Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

For the reaction,

H2O(l)H2O(g)

at 373 K and 1 atm pressure

  • H = 0

  • E = 0

  • H = TS

  • H = E


C.

H = TS

G=H-TS

For the reaction,

H2O(l)H2O(g)

At equilibrium,G = 0

Therefore,H=TS


2.

0.1 mol HCl is equal to

  • 3.65 g

  • 36.5 g

  • 18 g

  • 1.8 g


A.

3.65 g

Atomic weight of Chlorine = 35.5 gm

Atomic weight of Hydrogen = 1

Molecular mass of HCl = (35.5 + 1) gm = 36.5 gm

As 1 mole is the amount of the substance which has mass equal to gram molecular mass. Therefore, 1 mol HCl is equal to 36.5 gm and 0.1 mol HCl is equal to 3.65 gm.


3.

An sp3-hybrid orbital contains

  • 14 s- character

  • 12 s- character

  • 23 s-character

  • 34 s-character


A.

14 s- character

Each sp3- hybrid orbital has 25% or14s- character and 75% or34p- character.


4.

The geometry of XeF6 is

  • planar hexagon

  • regular octahedron

  • distorted octahedron

  • square bipyramid


C.

distorted octahedron

The geometry of XeF6 is distorted octahedral in which all the six positions are occupied by fluorine atoms and the lone pair of electrons of Xe atom is present at the comer of one of the triangular faces.


5.

Hybridisation states of C in CH3+ and CH3- are

  • sp2, sp3

  • sp3, sp2

  • sp, sp2

  • sp2, sp


A.

sp2, sp3

In CH3+= lp + bp = 0 + 3 = 3

So, the hybridisation is sp2.

InCH3-= lp + bp = 1+ 3 = 4

So, the hybridisation is sp3.

Therefore, the correct answer is a.


6.

10-6 M NaOH is diluted 100 times. The pH of the diluted base is

  • between 7 and 8

  • between 5 and 6

  • between 6 and 7

  • between 10 and 11


A.

between 7 and 8

[OH-] in the diluted base =10-6102= 10-8

Total [OH-] = 10-8 + [OH-] of water

= (10-8 + 10-7) M

= 10-8 [1 + 10] M

= 11 x 10-8 M

pOH = -log (11 x 10-8)

= - log 11 + 8 log 10

= 6.9586

pH = 14 - 6.9586 = 7.0414


7.

A solution made by dissolving 40 g NaOH in 1000 g of H2O is

  • 1 molar

  • 1 normal

  • 1 molal

  • None of these


C.

1 molal

Number of moles NaOH (n) =Weightingmolarmass=4040= 1

As molality of a solution is defined as the number of moles of the solute dissolved in 1000 g (1 kg) of the solvent, thus the given solution of NaOH is 1 molal solution.


8.

Which one of the following complexes is outer orbital complex?

  • [Co(NH3)6]3+

  • [Mn(CN)6]4+

  • [Fe(CN)6]4-

  • [Ni(NH3)6]2+


D.

[Ni(NH3)6]2+

Generally weaker field ligands form outer orbital complex. Between [Co(NH3)6]3+ and [Ni(NH3)6]2+, the later form outer orbital complex because of its d8configuration. (The configuration of Co3+ = d6).


9.

Which of the following has the maximum number of unpaired electrons?

  • V3+

  • Fe2+

  • Mn2+

  • Cu+


C.

Mn2+

The outer electronic configuration of the given ions is as


10.

The correct statement with regard toH2+andH2-is

  • both H2+ and H2- are equally stable

  • both H2+ and H2- do not exist

  • H2- is more stable than H2+

  • H2+ is more stable than H2-


D.

H2+ is more stable than H2-

The correct statement with regard toH2+andH2-is, former is more stable than later.

H2+:σ1s1,σ1s°Bondorder=12H2-:σ1s2,σ1s1Bondorder=2-12=12

The bond order of H2+ and H2- are same.It is due to the presence of one electron in the antibonding molecular orbital in H2-.