Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The order of bond length is

  • O2 < O3 < O22-

  • O2 < O22- < O3

  • O22- < O3 < O2

  • O2 = O22- > O3


A.

O2 < O3 < O22-

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).


2.

In P versus V graph, the horizontal line is found in which exists

  • Gas

  • Liquid

  • Equilibrium between gas and liquid

  • Super critical temperature


C.

Equilibrium between gas and liquid

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.


3.

Inert gases are mixed in iodine vapours. Then there are _____ between them.

  • H-bonding

  • van der Waals forces

  • electrostatic forces

  • metallic bonds


B.

van der Waals forces

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.


4.

Critical temperatures for A, B, C and D gases are 25°C, 10°C, -80°C and 15°C respectively. Which gas will be liquefied more easily?

  • A

  • B

  • C

  • D


A.

A

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

TC8a27Rb 

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


5.

Largest difference in radii is found in case of the pair

  • Li, Na

  • Na, K

  • K, Rb

  • Rb, Cs


B.

Na, K

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: 

Li152 < Na186 < K227 < Rb248 < Cs265


6.

X-rays are emitted during

  • α, n reaction

  • K-capture

  • n, α-reaction

  • β- emission 


B.

K-capture

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- n).


7.

During estimation of nickel, we prepare nickel dimethylglyoxime, a scarlet red solid. This compound is

  • ionic

  • covalent

  • metallic

  • non-ionic complex


D.

non-ionic complex

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


8.

During titration of acetic acid with aqueous NaOH solution, the neutralisation graph has a vertical line. This line indicates

  • alkaline nature of equivalence

  • acidic nature of equivalence

  • neutral nature of equivalence

  • depends on experimental proceeding


A.

alkaline nature of equivalence

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 → CH3COONaSodium acetate + H2O

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


9.

Calculate change in internal energy if H = - 92.2 kJ, P = 40 atm and V = -1L.

  • -42 kJ

  • -88 kJ

  • +88 kJ

  • +42 kJ


B.

-88 kJ

Change in internal energy

H = E + PV

E = H - PV = -92.2 - 40 × (-1) × 101 × 10-3

      = -92.2 + 4.04

      = -88.16 kJ ≈ -88 kJ


10.

The pH of the solution obtained on neutralisation of 40 mL 0.1 M NaOH with 40 mL 0.1 M CH3COOH is

  • 7

  • 8

  • 6

  • 3


B.

8

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.