Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The following equilibrium constants are given:

N2 + 3H2  ⇌ 2NH3; K1

N2 +O2  ⇌ 2NO; K2

H2 + 1/2O2  ⇌ H2O' K3

The equilibrium constants for the oxidation of NH3 by oxygen to give NO is:

  • K2K33 /K1

  • KK2K32 /K1

  • K22K3 /K1

  • K22K3 /K1


A.

K2K33 /K1

The required equation for the oxidation of NH3 by oxygen to give NO is:

4 NH subscript 3 space plus space 5 straight O subscript 2 space rightwards arrow from 800 to the power of straight o space straight C to Pt space left parenthesis gauze right parenthesis of space 4 space NO space plus space 6 straight H subscript 2 straight O
For space this space space straight K space equals space fraction numerator left square bracket NO right square bracket to the power of 4 left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket NH subscript 3 right square bracket to the power of 4 left square bracket straight O subscript 2 right square bracket to the power of 5 end fraction
For space the space equation space left parenthesis straight I right parenthesis space straight K subscript 1 space equals space fraction numerator left square bracket NH subscript 3 right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight H subscript 2 right square bracket cubed end fraction
For space the space equation space left parenthesis II right parenthesis space straight K subscript 2 space equals space fraction numerator left square bracket NO right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket end fraction
For space the space equation space left parenthesis III right parenthesis space straight K subscript 3 space equals space fraction numerator left square bracket straight H subscript 2 straight O right square bracket over denominator left square bracket straight H subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket to the power of 1 divided by 2 end exponent end fraction
For space getting space the space straight K space we space must space do
straight K subscript 1 squared space equals space fraction numerator left square bracket NH subscript 3 right square bracket to the power of 4 over denominator left square bracket straight N subscript 2 right square bracket squared left square bracket straight H subscript 2 right square bracket to the power of 6 end fraction comma space straight K subscript 2 squared space equals space fraction numerator left square bracket NO right square bracket to the power of 4 over denominator left square bracket straight N subscript 2 right square bracket squared left square bracket straight O subscript 2 right square bracket squared end fraction
straight K subscript 3 superscript 6 space equals space fraction numerator left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket straight H subscript 2 right square bracket to the power of 6 left square bracket straight O subscript 2 right square bracket to the power of begin display style 6 over 2 end style equals 3 end exponent end fraction
straight K space equals space fraction numerator straight K subscript 2 superscript 2 space straight x space straight K subscript 3 superscript 6 over denominator straight K subscript 1 superscript 2 end fraction space substituting space the space value space we space get comma
straight K space equals space fraction numerator left square bracket NO right square bracket to the power of 4 left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket NH subscript 3 right square bracket to the power of 4 left square bracket straight O subscript 2 right square bracket to the power of 5 end fraction space so space the space value space of space straight K space in space terms space of space straight K subscript 1 comma space straight K subscript 2 space and space straight K subscript 3 space is space
space straight K thin space equals space fraction numerator straight K subscript 2 straight K subscript 3 superscript 3 over denominator straight K subscript 1 end fraction


2.

Calculate the pOH of a solution at 250C that contains 1 x 10-10 M of hydronium ions.

  • 7

  • 4

  • 9

  • 9


B.

4

[H3O+] = [H+] = 10-10

pH + pOH = [14]

pH = - log [ H+]

pH = - log [10-10]

pH = 10

⇒ pOH + 10 = 14

⇒ pOH = 14-10 = 4


3.

Which of the following oxidation states are the most characteristics for lead and tin respectively? 

  • +4,+2

  • +2,+4

  • +4,+4

  • +4,+4


B.

+2,+4

The tendency to form +2 ionic state increase on moving down the group due to the inert pair effect.

Most characteristic oxidation state for lead and tin are +2,+4 respectively.


4.

Consider the following sets of quantum numbers:


  n l m s
i) 3 0 0 +1/2
ii) 2 2 1 +1/2
iii) 4 3 -2 -1/2
iv) 1 0 -1 -1/2
v) 3 2 3 +1/2

Which of the following sets of quantum number is not possible? 
  • ii, iii and iv

  • i, ii, iii and iv

  • ii, iv and v

  • ii, iv and v


C.

ii, iv and v

The value of l varies from 0 to n-i and the value of m varies from -l to +l through zero.
the value of 's' negative-or-plus 1 half Which signifies the spin of the electron. The correct sets of quantum number are following:

  n l m s
ii) 2 1 1 +1/2
iv 1 0 0 -1/2
v) 3 2 2 +1/2

5.

With which of the following electronic configuration an atom has the lowest ionisation enthalpy? 

  • 1s2 2s22p5

  • 1s2 2s22p3

  • 1s2 2s22p5 3s1

  • 1s2 2s22p5 3s1


C.

1s2 2s22p5 3s1

The electronic configuration 1s2 2s22p5 3s1 shows lowest ionisation energy because this configuration is unstable due to the presence of one electron is s- orbital. Hence, less energy is required to remove the electron.


6.

The correct order of C-O bond length among CO, CO32-, CO2 is,

  • CO2<  CO32-<  CO

  • CO <  CO32-<  CO2

  • CO32- <CO2 <CO

  • CO32- <CO2 <CO


B.

CO <  CO32-<  CO2

The bond length is the average distance between the centres of nuclei of two bonded atoms. Centres of nuclei of two bonded atoms. A multiple bonds (double or triple bond) is always shorter than the corresponding single bond. 

