$Cr ^{+2}< Mn ^{+2}< Fe ^{+2}$ - Paramagnetic behaviour
none of the above
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The Correct Option isD
Solution and Explanation
$\underset{24}{ Cr }[ Ar ] 3 d ^{5} 4 s ^{1} \,\,\, Cr ^{2+} 3 d ^{4} 4 s ^{0}=4$ unpaired electrons $Mn [ Ar ] 3 d ^{5} 4 s ^{2} Mn ^{2+} 3 d ^{5} 4 s ^{0}=5$ unpairedelectrons $Fe [ Ar ] 3 d ^{6} 4 s ^{2} \,\,\,Fe ^{2+} 3 d ^{6} 4 s ^{0}=4$ unpaired electrons Since both $Cr ^{2+}$ and $Fe ^{2+}$ contain $4$ unpaired electrons, the correct option is $C$. Option D is also correct because density order given is not correct.
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