Chemistry 2006 Jamb Examination

Across a period from left to right the atomic radius generally decreases while the effective nuclear charge (pull on valence electrons) increases. Stronger nuclear attraction holds electrons more tightly, so atoms are less likely to lose electrons — that is, they become less electropositive (less metallic).

Thus the combined effect (smaller size + larger nuclear charge) reduces electropositivity which matches option C. The second paragraph: this explains why metals on the left are electropositive while right-hand elements are not.

Correct method: sublimation → dissolution → filtration → evaporation. First gently heat the solid mixture: ammonium chloride (NH₄Cl) sublimes and leaves as gas, separating it from sand + NaCl. Collect or remove NH₄Cl (sublimation).

Then dissolve the remaining solid in water (dissolution) to dissolve NaCl but not sand; filter off the sand (filtration); finally evaporate the water to obtain solid NaCl (evaporation). This matches option C.

The planetary model of the atom (electrons orbiting the nucleus like planets around the sun) is attributed to Niels Bohr, who refined Rutherford's nucleus model by introducing quantized electron orbits (Bohr model) for hydrogen-like atoms.

Thomson proposed the "plum pudding" model; Rutherford discovered the nucleus; Dalton proposed atomic theory. The Bohr model is the standard answer (option C).

Using Charles's law (V ∝ T in Kelvin) the fractional increase = (V₂ − V₁)/V₁ = (T₂ − T₁)/T₁.

Compute with Kelvin:
T₁ = 50 + 273 = 323 K,
T₂ = 80 + 273 = 353 K.
Then fractional increase
= (353 − 323)/323 = 30/323 ≈ 0.09288 ≈ 0.093. \[ = \dfrac{353-323}{323} = \dfrac{30}{323} \approx 0.0929 \approx 0.093.\]

Therefore option B is correct under the 50°C→80°C interpretation.

Molar mass of ethyne (C₂H₂) = 2×12 + 2×1 = 26 g·mol⁻¹.
Moles of C₂H₂ burned = 78 / 26 = 3.0 mol.

Stoichiometry
1 mol C₂H₂ produces 2 mol CO₂ (each C₂H₂ has 2 C atoms → 2 CO₂).
So 3.0 mol C₂H₂ → 6.0 mol CO₂.
Mass of CO₂ = 6.0 × 44 g·mol⁻¹ = 264 g. 
Moles of C₂H₂\[= \dfrac{78}{26} = 3\ \text{mol};\quad n_{CO_2} = 2\times 3 = 6\ \text{mol}\]\[m_{CO_2}=6\times 44=264\ \text{g}\]

Hence option A (264 g) is correct. 

An oxidizing agent is a species that can accept electrons (i.e., become reduced). Among the options, ClO⁻ (hypochlorite) contains chlorine in a high oxidation state (+1), and it readily accepts electrons to be reduced to Cl⁻.

In contrast: Au³⁺ and Fe³⁺ can also be reduced, but they are less common oxidizers under normal conditions. Ca²⁺ is very stable and does not act as an oxidizing agent. Thus the best answer is option C.

For a reaction to be spontaneous, ∆G = - ve; ∆H = - ve; ∆S = + ve

 Where ∆G = ∆H - T∆S

Given data; ∆G = ?

∆H = - 72KJ/mol, 

T = (25+273)K = 298K

∆S = -106J/mol or - 0.106KJ/mol/K

: ∆G = -72 -(298 * -  0.106)

= - 72 + 31.588

∆G = - 40.412KJ/mol

One of the condition for spontaneity of a reaction is ∆G to be negative.

The traditional Leclanché cell uses a zinc rod as the anode (oxidation occurs) and a carbon rod surrounded by MnO₂ paste as the cathode. Hence the answer is option A.

Lead accumulator uses Pb/PbO₂ electrodes, Daniell cell uses Zn/Cu electrodes, and "voltaic cell" is a general term, not specific.

Electrical conductivity depends on the concentration of free ions in solution. Strong acids like HCl dissociate completely, whereas weak acids like CH₃COOH dissociate only partially.

Therefore, a concentrated solution of a strong acid will conduct the best. Option B (2.0 mol dm⁻³ HCl) gives the highest ion concentration and hence the highest conductivity.

In H₂S, each H has oxidation number +1. Let oxidation state of S = x:

\[2(+1) + x = 0 \implies x = -2.\]

In SO₃, S = +6; in SO₂, S = +4; in S₈, S = 0 (elemental state). Therefore only in H₂S is sulphur -2. Option C is correct.