For 1 mole of a real gas, the compressibility factor Z is defined as:

Z = pV / RT.

From the given graph (pV vs p), the value of pV at point A is less than the ideal gas value, implying Z < 1. According to the Van der Waals equation, for small values of p:

Z = 1 − ^a/_RT V.

Hence, at point A, the compressibility factor is 1 − a/(RT V).

The energy of a transition in a hydrogen-like atom is given by:

E = −(13.6 Z² / n²)

where Z is the atomic number, and n is the principal quantum number. The wavelength λ for a transition is related to the energy difference via the Rydberg formula:

1/λ = R Z² (1/n₁² − 1/n₂²).

Lyman Series (Hydrogen):
The shortest wavelength corresponds to the transition n₂ = ∞ to n₁ = 1 (for Z = 1). This gives λ = 1/R.

Balmer Series (He⁺):
For He⁺ (Z = 2), the longest wavelength in the Balmer series is the transition n₂ = 3 to n₁ = 2. Thus:

1/λ_Balmer = R (2)² (1/2² − 1/3²) = 4R (1/4 − 1/9) = 4R [(9−4)/36] = 4R (5/36) = 5R/9.

Hence λ_Balmer = (9/5)(1/R) = (9/5) λ.

Bond dissociation energy is the energy required to break the bond in a molecule. For halogens, the order of bond dissociation energy is:

Cl₂ > Br₂ > F₂ > I₂.

Although F₂ has a shorter bond length, its bond energy is lower due to strong lone pair-lone pair repulsions. Cl₂ has the highest bond dissociation energy because of an optimal bond length and fewer repulsions.

The Mond process is used for refining nickel. In this process, impure nickel reacts with CO at moderate temperatures to form volatile nickel tetracarbonyl:

Ni + 4CO Δ → Ni(CO)₄.

The volatile Ni(CO)₄ then decomposes at higher temperatures to yield pure nickel.

Hydrogen storage materials should show high absorption capacity and reversibility. NaNi₅ is a hydride-forming alloy that efficiently stores and releases hydrogen. Other substances listed do not meet these requirements as effectively.

Hydration enthalpy depends on the charge density of ions. Smaller ions with higher charges have stronger electrostatic interactions with water and thus higher hydration enthalpies.

1. Among the alkali metal ions, K+ is smaller than Rb+ and Cs+, so K+ has higher hydration enthalpy than Rb+ and Cs+. The order is K+ > Rb+ > Cs+ (A > B > D).

2. Among the alkaline earth metal ions, Mg2+ is smaller and has a higher charge density than Ca2+. So Mg2+ > Ca2+ (C > E).

Combining these trends, we get:

Mg2+ > Ca2+ > K+ > Rb+ > Cs+ (C > E > A > B > D).

Li2O contains O(2-) ions, which have fully paired electrons and thus is diamagnetic. Na2O2 contains O2(2-) (peroxide) ions, also with paired electrons, so it is diamagnetic. KO2 contains O2(-) (superoxide) ions, which have an unpaired electron and are therefore paramagnetic.

In the Ostwald process, ammonia (NH3) is oxidized by air:

4 NH3 + 5 O2 -> 4 NO + 6 H2O

"A" is NO. It is further oxidized to "B":

2 NO + O2 -> 2 NO2

"B" is NO2. On hydration:

3 NO2 + H2O -> 2 HNO3 + NO

This forms nitric acid (an oxoacid of nitrogen) and releases NO, which gives the brown ring test.

Negative standard reduction potentials (M3+/M2+) or positive oxidation potentials (M2+/M3+) mean that the metal ions can reduce H+ from dilute acids to liberate hydrogen. Both V2+ (-0.26 V) and Cr2+ (-0.41 V) can do this, whereas Mn2+ and Co2+ have positive values that do not favor hydrogen evolution under these conditions.

Classical smog is also known as reducing smog. It occurs in cool, humid conditions and primarily involves SO₂ and particulate matter. It does not contain strong oxidizing agents like NO₂.

Photochemical smog, on the other hand, forms in warm, sunny conditions. It results from the reaction of NO₂ and volatile organic compounds under sunlight, producing ozone (O₃) and other oxidizing agents such as peroxyacetyl nitrate (PAN).

Thus, photochemical smog contains a high concentration of oxidizing agents.

A coordination complex is chiral if it lacks a plane of symmetry. In the cis arrangement of [PtCl₂(en)₂]²⁺, the bidentate ethylenediamine (en) ligands create a three-dimensional arrangement without any plane of symmetry, making it chiral.

