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The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is     [JEE-2009]
  • a)
    0
  • b)
    2.84 
  • c)
    4.90 
  • d)
    5.92
Correct answer is option 'A'. Can you explain this answer?
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The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)...
Spin Only Magnetic Moment of Cr(CO)6

To calculate the spin-only magnetic moment of Cr(CO)6, we need to determine the number of unpaired electrons in the compound. The spin-only magnetic moment can be calculated using the formula:

μ = √(n(n+2)) BM

where μ is the spin-only magnetic moment in Bohr magneton units, n is the number of unpaired electrons, and BM is the Bohr magneton unit.

Step 1: Determine the number of valence electrons in Cr(CO)6

To determine the number of valence electrons in Cr(CO)6, we need to consider the electronic configuration of Cr.

The atomic number of Cr is 24, and its electronic configuration is [Ar]3d54s1. In Cr(CO)6, the oxidation state of Cr is zero, which means it has lost all its valence electrons.

Therefore, the number of valence electrons in Cr(CO)6 is 0.

Step 2: Determine the number of unpaired electrons

Since the number of valence electrons in Cr(CO)6 is 0, there are no unpaired electrons in the compound. This is because all the valence electrons have been lost by Cr.

Therefore, the number of unpaired electrons in Cr(CO)6 is 0.

Step 3: Calculate the spin-only magnetic moment

Using the formula μ = √(n(n+2)) BM, we can calculate the spin-only magnetic moment of Cr(CO)6.

Since the number of unpaired electrons is 0, the value of n is 0.

μ = √(0(0+2)) BM
= √(0) BM
= 0 BM

Therefore, the spin-only magnetic moment of Cr(CO)6 is 0 Bohr magneton units.

Conclusion

The correct answer is option A, 0. This is because there are no unpaired electrons in Cr(CO)6, resulting in a spin-only magnetic moment of 0 Bohr magneton units.
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The orbital and spin angular momentum of the atom influence its magnetic structure and these properties are most directly studied by placing the atom in a magnetic field. Also, a magnetic field can affect the wavelengths of the emitted photons.The angular momentum vector associated with an atomic state can take up only certain specified directions in space. This concept of space quantization was shown by Otto Stern and Walthor Gerlach in their experiment.In the experiment, silver is vapourized in an electric oven and silver atoms spray into the evacuated apparatus through a small hole in the oven wall. The atoms which are electrically neutral but have a magnetic moment, are formed into a narrow beam as they pass through a slit in a screen. The beam, thus collimated, then passes between the poles of an electromagnet and finally, deposits its silver atoms on a glass plate that serves as a detector. The pole faces of the magnet are shaped to make the magnetic field as nonuniform as possible.In a non-uniform magnetic field, there is a net force on a magnetic dipole. Its magnitude and direction depends on the orientation of the dipole. Thus the silver atoms in the beam are deflected up or down, depending on the orientation of their magnetic dipole moments with respect to the z–direction.The potential energy of a magnetic dipole in a magnetic field where is magnetic dipole moment of the atom. From symmetry, the magnetic field at the beam position has no x or y components i.e.The net force Fz on the dipole isThus, the net force depends, not on the magnitude of the field itself, but on its spatial derivative or gradient.The ResultsIf space quantization did not exist, then could take on any value from + to –, the result would be a spreading out of the beam when the magnet was turned ON. However, the beam was split cleanly into two subbeams, each subbeam corresponding to one of the two permitted orientations of the magnetic moment ofthe silver atom, as shown.In a silver atom, all the spin and orbital magnetic moments of the electrons cancel, except for those of the atoms single valance electron. For this electron the orbital magnetic moment is zero because orbital angular momentum is zero (because for electrons of s–orbit, L = 0), leaving only the spin magnetic moment. This can take up only two orientations in a magnetic field, corresponding to ms = +1/2 and ms = – 1/2. Hence there are two subbeams – and not some other number.Q.A hydrogen atom in ground state passes through a magnetic field that has a gradient of 16mT/m in the vertical direction. If vertical component magnetic moment of the atom is 9.3 × 10–24 J/T, then force on it due to the magnetic moment of the electron is

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The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6is [JEE-2009]a)0b)2.84c)4.90d)5.92Correct answer is option 'A'. Can you explain this answer?
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