Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

Which is correct regarding size of atom?

  • N < O

  • B < Ne

  • V > Ti

  • Na > K


B.

B < Ne

The atomic radii of noble gases are by far the largest in their respective periods. This is due to the reason that noble gases have only van der Waals radii.


2.

Choose the correctly paired gaseous cation and its magnetic (spin only) moment (in B.M.)

  • Ti2+,3.87 B.M.

  • Cr2+, 4.90 B.M.

  • Co3+,3.87 B.M.

  • Mn2+,4.90 B.M.


B.

Cr2+, 4.90 B.M.

 Using expression, µ = n(n+2 B.M. (where, n = no. of unpaired electrons)

Ion Outer configuration     n    μ
Ti2+22   3d2      2    2.84
C24r2+   3d4      4    4.90
C27o3+   3d6      4    4.90
M22n2+   3d5      5    5.92


3.

The equilibrium constant for the reaction,

12H2(g)+12I2(g)HI(g) is Kc

Equilibrium constant for the reaction 2HI(g) H2(g) + I2(g)  will be

  • 1/Kc

  • 1/(Kc)2

  • 2/Kc

  • 2/(Kc)2


B.

1/(Kc)2

The given reaction is,

12H2(g) +12I2(g) HI(g)          ...(i)

Hence, KC=[HI][H2]1/2[I2]1/2    ...(ii)

Now, reverse the equaion (i) and multiple by 2, we get

2HI H2(g)+I2(g)

      Hence, KC'=[H2] [I2][HI]2                    

 Equating equations (ii) and (iii), we get

             KC'=1(KC)2


4.

A diatomic gas at pressure P, compressed adiabatically to half of its volume, what is the final pressure?

  • (2)1.4P

  • P/(2)1.4

  • (2)5/3P

  • P/(2)5/3


A.

(2)1.4P

For adiabatic condition, PVγ = Constant 

P1V1γ = P2V2γ ; V212V1

P2 = P1V1V2γ    [For diantomic gas γ = 1.4]

P2 = P1V1 ×2V11.4

P2P1(2)1.4   = (2)1.4 P


5.

In Omolecule, the formal charge on the central O-atom is

  • 0

  • -1

  • -2

  • +1


D.

+1

Lewis gave the structure of O, molecule as

Using the relation,
Formal charge = [Total no. of valence electrons in the free atom] - [Total no. of non-bonding (lone pair) electrons]- 12 [Total no. of bonding (shared) electrons]
The formal charge on central O -atom i.e., no. 1

=6-2- -(6)=+1


6.

What is the oxidation number of Br in KBrO?

  • + 6

  • + 7

  • + 5

  • + 8


B.

+ 7

Let the oxidation no. of Br be x.

In KBrO4, + 1 + x + 4 (-2) =0,- 7 + x = 0, x=+7


7.

Which of the following pairs represent isotones?

  • As3377,Se34 78

  • Pt78195,Os76190

  • Ag47108,Cd48112

  • Hf72178,Ba56137


A.

As3377,Se34 78

Isotones have the same number of neutrons.
As = 77 - 33 = 44 ; Se = 78 - 34 = 44


8.

Which of the following is not sp2 hybridised?

  • Graphite

  • Graphene

  • Fullerene

  • Dry ice


D.

Dry ice

Solid CO2 is dry ice in which carbon atomundergoes sp-hybridisation.


9.

Which ofthe following statements is incorrect?

  • Li+ has minimum degree of hydration.

  • The oxidation state of K in KO2 is + 1.

  • Na is used to make a Na/Pb alloy.

  • MgSO4 is readily soluble in water.


A.

Li+ has minimum degree of hydration.

The hydration enthalpies of alkali metal ions decreases with increase in ionic sizes Hence, the order is Li+ > Na+> K> Rb+ > Cs+.Therefore, Li+ has maximum degree of hydration.


10.

Which of the following represents the correct bond order?

  • O2+<O2->O22-

  • O2->O22->O2+

  • O22->O2+>O2-

  • O2+>O2->O22-


D.

O2+>O2->O22-

Hence, the coorect Bond order is O2+>O2->O22-

Ion

Total no of electron

              MO configuration   Bond order
 o2+    15 KK σ2s2 σ*2s2  σ2px2 π2px2=π2py2π*2px1=π*2py0    2.5
 o2-    17 KK α2s2σ*2s2σ2pz2π2px2=π2py2π*2px2=π*2py1    1.5
 o22-    18 KKσ2s2 σ*2s2 σ2pz2 π2px2=π2py2 π*2px2=π*2py2    1.0