In contrast, the trans forms (for example, trans–[PtCl₂(en)₂]²⁺) are generally symmetrical and thus achiral. Similarly, cis–[PtCl₂(NH₃)₂] has a plane of symmetry, and trans–[Co(NH₃)₄Cl₂]⁺ is also symmetrical.

Based on analysis of boiling points, densities, and other physical properties of the given halogenated compounds, (A), (C), and (E) display correct trends in the data provided. (B) and (D) contradict known experimental results for the respective property.

Therefore, the valid statements are (A), (C), and (E) only.

Lower pKa values indicate stronger acids. Electron-withdrawing groups such as NO2 or Cl increase acidity (lower pKa) by stabilizing the phenoxide ion. Electron-donating groups like CH3 decrease acidity (higher pKa) by destabilizing the phenoxide ion.

1. 2,4-Dinitrophenol has two strongly electron-withdrawing NO2 groups, making it the most acidic (lowest pKa).
2. 4-Nitrophenol has one NO2 group, so it is less acidic than 2,4-Dinitrophenol but still more acidic than 3-Chlorophenol or phenol.
3. 3-Chlorophenol has a moderate electron-withdrawing Cl group, making it less acidic than nitrophenols but more acidic than phenol.
4. Phenol itself is less acidic than the above derivatives with electron-withdrawing groups.
5. 2,4,5-Trimethylphenol has electron-donating CH3 groups, making it the least acidic (highest pKa).

Final Order: 2,4-Dinitrophenol < 4-Nitrophenol < 3-Chlorophenol < Phenol < 2,4,5-Trimethylphenol

(A) Hoffmann Degradation uses Br2 and NaOH to convert amides into amines, so (A) → (III).
(B) Clemenson Reduction uses Zn-Hg and HCl to convert carbonyl compounds into alkanes, so (B) → (IV).
(C) Cannizzaro Reaction uses concentrated KOH (heat) to disproportionate aldehydes without alpha-hydrogen, so (C) → (I).
(D) Reimer-Tiemann Reaction uses CHCl3 and NaOH / acidic workup to introduce a -CHO group into phenols, so (D) → (II).

First Step (Clemenson Reduction with Zn/Hg in HCl):
A ketone group -CO- in the starting compound (Ph-CH2-CO-CH2-CH2-NH2) is reduced to a -CH2- group.
Result: Ph-CH2-CH2-CH2-CH2-NH2

Second Step (LiAlH4 Reduction):
If an amide group were present, LiAlH4 would reduce it to -CH2-NH2, but here the main functionality was already converted in the first step. Thus, no further significant change occurs to the -NH2 group.

Hence, the major product (P) is Ph-CH2-CH2-CH2-CH2-NH2.

1. In the borax bead test, borax (Na2B4O7 · 10H2O) decomposes on heating to give sodium metaborate (NaBO2) and boric anhydride (B2O3).

2. Copper ions (from CuSO4) combine with boric anhydride (B2O3) in the oxidizing flame to form copper metaborate, Cu(BO2)2.

3. The compound Cu(BO2)2 imparts a blue-green colour to the bead in the oxidizing flame.

(A) Narrow Spectrum Antibiotic: Penicillin-G (III) is effective against a limited range of bacteria.
(B) Antiseptic: Furacin (I) is used to prevent infections in wounds and burns.
(C) Disinfectants: Sulphur Dioxide (II) is used as a disinfectant due to its antimicrobial properties.
(D) Broad Spectrum Antibiotic: Chloramphenicol (IV) is effective against a wide range of bacteria.

A cyclic tripeptide contains three amino acids linked end-to-end in a ring. With two distinct amino acids (A and B), the following unique sequences can form (in a cyclic arrangement):

• A-A-A
• A-A-B
• A-B-A
• A-B-B

This gives a total of 4 unique cyclic tripeptides.

The Lassaigne's test is used to detect nitrogen, sulfur, and halogens in organic compounds. Among the given options, CH3NH2·HCl contains both nitrogen (in the NH2 group) and a halogen (Cl from HCl) in an organic framework, making it capable of giving a positive test for both elements.

Other choices either do not qualify as organic compounds (NH4Cl) or do not release the correct species on fusion (N2H4·HCl, NH2OH·HCl).