The C- atom is CO32- is sp2 hybridised as shown


The C- atom is CO2 is sp hybridised with a bond distance of carbon -oxygen is 122 pm.

straight O equals straight C equals straight O space left right arrow space to the power of plus straight O space identical to space straight C minus straight O to the power of minus space left right arrow space straight O to the power of minus minus straight C identical to straight O to the power of plus
The C- atom in CO is sp hybridised with C-O bond distance is 110 pm:
So the correct order is 
CO <  CO32-<  CO2


7.

A Weak acid, HA has a Ka of 1.00 x 10-5. If 0.100 mole of this percentage of acid dissociated at equilibrium is closest to:

  • 99.0%

  • 1.00%

  • 99.9 %

  • 99.9 %


B.

1.00%

HA space leftwards harpoon over rightwards harpoon space straight H to the power of plus space plus space straight A to the power of minus
straight K subscript straight a space space equals space fraction numerator left square bracket straight H to the power of plus right square bracket left square bracket straight A to the power of minus right square bracket over denominator left square bracket HA right square bracket end fraction space equals space fraction numerator left square bracket straight H to the power of plus right square bracket over denominator left square bracket HA right square bracket end fraction
left square bracket straight H to the power of plus right square bracket space equals space square root of straight K subscript straight a left square bracket HA right square bracket end root space equals space square root of 1 space straight x space 10 to the power of negative 5 end exponent space 0.1 end root
space equals square root of 1 space straight x 10 to the power of negative 6 end exponent end root space equals space 1 space straight x space 10 to the power of negative 3 end exponent

straight alpha space equals space fraction numerator actual space ionization over denominator molar space concentration end fraction space equals space fraction numerator 10 to the power of negative 3 end exponent over denominator 0.1 end fraction space equals space 10 to the power of negative 2 end exponent
percent sign space of space acid space dissociated space space equals space 10 to the power of negative 2 end exponent space straight x space 100
space equals space 1 percent sign

8.

Given that bond energy of H-H and Cl- Cl are 430 kJ mol-1 and 240 kJ mol-1 respectively and ΔHf for HCl is -90 kJ mol-1. Bond and ΔHf for HCl is -90 kJ mol-1.Bond enthalpy of HCl is:

  • 290 kJ mol-1

  • 380 kJ mol-1

  • 425 kJ mol-1

  • 425 kJ mol-1


B.

380 kJ mol-1

increment straight H subscript reaction space equals increment subscript straight H minus straight H end subscript space plus increment straight H subscript Cl minus Cl end subscript
minus 2 space increment straight H subscript HCl equals negative negative 90 space kJ
Or space increment straight H subscript straight H minus Cl end subscript space equals space fraction numerator 430 plus 240 minus left parenthesis negative 90 right parenthesis over denominator 2 end fraction
space equals space 760 over 2 space equals space 380 space kJ space mol to the power of negative 1 end exponent

9.

Consider the following reactions:

straight i right parenthesis space straight H to the power of plus space left parenthesis straight g right parenthesis thin space space plus space OH to the power of minus space left parenthesis aq right parenthesis space equals space straight H subscript 2 straight O space left parenthesis straight l right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space minus space straight x subscript 1 space kJ space mol to the power of blank to the power of negative 1 end exponent end exponent
ii right parenthesis space straight H subscript 2 space left parenthesis straight g right parenthesis thin space space plus space straight O subscript 2 space left parenthesis straight q right parenthesis space equals space straight H subscript 2 straight O space left parenthesis straight l right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space minus space straight x subscript 2 space kJ space mol to the power of blank to the power of negative 1 end exponent end exponent
iii right parenthesis space CO subscript 2 space left parenthesis straight g right parenthesis thin space space plus space straight H subscript 2 space left parenthesis straight g right parenthesis space equals space straight H subscript 2 straight O space left parenthesis straight l right parenthesis space plus space CO subscript 2 space left parenthesis straight g right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space minus space straight x subscript 3 space kJ space mol to the power of blank to the power of negative 1 end exponent end exponent
iv right parenthesis straight C subscript 2 straight H subscript 2 space left parenthesis straight g right parenthesis thin space space plus 5 over 2 space straight O subscript 2 space left parenthesis straight g right parenthesis space equals space straight H subscript 2 straight O space left parenthesis straight l right parenthesis space plus 2 CO subscript 2 space left parenthesis straight g right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space plus space straight x subscript 4 space kJ space mol to the power of blank to the power of negative 1 end exponent end exponent
Enthalpy of formation of H2O (l) is:

  • - x2 kJ mol-1

  • + x3 kJ mol-1

  • - x4 kJ mol-1

  • - x4 kJ mol-1


A.

- x2 kJ mol-1

Enthalpy of formation: The amount of heat evolved or absorbed during the formation of 1 mole of a compound from its constituent elements is known as heat of formation. SO, the correct answer is:

straight H subscript 2 space left parenthesis straight g right parenthesis space space plus space 1 half space straight O subscript 2 space left parenthesis straight g right parenthesis thin space rightwards arrow space straight H subscript 2 straight O space left parenthesis straight l right parenthesis space semicolon increment straight H equals negative straight x subscript 2 space kJ space mol to the power of negative 1 end exponent


10.

An element, X has the following isotopic composition;


200X: 90%

199X : 8.0%

202X ; 2.0 %

The weighted average atomic mass of the naturally -occurring element X is closet to:

  • 200 amu

  • 201 amu

  • 202 amu

  • 202 amu


A.

200 amu

Weight of  200X = 0.90 x 200 = 180.00 amu

Weight of 199X  = 8.08 x 199 = 15.92 amu

Weight 0f  202X =  0.02 x 202 = 4.04 amu

Total weight = 199.06 = 200 amu