more strongly associated, requiring more energy (higher temperature) to separate into the gas phase.


Step 3: Final Answer:

Both statements are true, and the presence of hydrogen bonding in alcohols explains why they have higher boiling points than isomeric or similarly sized ethers.
Quick Tip: Hydrogen bonding typically gives alcohols much higher boiling points compared to hydrocarbons, haloalkanes, and ethers of comparable molecular masses.

Step 1: Understanding the Concept:

Benzene Diazonium Chloride (\(C_6H_5N_2^+Cl^-\)) is a highly versatile intermediate in organic synthesis, undergoing coupling reactions and displacement of the diazonium group.


Step 2: Detailed Explanation:

1. Reaction with \(C_6H_5NH_2\): This is a coupling reaction. Aniline reacts with the diazonium salt to form p-aminoazobenzene (a yellow dye). (A \(\rightarrow\) III)

2. Reaction with \(HBF_4, \Delta\): This is the Balz-Schiemann reaction. It yields fluorobenzene. (B \(\rightarrow\) I)

3. Reaction with \(Cu, HCl\): This is the Gattermann reaction. It yields chlorobenzene. (C \(\rightarrow\) IV)

4. Reaction with \(CuCN/KCN\): This is the Sandmeyer reaction for cyanide. It yields benzonitrile. (D \(\rightarrow\) II)


Step 3: Final Answer:

The match is A-III, B-I, C-IV, D-II.
Quick Tip: Sandmeyer (using CuX) and Gattermann (using Cu powder/HX) reactions are the standard ways to substitute the diazonium group with Cl, Br, or CN.

Step 1: Understanding the Concept:

Proteins are composed of amino acids. Among the 20 standard amino acids, only two contain sulfur in their side chains.


Step 2: Detailed Explanation:

1. Cysteine (b): Contains a thiol group (\(-SH\)) in its side chain (\(-CH_2SH\)). It is crucial for forming disulfide bridges in protein structures.

2. Methionine (d): Contains a thioether group (\(-S-CH_3\)) in its side chain (\(-CH_2CH_2SCH_3\)).

The other listed amino acids:

- Isoleucine (a): Aliphatic side chain.

- Lysine (c): Basic side chain with amino group.

- Glutamic acid (e): Acidic side chain with carboxyl group.


Step 3: Final Answer:

The sulfur-containing amino acids are cysteine (b) and methionine (d).
Quick Tip: Mnemonic: "S-C-M" \(\rightarrow\) Sulfur is in Cysteine and Methionine. Cysteine has the -SH group while Methionine has -S-CH3.

Step 1: Understanding the Concept:

Artificial sweeteners are synthetic substitutes for sugar that provide sweetness without the calories. Each has different chemical stabilities and potencies.


Step 2: Detailed Explanation:

1. Saccharin (A): Discovered in 1879, it is the first popular artificial sweetening agent. (A \(\rightarrow\) II)

2. Aspartame (B): It is a dipeptide methyl ester. It decomposes at high temperatures, making it unsuitable for cooking. (B \(\rightarrow\) IV)

3. Alitame (C): It is a high-potency sweetener, being about 2000 times sweeter than sucrose. (C \(\rightarrow\) I)

4. Sucralose (D): It is a trichloro derivative of sucrose. It is stable at cooking temperatures. (D \(\rightarrow\) III)


Step 3: Final Answer:

The match is A-II, B-IV, C-I, D-III.
Quick Tip: Aspartame is for cold drinks only. Sucralose is for everything because it stays stable when heated.

Step 1: Understanding the Concept:

Qualitative analysis of organic compounds involves using specific reagents that produce a visible change (color, precipitate) with certain functional groups.


Step 2: Detailed Explanation:

1. Benedict's solution (A): (Alkaline copper sulphate and sodium citrate) is used to detect reducing sugars or aliphatic aldehydes. It gives a red precipitate of \(Cu_2O\). (A \(\rightarrow\) III)

2. Neutral \(FeCl_3\) (B): Reacts with phenols or enols to give characteristic colored complexes (usually violet, blue, or green). (B \(\rightarrow\) I)

3. Carbylamine reaction (C): (Alkaline chloroform) Aliphatic or aromatic primary amines react to form foul-smelling isocyanides. (C \(\rightarrow\) II)

4. Iodoform test (D): (Potassium iodide and sodium hypochlorite/NaOH) Detects methyl ketones (\(CH_3CO-\)) or alcohols oxidizable to them. It produces a yellow precipitate of \(CHI_3\). (D \(\rightarrow\) IV)


Step 3: Final Answer:

The match is A-III, B-I, C-II, D-IV.
Quick Tip: Benedict's and Fehling's tests are for aldehydes. \(FeCl_3\) is for phenols. Carbylamine is for \(1^\circ\) amines. Iodoform is for "methyl" groups next to a carbonyl.

