NEET Chemistry Solved Question Paper 2005

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

pK_a = -logK_a, pK_b = -logK_b

$$\begin{gathered} \mathrm{pH} = -\frac{1}{2}[\log {\mathrm{K}}_{\mathrm{a}} + \log {\mathrm{K}}_{\mathrm{w}} - \log {\mathrm{K}}_{\mathrm{b}}] \\ = -\frac{1}{2}[5 \log (1\times {10}^{-14})-(-5\left)\right] \\ = -\frac{1}{2}[5 - 14 - 5] \\ = -\frac{1}{2}(-14) = 7 \end{gathered}$$

F₂ is formed by reacting K₂MnF₆ with SbF₅.

K₂MnF₆ + 2SbF₅ → 2KSbF₆ + MnF₃ + $\frac{1}{2}$F₂

Among all the given options, option d is correct. All the 3 images have same number of lone pairs on Xe. Their structure are as follows- 

XeO₃ - 

XeOF₄ - 

XeF₆ - 

Among all the given options, option a is correct. [Ni(CN)₄]^2- is a dimagnetic species. [NiCl₄]^2-, [CoCl₄]^2-, [CoF₆]² are paramagnetic species.

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

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 

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) + 2H_2(g) → CH₃OH_(l)

The neutron/ proton ratio in the product nucleus is 62/41. ₉₂U²³⁸ $\underset{-6\mathrm{\beta }}{\overset{-8\mathrm{\alpha }}{\to }}$ 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 by using zinc sulphide screen. It is used as phosphor in detection.