Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The number of atoms in 0.1 moles of a triatomic gas is (NA = 6.02 x 1023 mol-1)

  • 6.026 x 1022

  • 1.806 x 1023

  • 3.600 x 1023

  • 3.600 x 1023


B.

1.806 x 1023

Number of atoms = number of moles x NA x atomicity

= 0.1 x 6.02 x 1023 x 3

= 1.806 x 1023 atoms

865 Views

2.

In which one of the following species the central atom has the type of hybridisation which si not the same as that present in the other three? 

  • SF4

  • I3-

  • SbCl52-

  • SbCl52-


C.

SbCl52-

Molecules having the same number of hybrid orbitals, have same hybridization and number of hybrid orbitals,


straight H space equals space 1 half left square bracket space straight V space plus space straight X space minus space straight C space plus straight A right square bracket space equals space 5
where comma space straight V space equals space no. space of space valence space electrons space of space central space atom
straight X space equals space no. space of space monovalent space atoms
straight C space equals space charge space on space cation
straight A space equals space Charge space on space anion

straight a right parenthesis space In space SF subscript 4
space straight H space equals space 1 half space left square bracket space 6 plus 4 minus 0 plus 0 right square bracket space equals space 5

straight b right parenthesis space In space straight I to the power of minus subscript 3
space straight H space equals space 1 half space left square bracket space 7 space plus space 2 space plus space 1 space right square bracket

straight c right parenthesis space In space SbCl subscript 5 to the power of negative 2 end exponent space
straight H space equals space 1 half left square bracket space 5 space plus 5 plus 2 right square bracket space equals space 6

straight d right parenthesis space In space PCl subscript 5
straight H space equals space 1 half space left square bracket space 5 plus space 5 plus 0 minus 0 right square bracket space equals space 5
Since, only SbCl52- has a different number of hybrid orbitals (i.e,6) from the other given species, its hybridization is different front the others, ie, sp3d2.

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3.

Which of the following pairs has the same size? 

  • Fe2+ , Ni2+

  • Zr4+, Ti4+

  • Zr4+, Hf4+

  • Zr4+, Hf4+


C.

Zr4+, Hf4+

In general, the atomic and ionic radii increases on moving down a group. But the elements of second transition series (eg, Zr, Nb, Mo etc) have the almost same radii as the elements of third transition series (eg Hf, Ta, W etc.) This is because of lanthanide contraction i.e imperfect shielding of one 4f- electron by another.

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4.

In which of the following pairs of molecules/ions, the central atoms have sp2 hybridization?

  • NO2- and NH3

  • BF3 and NO2-

  • NH2- and H2O

  • NH2- and H2O


B.

BF3 and NO2-

(i) NO2- ⇒ 2σ + 1 lp =3, ie, sp2 hybridisation
(ii) NH3 ⇒ 3σ + 1 lp = 4, ie, sp3 hybridisation
(iii) BF3 ⇒ 3σ + 0 lp = 3, i.e sp2 hybridisation
(iv) NH2- ⇒ 2σ + 2 lp = 4 i.e sp3 hybridisation
(v) H2O ⇒ 2σ + 2 lp = 4 i.e sp3 hybridisation

Thus, among the given pairs, only BF3 and NO2- have sp2 hybridizations.

1608 Views

5.

What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH? 
(Ka for CH3COOH = 1.8 x 10-5 )

  • 3.5 x 10-4

  • 1.1 x 10-5

  • 1.8 x 10-5

  • 1.8 x 10-5


D.

1.8 x 10-5

CH3COOH (weaK acid) and CH3COONa (conjugated salt) form acidic buffer and for acidic buffer,

pH space equals space pK subscript straight a space plus space log space fraction numerator open square brackets salt close square brackets over denominator left square bracket acid right square bracket end fraction
and
left square bracket straight H to the power of plus right square bracket space equals space minus space antilog space pH
pH space equals space minus space log space straight K subscript straight a space plus space log space fraction numerator left square bracket salt right square bracket over denominator left square bracket acid right square bracket end fraction space space space space space left square bracket therefore space pK subscript straight a space equals space minus space log space straight K subscript straight a right square bracket
space equals space minus log space left parenthesis 1.8 space straight x space 10 to the power of negative 5 end exponent right parenthesis space plus space log space fraction numerator 0.20 over denominator 0.10 end fraction
equals space 4.74 space plus space log space 2
equals space 4.74 space plus space 0.3010 space equals space 5.041
Now comma space left square bracket straight H to the power of plus right square bracket space equals antilog space left parenthesis negative 5.045 right parenthesis
equals space 9.0 space straight x space 10 to the power of negative 6 end exponent space mol divided by straight L

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6.

Standard entropies X2, Y2 and XY2 are 60, 40 and 50 J K-1 mol-1 respectively for the reaction

1 half space straight X subscript 2 space plus 3 over 2 space straight Y subscript 2 space leftwards harpoon over rightwards harpoon space XY subscript 3 space semicolon space increment straight H space equals space minus space 30 space kJ comma space to space be space at
equilibrium, the temperature should be

  • 750 K

  • 1000 K 

  • 1250 K

  • 1250 K


A.

