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Subject

Chemistry

Class

NEET Class 12

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NEET Chemistry Solved Question Paper 2005

Multiple Choice Questions

The isoelectronic pair is

Cl₂O, ICl₂^-
ICl₂^-, ClO₂
IF₂⁺, I₃^-
ClO₂^-, ClF₂⁺

ClO₂^-, ClF₂⁺

Among all the given options, isoelectronic pair is ClO₂^-, ClF₂⁺.

Number of electrons in ClO₂^- = 7 + 6 + 6 + 1 = 20

Number of electrons in ClF₂⁺ = 7 + 7 + 7 - 1 = 20

${}_{92}^{238}U$ emits 8 α-particles and 6 β-particles. The neutron/proton ratio in the product nucleus is

60/41
61/40
62/41
61/42

62/41

The neutron/ proton ratio in the product nucleus is 62/41. ₉₂U²³⁸ $\to_{- 6 β}^{- 8 α}$ ₈₂X²⁰⁶

Number of protons = 82

Number of neutrons = 124

Neutron/proton ratio in the product nucelus = $\frac{124}{82} = \frac{62}{41}$

α-Particles can be detected using

thin aluminium sheet
barium sulphate
zinc sulphide screen
gold foil

zinc sulphide screen

α-particles can be detected by using zinc sulphide screen. It is used as phosphor in detection.

An endothermic reaction with high activation energy for the forward reaction is given by the diagram

Endothermic reactions are those which involve absorption of heat. High activation energy means potential energy of product must be much greater than reactants.

In diborane, the two H - B - H angles are nearly

60°, 120°
95°, 120°
95°, 150°
120°, 180°

95°, 120°

In diborane, the two H - B - H angles are nearly 95°, 120°. The structure can be represented as -

$∆ H_{f}^{°}$ (298 K) of methanol is given by the chemical equation

CH_{4 (g)} + $\frac{1}{2}$ O_{2 (g)}→CH₃OH_(g)
C (graphite) + $\frac{1}{2}$ O_{2 (g)} + 2H₂ → CH₃OH_(l)
C (diamond) + $\frac{1}{2}$ O₂_(g) + 2H₂_(g) → CH₃OH_(l)
CO_(g) + 2H₂ → CH₃OH_(l)

C (graphite) + $\frac{1}{2}$ O_{2 (g)} + 2H₂ → CH₃OH_(l)

Methanol can be prepared synthetically by heating carbon monoide and hydrogen gases under pressure in the presence of a catalyst.

C (graphite) + $\frac{1}{2}$ O_2(g) → CO_(g) ...(i)

CO_(g) + 2H_2(g) → CH₃OH_(l) ...(ii)

From (i) and (ii)

C (graphite) + $\frac{1}{2}$ O_2(g) + 2H₂_(g) → CH₃OH_(l)

When 10 mL of 0.1 M acetic acid (pK_a= 5.0) is titrated against 10 ml of 0.1 M ammonia solution (pK_b= 5.0), the equivalence point occurs at pH

7.0

pK_a = -logK_a, pK_b = -logK_b

$pH = - \frac{1}{2} [\log K_{a} + \log K_{w} - \log K_{b}] = - \frac{1}{2} [5 \log (1 \times 10^{- 14}) - (- 5)] = - \frac{1}{2} [5 - 14 - 5] = - \frac{1}{2} (- 14) = 7$