Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

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

  • 60°, 120°

  • 95°, 120°

  • 95°, 150°

  • 120°, 180°


B.

95°, 120°

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


2.

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

  • 5.0

  • 6.0

  • 7.0

  • 9.0


C.

7.0

pKa = -logKa, pKb = -logKb

pH = -12[log Ka + log Kw - log Kb]      = -12[5 log (1×10-14)-(-5)]      = -12[5 - 14 - 5]       = -12(-14) = 7


3.

F2 formed by reacting K2MnF6 with

  • SbF5

  • MnF3

  • KSbF6

  • MnF4


A.

SbF5

F2 is formed by reacting K2MnF6 with SbF5.

K2MnF6 + 2SbF5 → 2KSbF6 + MnF312F2


4.

Among the following molecules

(i) X..eO3 (ii) X..eOF4 (iii) X..eF

those having same number of lone pairs on Xe are

  • (i) and (ii) only

  • (i) and (iii) only

  • (ii) and (iii) only

  • (i), (ii) and (iii)


D.

(i), (ii) and (iii)

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


5.

The diamagnetic species is

  • [Ni(CN)4]2-

  • [NiCl4]2-

  • [CoCl4]2-

  • [CoF6]2-


A.

[Ni(CN)4]2-

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.


6.

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


C.

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


7.

The isoelectronic pair is

  • Cl2O, ICl2-

  • ICl2-, ClO2

  • IF2+, I3-

  • ClO2-, ClF2+


D.

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 


8.

Hf° (298 K) of methanol is given by the chemical equation

  • CH4 (g)12O2 (g) →CH3OH(g)

  • C (graphite) + 12O2 (g) + 2H2 → CH3OH(l)

  • C (diamond) + 12O2(g) + 2H2(g) → CH3OH(l)

  • CO(g) + 2H2 → CH3OH(l)


B.

C (graphite) + 12O2 (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) + 12O2(g)  → CO(g)  ...(i)

CO(g) + 2H2(g) → CH3OH(l)         ...(ii)

From (i) and (ii)

C (graphite) + 12O2(g) + 2H2(g) → CH3OH(l)


9.

U92238 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


C.

62/41

The neutron/ proton ratio in the product nucleus is 62/41. 92U238 -6β-8α 82X206

Number of protons = 82

Number of neutrons = 124

Neutron/proton ratio in the product nucelus = 12482 = 6241


10.

α-Particles can be detected using

  • thin aluminium sheet

  • barium sulphate

  • zinc sulphide screen

  • gold foil


C.

zinc sulphide screen

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