Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

Which of the following is electron deficient?

  • (CH3)2

  • (SiH3)2

  • (BH3)2

  • PH3


C.

(BH3)2

Boron is an element of 13 group andcontains three electrons in its valence shell.When its compound BH3 dimerises, each boron atom carry only 6e-, that is their octet is incomplete. Hence, (BH3)2 is an electron deficient compound. In all other given molecules octet of central atom is complete.


12.

Which one of the following molecules contain no π-bond?

  • CO2

  • H2O

  • SO2

  • NO2


B.

H2O

All the molecules have O atom with lone pairs, but in H2O the H atom has no vacant orbital for π bonding.That's why it does not have any π bond. ln all other given molecules, the central atom because of the presence of vacant orbitals is capable to farm π bonds.


13.

Which of these is not a monomer for a high molecular mass silicone polymer?

  • MeSiCl3

  • Me2SiCl2

  • Me3SiCl

  • PhSiCl3


C.

Me3SiCl

Me3SiCl is not a monomer for a high molecular mass silicone polymer because it generates Me3SiOH when subjected to hydrolysis which contains only one reacting site. Hence, the. polymerisation reaction stops just after first step.

               


14.

Identify the correct order of solubility in aqueous medium.

  • CuS > ZnS > Na2S

  • ZnS > Na2S > CuS

  • Na2S > CuS > ZnS

  • Na2S > ZnS > CuS


D.

Na2S > ZnS > CuS

Ionic compounds are more soluble in water or in aqueous medium. Ionic character  size of cation (if anion is same).

The order of size of cation is

            Na+> Zn2+> Cu2+

The order of ionic character and hence, of solubility in water is as
           Na2S > ZnS > CuS


15.

Which of the following statements about the interstitial compounds is incorrect?

  • They retain metallic conductivity

  • They are chemically reactive

  • They are much harder than the pure metal

  • They have higher melting points than the pure metal


B.

They are chemically reactive

Interstitial compounds are obtained when small atoms like H, B, C, N etc., fit into the lattice of other elements. These retain metallic conductivity. These resemble the parent metal in chemical properties (reactivity) but differ in physical properties like hardness, melting point etc.


16.

The basic structural unit of silicates is

  • SiO-

  • SiO44-

  • SiO32-

  • SiO42-


B.

SiO44-

The basic building unit of all silicates is the tetrahedral SiO44-. It is represented as

                        

 


17.

Which of the following structure is similar to graphite?

  • BN

  • B

  • B4C

  • B2H6


A.

BN

Boron nitride, (BN)X resembles with graphite in structure as


18.

The structure of isobutyl group in an organic compound is

  • C2H6-CH-CH2-

  • CH3-CH|-CH2-CH3

  • CH3-CH2-CH2-CH2-

  • CH3-C(C2H6)-


A.

C2H6-CH-CH2-

'Iso' mean one Me group is present in six chain. Hence, the structure of iso-butyl group in an organic compound is

                      

('yl' suffix is used to represent one -H less than the parent hydrocarbon.)


19.

KMnO4 can be prepared from K2MnOas per reaction

  3MnO42- + 2H2O  2MnO4- + MnO2 + 4OH-

The reaction can go to completion by removing OH- ions by adding

  • HCl

  • KOH

  • CO2

  • SO2


D.

SO2

Since, OH- are generated from weak acid (H2O), a weak acid (like CO2) should be used to remove it because of strong acid (HCl) reverse the reaction. KOH increases the concentration of OH-, thus again shifts the reaction in backward side. CO2 combines with OH- to give carbonate which is easily removed. SO2, reacts with water to give strong acid, so it cannot be used.


20.

6.02 x 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is

  • 0.02 M

  • 0.01 M

  • 0.001 M

  • 0.1 M


B.

0.01 M

Given, number of molecules of urea = 6.02 × 1020therefore,         Number of moles =6.02 × 1020NA                                         = 6.02 × 10206.02 × 1023= 1 ×10-3 molVolume of the solution                        = 100mL = 1001000L = 0.1 LConcentration of urea solution( in mol L-1) = 1 × 10-30.1= mol L-1                      = 1 × 10-2 mol L-1= 0.01 mol L-1