Step 1: Understanding the Concept:

The number of d-electrons depends on the electronic configuration of the metal atom and its oxidation state in the compound.


Step 2: Detailed Explanation:

1. Chromyl Chloride (\(CrO_2Cl_2\)):

Oxidation state of \(Cr\): \(x + 2(-2) + 2(-1) = 0 \Rightarrow x = +6\).

Neutral \(Cr\): \([Ar] 3d^5 4s^1\).
\(Cr^{6+}\): \([Ar] 3d^0 4s^0\). Number of d-electrons = 0.

2. Check options for \(d^0\) configuration:

- Mn (VII): Neutral \(Mn\) is \([Ar] 3d^5 4s^2\). \(Mn^{7+}\) is \([Ar] 3d^0 4s^0\). Number of d-electrons = 0.

- Fe (III): \(Fe^{3+}\) is \(3d^5\).

- V (IV): \(V^{4+}\) is \(3d^1\).

- Ti (III): \(Ti^{3+}\) is \(3d^1\).


Step 3: Final Answer:

Chromyl chloride (\(Cr^{6+}\)) has the same number of d-electrons (zero) as Mn(VII).
Quick Tip: Highest oxidation states of 3d series elements (\(Sc^{+3}\) to \(Mn^{+7}\)) usually correspond to a \(d^0\) noble gas configuration.

Step 1: Understanding the Concept:

According to Dalton's Law of Partial Pressures, for a mixture of non-reacting gases at constant temperature and total volume, the total pressure is the sum of the partial pressures of individual gases.

Since the bulbs are connected, the final volume is the sum of the individual volumes of the bulbs.


Step 2: Key Formula or Approach:

The final pressure \(P_f\) of the system is given by:
\[ P_f = \frac{P_1V_1 + P_2V_2 + P_3V_3}{V_1 + V_2 + V_3} \]


Step 3: Detailed Explanation:

Given data:

Bulb 1: \(P_1 = 2 atm, V_1 = 2 L \)

Bulb 2: \(P_2 = 4 atm, V_2 = 3 L \)

Bulb 3: \(P_3 = 3 atm, V_3 = 4 L \)

Total Volume \(V_{total} = V_1 + V_2 + V_3 = 2 + 3 + 4 = 9 L \).

Now, calculating the final pressure:
\[ P_f = \frac{(2 \times 2) + (4 \times 3) + (3 \times 4)}{9} \]
\[ P_f = \frac{4 + 12 + 12}{9} = \frac{28}{9} \approx 3.111 atm \]

The nearest integer value is 3.


Step 4: Final Answer:

The final pressure of the system is 3 atm.
Quick Tip: For non-reacting gases, Boyle's Law can be extended to mixtures as \(P_{total}V_{total} = \sum P_iV_i\). Always ensure the total volume accounts for all connecting parts if their volumes are not negligible.

Step 1: Understanding the Concept:

Atomic spectra are of two types: emission and absorption. Atomic spectra of atoms in the gas phase are typically discontinuous (line) spectra, unlike continuous spectra (like sunlight or white light passing through a prism).


Step 3: Detailed Explanation:

1. Statement (A): Correct. Line spectra provide specific wavelengths corresponding to energy differences between quantized electronic states, helping determine electronic structures.

2. Statement (B): Incorrect. Atomic emission spectra of elements in the gas phase are line spectra (discrete wavelengths), not continuous. Continuous spectra are observed from incandescent solids or liquids.

3. Statement (C): Correct. Absorption spectra show dark lines at precisely the same wavelengths where emission spectra show bright lines for the same element.

4. Statement (D): Correct. Helium was first detected in the solar spectrum during the 1868 eclipse before it was found on Earth.

Only statement (B) is incorrect.


Step 4: Final Answer:

The number of incorrect statements is 1.
Quick Tip: Remember: Atoms show line spectra (fingerprints of elements), while molecules often show band spectra. Sunlight or a heated filament produces a continuous spectrum.

Step 1: Understanding the Concept:

Fajans' Rules describe the factors that determine the degree of covalent character in an ionic bond. Covalent character is introduced via polarization.


Step 2: Detailed Explanation:

According to Fajans' Rules:

1. Polarising power of cation (A): This is a fundamental factor. Small, highly charged cations have high polarizing power and increase covalent character.

2. Polarisability of the anion (C): This is another fundamental factor. Large, highly charged anions are easily polarized, increasing covalent character.

