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JEE Chemistry

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Question
CBSEENCH11008090
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Among the following, the maximum covalent character is shown by the compound

  • FeCl2

  • SnCl2

  • AlCl3

  • MgCl2

Solution

C.

AlCl3

The covalent character in ionic compounds is governed by Fazanís Rule.
(i) larger the charge on the ions.
(ii) smaller the size of anions.
(iii) larger the size of anion.
(iv) larger the polarizing power larger the covalent character.
AlCl3 will show Maximum covalent character on account of the higher polarising power of Al3+ because of its having higher positive charge and smaller size.

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Question
CBSEENCH11008091
Wired Faculty App

The hybridization of orbitals of N atom in NO3-, NO2+ and NH4+ are respectively

  • sp, sp2,sp3

  • sp2 , sp, sp

  • sp, sp3 , sp2

  • Sp2, sp3, sp

Solution

B.

sp2 , sp, sp

NO2+
Number of electron pairs = 2
Number of bond pairs = 2
Number of lone pair = 0
So, the species is linear with sp hybridisation.
NO3-
Number of electron pairs = 3
Number of bond pairs = 3
Number of lone pair = 0
So, the species is trigonal planar with sp2 hybridisation

NH4+
Number of electron pairs = 4
Number of bond pairs = 4
Number of lone pair = 0
So, the species is tetrahedral with sp3 hybridisation.

Question
CBSEENCH11008092
Wired Faculty App

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at

  • 1035 nm

  • 325 nm

  • 743 nm 

  • 518 nm

Solution

C.

743 nm 

E = E1 + E2
hc over straight lambda space equals space hc over straight lambda subscript 1 space plus hc over straight lambda subscript 2 1 over straight lambda space equals space 1 over straight lambda subscript 1 space plus 1 over straight lambda subscript 2 1 over 355 space equals space 1 over 680 space plus 1 over straight lambda subscript 2 straight lambda subscript 2 space equals space 742.76 space nm

Question
CBSEENCH12010607
Wired Faculty App

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27°C is

  • 38.3 J mol-1 K-1

  • 35.8 J  mol-1 K-1

  • 32.3 J  mol-1 K-1

  • 42.3 J mol-1 K-1

Solution

A.

38.3 J mol-1 K-1

increment straight S space equals space nR space In space straight V subscript 2 over straight V subscript 1 equals space 2.303 space nR space log space straight V subscript 2 over straight V subscript 1 space equals space 2.303 space straight x space 2 space straight x space 8.314 space straight x space log space 100 over 10 equals space 38.3 space straight J space mol to the power of negative 1 end exponent straight k to the power of negative 1 end exponent

Question
CBSEENCH11008094
Wired Faculty App

'a' and 'b' are van der Waals constants for gases. Chlorine is more easily liquefied than ethane because :

  • a and b for Cl2 > a and b for C2H6

  • a and b for Cl2 < a and b for C2H6

  • a and Cl2 < a for C2H6 but b for Cl2 > b for C2H6

  • a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6

Solution

D.

a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6

Vander Waals, constant a is due to force of attraction and b due to the infinite size of molecules. Thus, greater the value a and smaller the value b, larger the liquefaction.

  a b
Cl2  6.579 L2 bar mol-2 0.05622 L bar mol-2
C2H5 5.562 L2 bar mol-2 0.06380 L mol-1

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