The angular momentum of electrons in d orbital is equal to

  • square root of 6 space h
  • square root of 2 straight h
  • 2 square root of 3 straight h
  • 2 square root of 3 straight h

A.

square root of 6 space h

Angular momentum of electrons in d- orbital is 

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Magnetic moment 2.84 BM is given by (At. no. Ni= 28, Ti= 22, Cr=24, Co = 27)

  • Ni2+

  • Ti3+

  • Cr3+

  • Cr3+


A.

Ni2+



The electronic configuration of the given ions,
Ni2+ = [Ar]3d8 4s0 (two unpaired electron)
Ti3+ = [Ar] 3d1 4s0 ( one unpaired electrons)
Cr3+ = [Ar] 3d3 ( three unpaired electrons)
Co2+ = [Ar],3d7, 4s0 (three unpaired electrons)
So, only Ni2+ has 2 unpaired of electrons.
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The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are:

  • Paschen and Pfund

  • Lyman and Paschen

  • Brackett and Pfund

  • Balmer and Brackett


B.

Lyman and Paschen

Shortest wavelength of lymen series:

1λ1= RH112-121λ1= R1 =λ1=1RHShortest wavelength of paschan series:1λ2= RH=19-12=1λ2= RH91λ2= RH9= λ2= 9RH


Two electrons occupying the same orbital are distinguished by

  • Magnetic quantum number

  • Azimuthal quantum number

  • Spin quantum number

  • Spin quantum number


C.

Spin quantum number

Two electrons occupying the same orbital has equal spin but the directions of their spin are opposite. Hence, spin quantum number, s(represented +1/2 and -1/2) distinguishes them.

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What is the maximum number of orbitals that can be identified with the following quantum numbers?
n=3, l =1, m1 = 0

  • 1

  • 2

  • 3

  • 3


A.

1

The value of n=3 and l =1 suggest that it is a 3p orbital while the value of m1 = 0 [magnetic quantum number] shows that the given 3p orbital is 3pz in nature.
Hence, the maximum number of orbitals identified by the given quantum number is only 1, i.e. 3pz.

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