3. Extent of distortion of anion (B): This is the \textit{result of polarization (covalent character itself), not a factor \textit{affecting it.

4. Polarising power of anion (D): This is an incorrect term. Anions have polarisability, not polarising power.

The factors affecting the covalent character are (A) and (C).


Step 3: Final Answer:

The number of factors is 2.
Quick Tip: Polarization is a two-way street involving a cation "attacking" and an anion being "distorted". Cation = Polarizing power; Anion = Polarisability.

Step 1: Understanding the Concept:

Heat capacity (\(C\)) is the amount of heat energy required to raise the temperature of a substance by \(1 K\) (or \(1^\circ C\)). In an adiabatic container at constant volume, all the electrical energy supplied by the heater is converted into the internal energy (heat) of the gas.


Step 2: Key Formula or Approach:

Total heat supplied \(Q = Power \times Time\).

Heat capacity \(C = \frac{Q}{\Delta T} \).


Step 3: Detailed Explanation:

Given data:

Power \(P = 60 W = 60 J/s \).

Time \(t = 100 s \).

Temperature rise \(\Delta T = 5^\circ C = 5 K \).

Total heat energy supplied \(Q = P \times t = 60 \times 100 = 6000 J \).

Heat capacity \(C = \frac{6000}{5} = 1200 J \cdot K^{-1} \).


Step 4: Final Answer:

The heat capacity of the given gas is 1200 \(J \cdot K^{-1}\).
Quick Tip: For numeric type questions in thermodynamics, check the unit carefully. Here, the result is required in \(J \cdot K^{-1}\), which is simply the total heat divided by change in temperature.

Step 1: Understanding the Concept:

The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure (\(1 atm\)). On a vapour pressure vs. temperature graph, the curve for a pure solvent always lies above the curve for a solution (due to relative lowering of vapour pressure).


Step 2: Detailed Explanation:

1. From the graph, we observe three curves. The curve to the extreme left represents the pure solvent because at any given temperature, it has the highest vapour pressure.

2. To find the boiling point, we look at the point where the vapour pressure equals \(1 atm\) on the y-axis.

3. Following the horizontal dashed line at \(1 atm\), it intersects the solvent curve (left-most curve) at a specific temperature.

4. Dropping a perpendicular to the x-axis from this intersection point, we find the temperature corresponds to \(80^\circ C\).

5. The subsequent curves (solutions) intersect the \(1 atm\) line at \(81^\circ C, 82^\circ C, \) and \(83^\circ C\), showing elevation in boiling point.


Step 3: Final Answer:

The boiling point of the solvent is 80 \(^\circ C\).
Quick Tip: In colligative property graphs: Top curve = Solvent; Curves below it = Solutions with increasing concentration of non-volatile solute. Boiling point increases from left to right.

Step 1: Understanding the Concept:

Phenolphthalein is a synthetic pH indicator used primarily for strong base titrations. Indicators are themselves weak organic acids or bases that change colour over a specific pH range.


Step 2: Key Formula or Approach:

The theoretical pH at which an indicator starts to change colour is related to its dissociation constant: \(pH = pK_{In} \pm 1\).
\(pK_{In} = -\log(K_{In})\).


Step 3: Detailed Explanation:

1. Statement A: Incorrect. For weak acid-weak base titrations, there is no sharp change in pH at the equivalence point, making indicators like phenolphthalein unsuitable.

2. Statement B: Correct. Given \(K_{In} = 4 \times 10^{-10}\), \(pK_{In} = -\log(4 \times 10^{-10}) = 10 - \log 4 = 10 - 0.6 = 9.4\). Phenolphthalein usually changes colour in the range \(pH \approx 8.3\) to \(10\). Standard textbook values often cite colour change beginning around \(8.2--8.4\).

3. Statement C: Incorrect. Phenolphthalein is a weak organic acid.

4. Statement D: Correct. Phenolphthalein is colourless in acidic and neutral media (up to \(pH \approx 8\)) and turns pink/magenta in basic media.

Statements B and D are correct.


Step 4: Final Answer:

The number of correct statements is 2.
Quick Tip: An indicator is suitable for a titration only if its colour change pH range (\(pK_{In} \pm 1\)) falls within the steep vertical portion of the titration curve.

Step 1: Understanding the Concept:

The Arrhenius equation \(k = Ae^{-E_a/RT}\) quantitatively describes the effect of temperature on the rate of a chemical reaction.


Step 2: Detailed Explanation:

1. Statement (A): Correct. Differentiating Arrhenius equation gives \(\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}\). For a fixed temperature change, the change in rate constant (\(k\)) is directly proportional to \(E_a\). Thus, higher \(E_a\) means a more sensitive temperature dependence.