1. Given pH = 12
 → pOH = 14 − 12 = 2
 → [OH⁻] = 10⁻² M

2. Calcium hydroxide (Ca(OH)₂) dissociates as:
Ca(OH)₂ → Ca²⁺ + 2 OH⁻
 [Ca(OH)₂] = [OH⁻]/2 = 10⁻²/2 = 5 × 10⁻³ M

3. Number of millimoles of Ca(OH)₂ in 100 mL = Molarity × Volume (in mL):
= 5 × 10⁻³ × 100 = 5 × 10⁻¹

4. Comparing with x × 10⁻¹, we see x = 5.

Calculate total valence electrons for each species:

• (A) NO₂: N (5) + 2 × O (2 × 6) = 5 + 12 = 17 (odd)
• (B) ICl₄⁻: I (7) + 4 × Cl (4 × 7) + 1 extra electron = 7 + 28 + 1 = 36 (even)
• (C) BrF₃: Br (7) + 3 × F (3 × 7) = 7 + 21 = 28 (even)
• (D) ClO₂: Cl (7) + 2 × O (2 × 6) = 7 + 12 = 19 (odd)
• (E) NO₂⁺: N (5) + 2 × O (2 × 6) − 1 electron (positive charge) = 5 + 12 − 1 = 16 (even)
• (F) NO: N (5) + O (6) = 11 (odd)

Species with an odd number of electrons: NO₂, ClO₂, NO.
Species with an even number of electrons: ICl₄⁻, BrF₃, NO₂⁺.

Hence, the count of species without odd electrons is 3.

1. For the reaction A + B → C + D, the net equilibrium constant K is given by:
K = K_f / K_r = 10³ / 10² = 10

2. Standard Gibbs free energy change is related to K by:
ΔrG° = – RT ln(K)

3. At T = 27°C (≈ 300 K) and R = 8.3 J K⁻¹ mol⁻¹:
ΔrG° = – (8.3 J mol⁻¹ K⁻¹) × (300 K) × ln(10)
ln(10) = 2.3
⇒ ΔrG° = – 8.3 × 300 × 2.3 J mol⁻¹
⇒ ΔrG° ≈ – 5727 J mol⁻¹ = – 5.727 kJ mol⁻¹

4. Rounding to the nearest integer, we get – 6 kJ mol⁻¹. The question format asks for the magnitude (–) —— kJ mol⁻¹, so the answer is 6.

Step 1: Determine the molality from boiling point elevation.
The elevation in boiling point (ΔT_b) is given by:
ΔT_b = K_b × m
Here, the boiling point elevation is 0.15°C, and K_b = 0.5 K kg mol⁻¹.
So, 0.15 = 0.5 × m ⇒ m = 0.15 / 0.5 = 0.3 mol/kg.

Step 2: Total molality after adding NaCl.
NaCl dissociates into two ions (Na⁺ and Cl⁻). When 0.2 mol of NaCl is added to 1 kg of solution, the contribution to molality is 0.2 × 2 = 0.4 mol/kg.
Initial molality from lead nitrate solution = 0.3 mol/kg
Total molality (m_total) = 0.3 + 0.4 = 0.7 mol/kg

Step 3: Use freezing point depression to find the new total molality.
The observed freezing point depression (ΔT_f) is 0.8°C, and K_f = 1.8 kg mol⁻¹.
ΔT_f = K_f × m_total ⇒ 0.8 = 1.8 × m_total
m_total = 0.8 / 1.8 ≈ 0.444 mol/kg

Notice there is a discrepancy: initially, we calculated total molality as 0.7 from boiling point data plus added salt, but from freezing point data, the measured total molality is 0.444. This indicates that PbCl₂ has precipitated due to the common ion effect of added Cl⁻. The effective molality in the final solution is 0.444 mol/kg.

Step 4: Relate to the solubility of PbCl₂ (K_sp).
Lead chloride dissociates as:
PbCl₂ ⇌ Pb²⁺ + 2Cl⁻
Let the molarity (or effective concentration) of PbCl₂ in solution be s mol/L. Then [Pb²⁺] = s and [Cl⁻] = 2s.
K_sp = [Pb²⁺] × [Cl⁻]² = s × (2s)² = 4s³.

From the freezing point measurement, the total ionic molality is 0.444 mol/kg. For every 1 mol of PbCl₂ dissolved, we get 1 mol of Pb²⁺ and 2 mol of Cl⁻, i.e. 3 total moles of ions per mole of PbCl₂.
So, if s is the concentration of PbCl₂ in mol/kg, then 3s = 0.444 ⇒ s = 0.444 / 3 = 0.148 mol/kg (approximating kg water ≈ 1 L for dilute solutions).

