Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

Which of the following structure is similar to graphite?

  • CO

  • O2-

  • CN-

  • NO+


B.

O2-

Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration.
Molecular orbital configuration of the given species is as

CO(6+8=14)=σ1s2,σ*1s2,σ2s2,σ*2s2,π2px2π2py2,σ2pz2

(All the electrons are paired so it is diamagneitc.)

O2-(8+8+1=17)=σ1s2,σ*1s2,σ2s2,σ*2s2,σ2pz2,π2px2,π2py2,π*2px2π*2py1

It contains one unpaired electron so it is paramagnetic.
CN- (6+ 74 1= 14)= same as CO
NO+ (7+ 8- 1)= same as CO
Thus, among the given species onlyO2- is paramagnetic.


2.

What is the maximum numbers of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m =-1.

  • 10

  • 6

  • 4

  • 2


D.

2

The orbital of the electron having n= 3, l=1 and m = -1 is 3pz(as nlm) and an orbital can have a maximum of two electrons with opposite spins.
:. 3pz orbital contains only two electrons or only 2 electrons are associated with n= 3 , l = 1, m =-1.


3.

Maximum deviation from ideal gas is expected from

  • H2(g)

  • N2(g)

  • CH4(g)

  • NH3(g)


D.

NH3(g)

Easily liquifiable gases like NH3, SO2etc. exhibit maximum deviation from ideal gas as for them Z <<<1.
CH4 also exhibits deviation but it is less as compared to NH3.


4.

Which ofthese is least likely to act as a Lewis base?

  • CO

  • F-

  • BF3

  • PF3


C.

BF3

Electron rich species are called Lewis base. Among the given, BF3 is an electron deficient species, so have a capacity of electron accepting instead of donating that's why it is least likely to act as a Lewis base. It is a Lewis acid.


5.

XeF2 is isostructural with

  • TeF2

  • ICl2-

  • SbCl3

  • BaCl2


B.

ICl2-

Species having the same number of bond pairs and lone pairs are isostructural (have same structure).

Species lp + bp Structure
XeF2 4lp + 2bp

Linear

TeF2 2lp + 2bp

Angular or V - Shaped

ICl2- 4lp + 2bp

Linear

BaCl2 0lp + 2bp Cl-Ba-Cl (linear)

Thus, XeF2 is isostructural with ICl2- and BaCl2.


6.

Based on equationE=-2.178×10-18JZ2n2certain conclusions are written. Which of them is not correct?

  • The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.

  • Larger the value of n, the larger is the orbit radius

  • Equation can be used to calculate the change in energy when the electron changes orbit.

  • For n = 1 the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.


D.

For n = 1 the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.

Ifn=1E1=-2.178×10-18Z2JIfn=6E6=-2.178×10-18Z236J=6.05×10-20Z2J

From the above calculation, it is obvious that electron has a more negative energy than it does for n=6. It means that electron is more strongly bound in the smallest allowed orbit.


7.

An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitate would be

  • 0.001

  • 0.002

  • 0.003

  • 0.01


A.

0.001

The formula of dichlorotetraqua chromium (lll) chloride is [Cr(H2O)4Cl2] Cl.
On ionisation it generates only one Cl- ion

[Cr(H2O)4Cl2]Cl[Cr(H2O)4Cl2]++Cl-Intial100×0.0100mmol=1mmolAfter01mmol1mmolinonisation

One mole of Cl- ions react with on only 1 mole of AgNO3, molecule to produce 1 mole of AgCl.
.:. 1 mmol or 1 x 10-3 mole reacts with AgNO3to give AgCl

=1×1×10-31=10-3or0.001molAgCl


8.

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0M HNO3 ? The concentrated acid is 70% HNO3.

  • 45.0 g conc. HNO3

  • 90.0 g con. HNO3

  • 70.0 g conc. HNO3

  • 54.0 g conc. HNO3


A.

45.0 g conc. HNO3

Given,molairtyofsolution=2Volumeofsolution=250mL=2501000=14LMolarmassofHNO3=1+14+3×16=63gmol-1therefore,Molarity=weightofHNO3massofHNO3×volumeofsolution(L)therefore,weightofHNO3=molarity×mol.mass×volume(L)=2×63×14g=31.5gItistheweightof100%HNO3.Butthegivenacidis70%HNO3therefore,Itsweight=31.5×10070g=45g


9.

The value of Planck's constant is 6.63 x 10-34Js. The speed of light is 3 x 1017 nms-1. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6 x 1015 s-1?

  • 10

  • 25

  • 50

  • 75


C.

50

Given, Plank's constant: h= 6.63 x 10-34 Js

Speed of light, c= 3 x 1017 nms-1

Frequency of quanta, v= 6 x 1015s-1

Wavelength,λ= ?

We know that,v=λcorλ=cv=3×10176×1015=0.5×102nm=50nm


10.

Dipole-induced dipole interactions are present in which of the following pairs?

  • H2O and alcohol

  • Cl2 and CCl4

  • HCl and He atoms

  • SiF4 and He atoms


C.

HCl and He atoms

Dipole-induced dipole interaction are present in the pair in which the first species is polar and the other is non-polar.H2O and alcohol both are polar molecules, so there exists dipole-dipole interactions in between them. Cl2 and CCl4 both are non-polar so there exists induced dipole-induced dipole interactions in between them. Similar is true for SiCl4 and He atoms pair. HCl is a polar molecule, whereas He atoms are non-polar, so in between them dipole-induced dipole interactions exist.