2. Statement (B): Correct. If \(E_a = 0\), the equation becomes \(k = A \cdot e^0 = A\). Since \(A\) (the pre-exponential factor) is roughly constant, the rate constant \(k\) becomes independent of temperature.

3. Statement (C): Incorrect. It is the opposite of statement (A).

4. Statement (D): Incorrect. No correlation doesn't imply negative activation energy (which is physically rare in elementary steps); it implies zero activation energy or experimental error. Negative activation energy usually suggests a complex mechanism where the rate \textit{decreases as temperature increases.

Statements (A) and (B) are correct.


Step 3: Final Answer:

The number of correct statements is 2.
Quick Tip: A plot of \(\ln k\) vs \(1/T\) gives a slope of \(-E_a/R\). A steeper slope indicates higher activation energy, meaning the reaction rate changes rapidly with temperature.

Step 1: Understanding the Concept:

Noble gas fluorides like \(XeF_4\) can act as fluoride ion donors when reacting with strong Lewis acids (fluoride acceptors) such as \(SbF_5, PF_5, AsF_5\).


Step 2: Detailed Explanation:

When \(XeF_4\) reacts with \(SbF_5\), it donates a fluoride ion (\(F^-\)) to form a cationic species and an anionic species:
\[ XeF_4 + SbF_5 \rightarrow [XeF_3]^+ + [SbF_6]^- \]

Comparing this with the given general formula \([XeF_m]^n [SbF_y]^z\):
\(m = 3\)
\(n = +1\)
\(y = 6\)
\(z = -1\)

Calculating the sum:
\(m + n + y + z = 3 + 1 + 6 + (-1) = 9 \).


Step 3: Final Answer:

The value of \(m + n + y + z\) is 9.
Quick Tip: \(XeF_2\) and \(XeF_4\) are good \(F^-\) donors. They form \([XeF]^+\) and \([XeF_3]^+\) respectively with fluoride ion acceptors like \(SbF_5\).

Step 1: Understanding the Concept:

Ozonolysis is a reaction where an alkene's double bond is cleaved to form carbonyl compounds. If 1 mole of hydrocarbon consumes 1 mole of \(O_3\), the hydrocarbon must contain exactly one carbon-carbon double bond (it is an alkene).


Step 2: Detailed Explanation:

The products are:

1. Ethanal: \(CH_3CHO\) (contains 2 Carbons)

2. Propanone: \(CH_3COCH_3\) (contains 3 Carbons)

To find the structure of the alkene, we join the two carbonyl groups at the oxygen atoms:
\(CH_3-CH=O + O=C(CH_3)_2 \rightarrow CH_3-CH=C(CH_3)_2 \) (2-methylbut-2-ene)

Molecular formula of X is \(C_5H_{10} \).

Calculating molar mass:

Molar mass \(= (5 \times 12) + (10 \times 1) = 60 + 10 = 70 g/mol \).


Step 3: Final Answer:

The molar mass of X is 70 \(g \cdot mol^{-1}\).
Quick Tip: Structure of alkene from ozonolysis products: Remove the oxygen atoms from the two carbonyl products and connect the carbon atoms with a double bond.

Step 1: Understanding the Concept:

During complete combustion of an organic compound, all the carbon present in the compound is converted into \(CO_2\). The mass of carbon in \(CO_2\) is \(\frac{12}{44}\) of the total mass of \(CO_2\).


Step 2: Key Formula or Approach:
\(Percentage of Carbon = \frac{12}{44} \times \frac{Mass of CO_2}{Mass of compound} \times 100 \).


Step 3: Detailed Explanation:

Given:

Mass of organic compound \(= 0.5 g \).

Percentage of Carbon \(= 60% \).

Mass of Carbon in the compound \(= 0.60 \times 0.5 = 0.3 g \).

Moles of Carbon \(= \frac{0.3}{12} = 0.025 mol \).

Since 1 mole of Carbon produces 1 mole of \(CO_2\),

Moles of \(CO_2\) produced \(= 0.025 mol \).

Mass of \(CO_2\) produced \(= 0.025 \times 44 = 1.1 g \).

Expressing in terms of \(10^{-1} g\):
\(1.1 g = 11 \times 10^{-1} g \).


Step 4: Final Answer:

The amount of \(CO_2\) produced is \(11 \times 10^{-1} g\).
Quick Tip: Remember: \(44 g\) of \(CO_2\) contains \(12 g\) of Carbon. Use stoichiometric ratios to find the mass of \(CO_2\) once the mass of carbon is known.