750 K

1 half space straight X subscript 2 space plus space 3 over 2 space straight Y subscript 2 space bold a bold space XY subscript 3 space semicolon space increment straight H space equals space minus space 30 space kJ
increment straight S to the power of straight o space equals space straight S subscript left parenthesis XY subscript 3 right parenthesis end subscript superscript straight o space minus space open square brackets 1 half straight S to the power of straight o subscript straight x subscript 2 end subscript plus 3 over 2 space straight S subscript straight Y subscript 2 end subscript superscript straight o close square brackets

equals space 50 space minus space open square brackets 1 half space straight x space 60 space plus space 3 over 2 space straight x space 40 close square brackets
space equals space 50 minus left square bracket 30 plus 60 right square bracket space equals space 50 minus 90
equals negative 40 space JK to the power of minus space mol to the power of negative 1 end exponent
we space know space that
increment straight G to the power of straight o space equals space increment straight H to the power of straight o minus straight T increment straight S to the power of straight o
At space equilibrium comma
increment straight G to the power of straight o space equals 0
increment straight H space equals space straight T increment straight S to the power of straight o
straight T space equals space fraction numerator increment straight H over denominator increment straight S to the power of straight o end fraction space equals space fraction numerator negative 30 space straight x space 10 cubed over denominator negative 40 space straight J space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent space end fraction space equals space 750 space straight K
581 Views

7.

Which one of the following species does not exist under normal conditions?

  • Be2+

  • Be2

  • B2

  • B2


B.

Be2

Molecules with zero bond order do not exist.

a) Be2+ (4 + 4 -1 = 7) = σ1s2 σ*1s2, σ2s2, σ2s1

BO = 4 - 3 / 2  = 0.5

b) Be2 (4 + 4 = 8 )  = σ1s2, σ*1s2, σ2s2, σ*2s2


BO = 4 - 4 / 2 = 0

c) B2 = (5 + 5) = 10

= σ1s2 , σ*1s2, 2σs2, σ*2s2, π2px1 = π2py1

BO = 6 - 4 /2 = 1

d) Li2 (3 + 3) = 6

= σ1s2 , σ*1s2,σ2s2

BO = 4 - 2 / 2 = 1

Thus, bond order of Be2 does not exist under normal conditions.


1091 Views

8.

The correct order of the decreasing ionic radii among the following isoelectronic species is

  • Ca2+ > K+ >S2- > Cl-

  • Cl- > S2- > Ca2+ > K+

  • S2- > Cl- > K+ > Ca2+

  • S2- > Cl- > K+ > Ca2+


C.

S2- > Cl- > K+ > Ca2+

Ionic space radii space proportional to space charge space on space anion space proportional to space fraction numerator 1 over denominator charge space on space cation end fraction
During the formation of a cation, the electrons are lost from the outer shell and the remaining electrons experience a great force of attraction by the nucleus, ie, attracted more towards the nucleus. In other words, nucleus holds the remaining electrons more tightly and result in decreased radii.
However, in the case of anion formation, the addition of an electron (s) takes place in the same outer shell, thus the hold of the nucleus on the electrons of outer shell decreases and this result in increased ionic radii.

Thus, the correct order of ionic radii is 

S2- > Cl- > K+ > Ca2+
1046 Views

9.

Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl?

  • Cl < F < O < S

  • O < S < F< Cl

  • F < S < O < Cl

  • F < S < O < Cl


B.

O < S < F< Cl

Electron gain enthalpy, generally, increases in a period from left to right and decreases in a group on moving downwards. However, members of III periods have somewhat higher electron gain enthalpy as compared to the corresponding members of the second period because of their small size.

O and S belong to VI A (16) group and Cl and F belong to VII-A (17) group. Thus, the electron gain enthalpy of Cl and F is higher as compared to O and S.

Cl and F > O and S

Between Cl and F, Cl has higher electron gain enthalpy as in F, the incoming electron experiences a greater force of repulsion because of the small size of F atom. Similarly is true is a case of O and S ie, the electron gain enthalpy of S is higher as compared to O due to its small size. Thus, the correct order of electron gain enthalpy of given elements is 

O < S < F < Cl

812 Views

10.

If pH ofa saturated solution of Ba(OH)2 is 12, the value of its Ksp is 

  • 4.00 x 10-6 M3

  • 4.00 x 10-7 M3

  • 5.00 x 10-6 M3

  • 5.00 x 10-6 M3


C.

5.00 x 10-6 M3

Given pH of Ba(OH)2 = 12
therefore, [H+] = [1 x 10-12]
and [OH-] = 1 x 10-14 / 1 x 10-12   {[H+][OH-] = 1 x 10-14}

Ba left parenthesis OH right parenthesis subscript 2 space equals space stack Ba to the power of 2 plus end exponent space with straight s below plus space stack 2 OH to the power of minus with 2 straight s below
Ksp space equals space left square bracket Ba to the power of 2 plus end exponent right square bracket left square bracket OH to the power of minus right square bracket squared
space equals space left square bracket straight s right square bracket left square bracket 2 straight s squared right square bracket

equals space open square brackets fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction close square brackets left parenthesis 1 space straight x space 10 to the power of negative 2 end exponent right parenthesis squared

equals space 0.5 space straight x space 10 to the power of negative 6 end exponent space equals space 5.0 space straight x space 10 to the power of negative 7 end exponent space straight M cubed

435 Views