Step 5: Calculate K_sp.
K_sp = 4 (0.148)³.
(0.148)³ ≈ 0.00323
4 × 0.00323 ≈ 0.01292
This can be written as about 1.29 × 10⁻² = 12.9 × 10⁻³ = 13 × 10⁻³ which can be adjusted to 13 × 10⁻⁶ depending on how the question formats the power of ten notation.
Thus, the final nearest integer for the value of K_sp is 13 when expressed as (some number) × 10⁻⁶.

Consider the decomposition: 2 H2O(g) → 2 H2(g) + O2(g).
Let alpha be the fraction of water that decomposes.

If we start with 1 bar of H2O, then at equilibrium:
Partial pressure of H2O ≈ 1 - alpha
Partial pressure of H2 = 2 × (alpha/2) = alpha (because for every 2 H2O, we get 2 H2)
Partial pressure of O2 = alpha/2 (because for every 2 H2O, we get 1 O2)

The equilibrium constant Kp is 2 × 10^-3 = (PH2 × (PO2)^(1/2)) / PH2O
Substituting approximate partial pressures and assuming alpha is small, 1 - alpha ≈ 1:
Kp ≈ (alpha × (alpha/2)^(1/2)) / 1 = 2 × 10^-3

Simplifying gives alpha ≈ 2 × 10^-2, which is 2%. Thus, about 2% of water decomposes.

Two electrolytes are shown: A (typically a strong electrolyte) and B (typically a weak electrolyte). For a strong electrolyte, molar conductance vs. sqrt(c) is approximately linear, and we can extrapolate to get the limiting molar conductance. For a weak electrolyte, the plot is not linear and tends to level off, indicating incomplete dissociation at higher concentrations.

To check each statement:

• (A) For a strong electrolyte (A), Lambda°m can indeed be obtained by straight-line extrapolation. This is typically correct.
• (B) For a weak electrolyte (B), the plot of sqrt(c) vs. Lambda(m) is generally not a perfect straight line. So claiming it has a straight-line intercept equal to Lambda°m can be incorrect.
• (C) At infinite dilution, a weak electrolyte (B) is almost fully dissociated, so the degree of dissociation approaches 1, not zero. Hence this statement can be incorrect.
• (D) One can calculate limiting molar conductivity by summing the limiting ionic conductivities, which is typically correct for both strong and weak electrolytes.

Therefore, two statements are incorrect (commonly (B) and (C) in this context), giving us 2 as the number of incorrect statements.

From the rate vs. time plot:

• Region-I shows a linear dependence on concentration, typical of first-order behavior.
• Region-II shows non-linear behavior, suggesting a fractional order between 0 and 1 (for example, 0.1 to 0.9).
• Region-III levels off (rate becomes constant), consistent with zero-order kinetics.

Checking statements:

• (A) is not necessarily correct, because the overall order changes with region.
• (B) is not correct; we can determine or at least approximate the order in each region.
• (C) is correct: region-I is first order, region-III is zero order.
• (D) is incorrect: region-II is not first order.
• (E) is correct: region-II shows fractional (0.1 to 0.9) order behavior.

Hence, (C) and (E) are the correct statements, giving a total of 2 correct statements.

1. W(CO)6 (Tungsten hexacarbonyl) has an octahedral geometry with all carbonyl ligands terminal, so it has 0 bridging carbonyls.

2. Mn2(CO)10 (Manganese decacarbonyl) features a Mn–Mn bond, but all 10 carbonyl ligands are also terminal, giving 0 bridging carbonyls.

Therefore, the sum of bridging carbonyls is 0 + 0 = 0.

1. The retardation factor (Rf) is given by (distance traveled by solute) / (distance traveled by solvent front).

2. If the solvent front traveled 6 cm and compound A traveled 3.6 cm, then
Rf = 3.6 / 6 = 0.6.

3. Converting 0.6 to the form (x × 10⁻¹) means 0.6 = 6 × 10⁻¹.
Hence, the value is 6.

1. The molecular formula is C10H16. The degree of unsaturation (or index of hydrogen deficiency) is [2C + 2 - H]/2 = [2(10) + 2 - 16] / 2 = 3. This suggests three rings or double bonds in total.

2. The hydrocarbon absorbed 8.40 mL of H2 at standard conditions (approx. 22.4 L per mole). This supports the presence of multiple pi bonds. Ozonolysis data also indicates 3 double bonds.

Therefore, the hydrocarbon has 3 double bonds.