NEET Chemistry Solved Question Paper 2007

Ozone molecule is V-shaped with O-O bond length 1.278 Å. It is intermediate between that for a single bond (1.48 Å as in H₂O₂) and for a double bond (1.21 Å as in O₂).

In the graph, at point A, CO₂ exists as a gas, as pressure is increased, the volume of the gas decreases along the curve AB. At B liquefaction of the gas starts. Hence, volume decreases rapidly along BC because liquid has much less volume than the gas. At point C, liquefaction is complete. Thus, along horizontal line, gas converts into liquid.

When inert gases are mixed in iodine vapours, they have van der Waals forces between them. These forces are strong but are only short range forces, for e.g. for surface of neutral molecules or inert gases. It increases with the size or surface are of the molecule. In halogens, it increases as F₂ > Cl₂ < Br₂ < I₂.

Critical temeperature of gas is defined as the temperature above which it cannot be liquefied, howsoever high pressure may be applied on the gas.

T_C = $\frac{8\mathrm{a}}{27\mathrm{Rb}}$ 

(where a = Van der Waal's constant. Greater the value of a more easily the gas can be liquefied and hence larger T_C means larger the value of a).

Atomic radii as well as ionic radii increases from Lithium (Li) to Francium (Fr) due to the presence of one extra shell of electron.

Metallic radii (in pm) of elements is shown as: 

$\underset{152}{\mathrm{Li}} < \underset{186}{\mathrm{Na}} < \underset{227}{\mathrm{K}} < \underset{248}{\mathrm{Rb}} < \underset{265}{\mathrm{Cs}}$

In a certain nucleus, the nucleus captures an electron from the K-shell (as it is near to the nucleus). The vacancy created is filled up with the electron from the higher shells thereby emitting X-rays. As a result of K-electron capture a proton in the nucleus is converted into a neutron (P⁺ + e^- $\to$n).

The planar complex is very poorly soluble and is used for the gravimetric determination of nickel e.g. ores.

In any type of acid-base titration, there is a sudden change in the pH value at the end point. During the titration of a weak acid with a strong base, beyond the equivalent point, the solution will contain the salt and excess of free base.

CH₃COOH + NaOH → $\underset{\mathrm{Sodium} \mathrm{acetate}}{{\mathrm{CH}}_3\mathrm{COONa}}$ + H₂O

Hence, due to the presence of free base, solution becomes alkaline in nature.

Change in internal energy

$∆\mathrm{H} = ∆\mathrm{E} + \mathrm{P}∆\mathrm{V}$

$∆\mathrm{E} = ∆\mathrm{H} - \mathrm{P}∆\mathrm{V}$ = -92.2 - 40 $\times$ (-1) × 101 × 10^-3

      = -92.2 + 4.04

      = -88.16 kJ ≈ -88 kJ

On neutralisation, the pH of 40 mL of 0.1 M NaOH with 40 mL of 0.1 M CH₃COOH gives us a basic solution having pH >7 as it forms CH₃COONa.