Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in gram) of the remaining product at STP will be

  • 1.4

  • 3.0

  • 4.4

  • 2.8


D.

2.8

HCOOH Dehydrating AgentH2SO4 CO +   H2O Water is abosrbed by sulphuric acid(Moles)i = 2.346 = 120          0           0(moles)f =  0                         120     120H2C2O4        H2SO4         CO + CO2 +H2OWater is abosrbed by sulphuric acid(Moles)i = 4.590 = 120          0           0       0(moles)f =  0                         120     120        120

CO2 is absorbed by KOH.
So the remaining product is only CO.
moles of CO formed from both reaction
=120+120 = 110

Let mass of CO = moles x molar mass

 =  = 110 x  28


12.

Among CaH2, BeH2, BaH2, the order of ionic character is

  • BeH2< CaH2<BaH2

  • CaH2<BeH2<BaH2

  • BaH2<BeH2<CaH2

  • BeH2 < BaH2 < CaH2


A.

BeH2< CaH2<BaH2

Smaller the size cation, more will be its polarising power. Hence, BeH2 will be least ionic.

Or

On moving down the group metallic character of metals increases so ionic character of metal hydride increases.


13.

Which of the following statements is not true for halogens?

  • All form monobasic oxyacids

  • All are oxidizing agents

  • Chlorine has the highest electron-gain enthalpy

  • All but fluorine show positive oxidation states


D.

All but fluorine show positive oxidation states

Due to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluoric (I) acid the oxidation number of F is +1 in HOF.


14.

For the redox reaction

MnO4- + C2O4-2 + H+    Mn2+ + CO2 + H2O

The correct coefficients of the reactants for the balanced equation are

  • MnO4- C2O42- H+
    16 5 2
  • MnO4- C2O42- H+
    2 5 16
  • MnO4- C2O42- H+
    5 16 2
  • MnO4- C2O42- H+
    2 16 5

B.

MnO4- C2O42- H+
2 5 16

Mn+7O4-    M+2n2+ ; 5e- gain ... (i)C+32O42-    C+4O2 ;2e- loss ... (ii)Multiplying (i) and 2 and (ii)by 5 to balance e-2MnO4- + 5C2O42-  2Mn2+ + 10CO2On balancing charge;2MnO4- + 5C2O42-  + 16H+  2Mn2+ + 10 CO2 + 8 H2O


15.

Considering the Ellingham diagram, which of the following metals can be used to reduce alumina?

  • Fe

  • Zn

  • Cu

  • Mg


D.

Mg

The metal which is more reactive than 'Al' can reduce alumina. Also, according to the Ellingham diagram, Mg has more –ΔG value then alumina. Metals which has more –ΔG value can reduce those metals oxide which has less –ΔG value.


16.

Which one of the following elements is unable to form MF6-3 ion?

  • Ga

  • Al

  • In

  • B


D.

B

The element M in the complex-ion MF6-3 has a coordination number of six.

B has only s  and p-orbitals and no d – orbitals available, therefore, at the maximum it can show a coordination number of 4. Thus, B cannot form the complex of the type MF6-3


17.

Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :

Column I Column II
a. Co3+ i. 8 BM
b. Cr3+ ii. 35 BM
c. Fe3+ iii. 3 BM
d. Ni2+ iv. 24 BM
    v. 15 BM

  • a b c d
    iv v ii i
  • a b c d
    i ii iii iv
  • a b c d
    iii v i ii
  • a b c d
    iv i ii iii

A.

a b c d
iv v ii i

Co3+ = [Ar] 3d6, Unpaired e(n) = 4
Spin magnetic moment 4(4+2) = 24 BM

Cr3+ = [Ar] 3d3, Unpaired e(n) = 3
Spin magnetic moment = 3(3+2) = 15 BM

Fe3+ = [Ar] 3d5, Unpaired e–(n) = 5
Spin magnetic moment =5(5+2) = 35 BM

Ni2+ = [Ar] 3d8, Unpaired e(n) = 2
Spin magnetic moment = 2(2+2) = 8 BM 


18.

Which of the following is correct with respect to – I effect of the substituents? (R = alkyl)

  • -NH2 <- OR < - F

  • -NR2 <- OR <-F

  • -NR2 >-OR>-F

  • -NH2>-OR>-F


A.

-NH2 <- OR < - F

-I effect increases on increasing electronegativity of the atom. So, the correct order of -I effect is

-NH2 <-OR < -F

Also, 

-NR2<-OR<-F

 


19.

In the structure of ClF3, the number of lone pair of electrons on central atom ‘Cl’ is

  • One

  • Two

  • Three

  • Four


B.

Two

The structure of ClF3 is

The number of lone pair of electrons on central Cl is 2.


20.

In which case is a number of molecules of water maximum?

  • 18 mL of water

  • 0.18 g of water

  • 10-3 mol of water

  • 0.00224 L of water vapours at 1 atm and 273K


A.

18 mL of water

a) Mass of water = 18 x 1 = 18 g

Molecules of water = mole x NA

=1818NA = NA

b) Molecules of water = mole x NA

=0.1818 NA =10-2 NA

c) Molecules of water = mole x NA = 10-3 NA
d) Moles of water = 0.00224/22.4 = 10-4

Molecules of water = mole x NA = 10-4 NA