Subject

Chemistry

Class

NEET Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

11.

Which of the following, not a correct statement? 

  • The electron -deficient molecules can act as Lewis acids

  • The canonical structures have no real existence

  • Every AB5 molecules does, in fact, have square pyramid structure

  • Every AB5 molecules does, in fact, have square pyramid structure


C.

Every AB5 molecules does, in fact, have square pyramid structure

Every AB5 molecule does not, in fact, have square pyramidal structure but AB5 molecules have trigonal bipyramidal structures due to sp3d hybridisation.

1146 Views

12.

CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs - 133 and that of Br = 80 amu and Avogadro number being
6.02 x 1023 mol-1, the density of CsBr is:

  • 42.5 g/cm3

  • 0.425 g/cm3

  • 8.25 g/cm3

  • 8.25 g/cm3


D.

8.25 g/cm3

Density space of space CsBr space equals space fraction numerator straight Z space straight x space straight M over denominator straight a cubed space straight x space straight N subscript straight o end fraction
straight Z space rightwards arrow space no. space of space atoms space in space the space bcc space unit space cell space equals 2
straight M space rightwards arrow space molar space mass space of space CsBr space equals space 133 space plus space 80 space equals space 213
straight a space rightwards arrow space edge space length space of space unit space cell space equals space 436.6 space pm
space equals space 436.6 space straight x space 10 to the power of negative 10 end exponent space cm
Therefore comma space
Density space equals space fraction numerator 2 space straight x space 213 over denominator left parenthesis 436.6 space straight x space 10 to the power of negative 10 end exponent right parenthesis cubed space straight x space 6.02 space straight x space 10 to the power of 23 end fraction
space equals space 8.50 space straight g space divided by cm cubed
For space straight a space unit space cell space equals space fraction numerator 8.50 over denominator 2 end fraction space equals space 4.25 space straight g divided by space cm cubed
1080 Views

13.

Which of the following is not isostructural with SiCl4?

  • SCl4


  • SO42-
  • PO43- 

  • PO43- 


A.

SCl4

SCl4 is not isostructural of SiCl4 because it shows square planar structure due to the involvement of repulsion between lone pair and bond pair of electrons. 

SO42- shows tetrahedral structural due to sp3 hybridization
PO43- shows tetrahedral structural due to sp3 hybridization
NH4+ shows tetrahedral structural due to sp3 hybridisation

1738 Views

14.

A plot of log x/m versus log p for the adsorption of a gas on a solid gives a straight line with slope equal to:

  • -log k

  • n

  • 1/n

  • 1/n


C.

1/n

The empirical relation x/m = kp1/n put forward by Freundlich is known as Freundlich adsorption isotherm. Taking logarithm
log space straight x over straight m space equals space log space straight k space plus 1 over straight n space log space straight p
If the following curve is plotted

555 Views

15.

More the number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is:

  • more energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals

  • the lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals

  • the greater metallic character of the lanthanoids than that of the corresponding actinoids

  • the greater metallic character of the lanthanoids than that of the corresponding actinoids


B.

the lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals

A number of oxidation states are exhibited by the actinoids than by the lanthanoids due to the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals.

503 Views

16.

The correct regarding the electronegativity of hybrid orbitals of carbon is: 

  • sp> sp2<sp3

  • sp> sp2 > sp3

  • sp< sp2 > sp3

  • sp< sp2 > sp3


B.

sp> sp2 > sp3

The correct  order regarding the electronegativity of hybrid orbitals of carbon is sp> sp2 > sp3 because in sp> sp2 > and sp3 hybrid orbitals s- orbital character is 50%, 33.3% and 25% respectively and due to higher s-orbital character electron attraction tendency i,e, electronegativity increases.

1125 Views

17.

The correct order of the mobility of the alkali metal ions in aqueous solution is:

  • Li+>Na+ > K+ > Rb+

  • Na+>K+ > Rb+  > Li+

  • K+ > Rb+ >Na+ > Li+

  • K+ > Rb+ >Na+ > Li+


D.

K+ > Rb+ >Na+ > Li+

The correct order of the mobility of the alkali metal ions in aqueous solutions 
Rb+ > K+ > Na+ > Lidue to following order of hydration energy of these ions Li+ > Na+ > K+ > Rb+ and due to hydration of ion, mobility decreases.

1091 Views

18.

The IUPAC name of is:

  • 3,4-dimethylpentanoyl chloride

  • 1-chloro-1-oxo-2, 3-dimethyl pentane

  • 2-ethyl -3-ethylbutanonyl chloride

  • 2-ethyl -3-ethylbutanonyl chloride


D.

2-ethyl -3-ethylbutanonyl chloride

666 Views

19.

The general molecular formula, which represents the homologous series of alkanols is:

  • CnH2nO2

  • CnH2nO

  • CnH2n+1O

  • CnH2n+1O


D.

CnH2n+1O

Alkanols are the derivatives of alkanes which are derived from the replacement of -H of alkanes with -OH (hydroxyl groups)

722 Views

20.

A solution containing 10 g per dm3 is urea (molecular mass = 60 g mol-1) is isotonic with a molecular mass of this non-volatile solute. The molecular mass of this of this non-volatile solute is:

  • 250 g mol-1

  • 300 g mol-1

  • 350 g mol-1

  • 350 g mol-1


B.

300 g mol-1

10 g per dm3 of urea is isotonic with 5% solution of a non-volatile solute. Hence, between this solution osmosis is not possible so, their molar concentrations are equal to each other,
Thus, molar concentration of urea solution

space equals space fraction numerator 10 straight g divided by dm cubed over denominator Mol. space wt. space of space urea end fraction
space equals 10 over 60 straight M space equals space 1 over 6 straight M
Molar space concentration space of space 5 percent sign space non minus volatile space solute

equals space fraction numerator 50 space straight g divided by dm cubed over denominator mol. wt. space of space non minus volatile space solute end fraction
space equals space 50 over straight m space straight M
Both space solution space are space isotonic space to space each space other comma space therefore comma space
1 over 6 space equals space 5 over straight m
straight m space equals space 50 space straight x space 6 space equals space 300 space straight g space mol to the power of negative 1 end exponent

3485 Views