The 15 one-mark questions from the March 2026 12th Standard Chemistry public exam, in paper order, with the correct option marked and a short explanation.
Q1
Electrochemical process is used in the extraction of which of the following element ?
- A. SodiumCorrect
- B. Iron
- C. Silver
- D. Lead
Explanation. Sodium is a highly electropositive metal that cannot be reduced by carbon, so it is extracted by electrolysis of fused NaCl (Down's process). Iron and lead are obtained by carbon reduction and silver by cyanide leaching.
Q2
The stability of \(+1\) oxidation state increases in the sequence ________.
- A. In < Tl < Ga < Al
- B. Al < Ga < In < TlCorrect
- C. Ga < In < Al < Tl
- D. Tl < In < Ga < Al
Explanation. Down group 13 the inert pair effect increases, so the ns² electrons are less readily used in bonding. Hence the stability of the +1 state increases Al < Ga < In < Tl.
Q3
\(XeF_6\) on complete hydrolysis produces :
- A. \(XeO_3\)Correct
- B. \(XeOF_4\)
- C. \(XeO_2\)
- D. \(XeO_2F_2\)
Explanation. Partial hydrolysis gives XeOF₄ and XeO₂F₂, but complete hydrolysis gives xenon trioxide: XeF₆ + 3H₂O → XeO₃ + 6HF.
Q4
Which of the following statements is incorrect ?
- A. \(Co^{3+}\) and \(Fe^{2+}\) have same number of unpaired electrons.
- B. Of all the known elements, Silver has the highest electrical conductivity at room temperature.
- C. To form a substitute alloy, the difference between the atomic radii of solvent and solute should be less than 15%.
- D. If the standard electrode potential (E°) of a metal is large and negative, the metal is a powerful oxidising agent.Correct
Explanation. Co³⁺ and Fe²⁺ are both 3d⁶ with 4 unpaired electrons, silver is the best conductor, and the 15% rule (Hume-Rothery) applies to substitutional alloys, so (a)-(c) are correct. A metal with a large negative E° loses electrons easily and is a powerful reducing agent, not an oxidising agent, so (d) is incorrect.
Q5
How many geometrical isomers are possible for \([Pt(Py)(NH_3)(Br)(Cl)]\) ?
- A. 0
- B. 3Correct
- C. 15
- D. 4
Explanation. Pt(II) complexes are square planar. A complex of type [Mabcd] has three geometrical isomers, depending on which ligand is trans to a chosen ligand (e.g. Py trans to NH₃, Br or Cl).
Q6
Assertion : Due to Frenkel defect, density of the crystalline solid decreases.
Reason : In Frenkel defect, cation and anion leaves the crystal.
- A. Assertion is true but Reason is false.
- B. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
- C. Both Assertion and Reason are false.Correct
- D. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Explanation. In Frenkel defect a cation only moves from its lattice site to an interstitial site; no ion leaves the crystal, so the density does not change. Both the assertion and the reason are false (loss of cation-anion pairs is Schottky defect).
Q7
The half life period of a radioactive element is 140 days. After 560 days 1 g of element will be reduced to :
- A. \(\left(\frac{1}{8}\right) g\)
- B. \(\left(\frac{1}{2}\right) g\)
- C. \(\left(\frac{1}{16}\right) g\)Correct
- D. \(\left(\frac{1}{4}\right) g\)
Explanation. Number of half-lives n = 560/140 = 4. Amount left = 1 g × (1/2)⁴ = 1/16 g.
Q8
The pH of sodium formate solution in terms of \(K_a\) and the concentration of the electrolyte is :
- A. \(pH = 7 - \frac{1}{2} pK_a - \frac{1}{2} \log C\)
- B. \(pH = 7 - \frac{1}{2} pK_a + \frac{1}{2} \log C\)
- C. \(pH = 7 + \frac{1}{2} pK_a + \frac{1}{2} \log C\)Correct
- D. \(pH = 7 + \frac{1}{2} pK_a - \frac{1}{2} \log C\)
Explanation. Sodium formate is a salt of a weak acid and a strong base; its hydrolysis gives a basic solution with pH = 7 + ½pKa + ½log C.
Q9
The molar conductivity of a \(0.5\ mol\ dm^{-3}\) solution of \(AgNO_3\) with specific conductance \(5.76 \times 10^{-3}\ S\,cm^{-1}\) at 298 K is :
- A. \(0.086\ S\,cm^2\,mol^{-1}\)
- B. \(2.88\ S\,cm^2\,mol^{-1}\)
- C. \(28.8\ S\,cm^2\,mol^{-1}\)
- D. \(11.52\ S\,cm^2\,mol^{-1}\)Correct
Explanation. Λm = κ × 1000 / C = (5.76 × 10⁻³ S cm⁻¹ × 1000 cm³ dm⁻³) / 0.5 mol dm⁻³ = 11.52 S cm² mol⁻¹.
Q10
On which of the following properties does the coagulating power of an ion depend ?
- A. The magnitude of the charge on the ion alone.
- B. Both magnitude and sign of the charge on the ion.Correct
- C. The sign of charge on the ion alone.
- D. Size of the ion alone.
Explanation. By the Hardy–Schulze rule, only an ion with charge opposite to that of the colloidal particles causes coagulation (sign), and its coagulating power increases with the magnitude of its charge.
Q11
Propan-2-ol on reaction with anhydrous \(ZnCl_2\) and concentrated HCl gives ________.
- A. 1-chloropropane
- B. Ethyl chloride
- C. 2-chloro-2-methyl propane
- D. 2-chloropropaneCorrect
Explanation. Lucas reagent replaces the –OH group by –Cl without changing the carbon skeleton: CH₃CH(OH)CH₃ + HCl → CH₃CHClCH₃ (2-chloropropane) + H₂O.
Q12
\(CH_2{=}CH_2 \xrightarrow[(ii)\ Zn/H_2O]{(i)\ O_3} X \xrightarrow{NH_3} Y\)
'Y' is :
- A. Hexamethylene tetramineCorrect
- B. Formaldehyde
- C. Oxime
- D. Diacetone ammonia
Explanation. Reductive ozonolysis of ethene gives two molecules of formaldehyde (X). Formaldehyde condenses with ammonia: 6HCHO + 4NH₃ → (CH₂)₆N₄ (hexamethylene tetramine, urotropine) + 6H₂O.
Q13
\(C_6H_5NO_2 \xrightarrow{Sn/HCl} A \xrightarrow[273\,K]{NaNO_2/HCl} B \xrightarrow[283\,K]{H_2O} C\)
'C' is :
- A. \(C_6H_5CHO\)
- B. \(C_6H_5OH\)Correct
- C. \(C_6H_5NH_2\)
- D. \(C_6H_5CH_2OH\)
Explanation. Sn/HCl reduces nitrobenzene to aniline (A); diazotisation at 273 K gives benzene diazonium chloride (B); warming with water at 283 K hydrolyses it to phenol (C), C₆H₅OH.
Q14
\(\alpha\)-D(+) Glucose and \(\beta\)-D(+) Glucose are :
- A. Enantiomers
- B. Epimers
- C. Conformational isomers
- D. AnomersCorrect
Explanation. α- and β-D-glucose differ only in the configuration at C-1, the anomeric (hemiacetal) carbon of the cyclic form, so they are anomers.
Q15
Drugs that bind to the receptor site and inhibit its natural function are called ________.
- A. enzymes
- B. antagonistsCorrect
- C. molecular targets
- D. agonists
Explanation. Antagonists bind to the receptor site and block its natural function, whereas agonists mimic the natural messenger and activate the receptor.
