JEE Chemistry

Question 1

Which of the following sets of quantum numbers is correct for an electron in 4f orbital?

  • n = 4, I =3, m = +4, s = + 1/ 2

  • n = 3, I = 2, m = -2, S = + 1/2

  • n =4, I = 3, m = +1, s = + 1/ 2

  • n =4, I = 4, m -4, s = -1/ 2

Solution

C.

n =4, I = 3, m = +1, s = + 1/ 2

For 4f orbital electrons, n = 4

                       s p d f

l = 3 (because 0 1 2 3 )

m = +3, +2, +1, 0, –1, –2, –3

s = ±1/2.

Question 2

Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers I =1 and 2 are respectively

  • 12 and 4

  • 16 and 5

  • 16 and 4

  • 12 and 5

Solution

D.

12 and 5

Question 3

Which one the following ions has the highest value of ionic radius?

  • Li+

  • F-

  • O2-

  • B3+

Solution

C.

O2-

Question 4

The wavelength of the radiation emitted, when in hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m-1)

  • 91 nm

  • 9.1×10-8 nm

  • 406 nm

  • 192 nm

Solution

A.

91 nm

1 over straight lambda space equals space top enclose straight V subscript straight H space equals space top enclose straight R subscript straight H end enclose open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets
space equals space 1.097 space straight x space 10 to the power of 7 space open square brackets 1 over 1 squared minus 1 over infinity close square brackets
therefore space straight lambda space equals space fraction numerator 1 over denominator 1.097 space straight x space 10 to the power of 7 end fraction space straight m space equals space 9.11 space straight x space 10 to the power of negative 8 end exponent space straight m
space equals space 91.1 space straight x space 10 to the power of negative 9 end exponent space straight m space equals space 91.1 space nm space left parenthesis 1 nm space equals space 10 to the power of negative 9 end exponent space straight m right parenthesis
Question 5

The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is

  • H2S < SiH4 < NH3 < BF3

  • H2S < NH3 < BF3 < SiH4

  • NH3 < H2S < SiH4 < BF3

  • H2S < NH3 < SiH4 < BF3

Solution

D.

H2S < NH3 < SiH4 < BF3