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The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is
  • 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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Verified Answer
The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)...
Chromium in Cr (CO)6 is in zero oxidation state and has [Ar]18 3d54s1 as the electronic configuration. However, CO is a strong ligand, hence pairing up of electrons takes place leading to following configuration in Cr(CO)6.
Since the complex has no unpaired electron, its magnetic moment is zero.
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The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)...
Understanding the Magnetic Moment in Cr(CO)6
The magnetic moment of a complex is determined by the presence of unpaired electrons. In the case of Cr(CO)6, we need to analyze its electronic configuration and the effect of the ligands.
Electronic Configuration of Chromium
- Chromium (Cr) has an atomic number of 24.
- Its ground state electronic configuration is [Ar] 3d5 4s1.
- In Cr(CO)6, chromium is in a +3 oxidation state, which results in an electronic configuration of 3d4 (after losing 3 electrons).
CO as a Ligand
- Carbon monoxide (CO) is a strong-field ligand and causes pairing of electrons.
- In an octahedral field created by six CO ligands, the 3d orbitals split into two groups: lower-energy t2g and higher-energy eg orbitals.
Electron Configuration in Cr(CO)6
- The 3d4 configuration under the influence of CO will lead to all four electrons being paired in the t2g orbitals.
- Thus, there are no unpaired electrons in Cr(CO)6.
Calculating the Magnetic Moment
- The spin-only magnetic moment formula is given by the equation: μ = √(n(n+2)), where n is the number of unpaired electrons.
- Since Cr(CO)6 has 0 unpaired electrons (n=0), substituting into the formula gives: μ = √(0(0+2)) = 0.
Conclusion
- Therefore, the spin-only magnetic moment of Cr(CO)6 is 0 Bohr magnetons, confirming that the correct answer is option 'A'.
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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)6 isa)0b)2.84c)4.90d)5.92Correct answer is option 'A'. Can you explain this answer?
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