The calculated spin only magnetic moment of $Cr^{2+}$ ion is :
- 3.87 BM
- 4.90 BM
- 5.92 BM
- 2.84 BM
The Correct Option is B
Solution and Explanation
Electronic configuration of $Cr^{+2}=[Ar]3d^{4}$
Number of unpaired electrons $= 4$
Spin only magnetic moment $=\sqrt{n\left(n+2\right)}BM $ as $n=4$
$\therefore\mu=\sqrt{4\left(4+2\right)}$
$=\sqrt{24}$
$=4.90\,BM$
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Concepts Used:
Properties of D Block Elements
- Multiple oxidation states- The oxidation states of d block elements show very few energy gaps; therefore, they exhibit many oxidation states. Also, the energy difference between s and d orbital is very less. Therefore both the electrons are involved in ionic and covalent bond formation, which ultimately leads to multiple oxidation states.
- Formation of complex compounds- Ligands show a binding behavior and can form so many stable complexes with the help of transition metals. This property is mainly due to:
- Availability of vacant d orbitals.
- Comparatively small sizes of metals.
- Hardness- Transition elements are tough and have high densities because of the presence of unpaired electrons.
- Melting and boiling points- Melting and boiling points of transition are very high because of the presence of unpaired electrons and partially filled d orbitals. They form strong bonds and have high melting and boiling points.
- Atomic radii- The atomic and ionic radius of the transition elements decreases as we move from Group 3 to group 6. However, it remains the same between group 7 and group 10, and from group 11 to group 12 increases.
- Ionization enthalpy- The ionization enthalpies of the transition elements are generally on the greater side as compared to the S block elements