Answer
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Hint: This $\text{Mn}$ is a transition metal. It has the electronic configuration\[\left[ \text{Ar} \right]\text{3}{{\text{d}}^{\text{5}}}\text{4}{{\text{s}}^{\text{2}}}\]. It is a half-filled d-orbital and has extraordinary stability. Thus even though it has a maximum number of electrons, it cannot form a metallic bond. Thus it has a low melting point.
Complete step by step solution:
The transition metals have very high melting and boiling points. The melting point of the metals rises and reaches the maximum and then decreases with an increase in atomic number.
The high boiling or melting point of transition metal is due to the strong metallic bond between the atoms of these elements. This is also evident from the fact that these metals have high enthalpies of atomization. The metallic bond is formed due to the interaction of electrons in outermost orbitals. The strength of bonding is roughly related to the number of unpaired electrons. In general, the greater is the number of valence electrons, the higher is the melting point. Therefore, as we move along a 3d series, the metallic strength increases up to the middle with increasing the availability of unpaired electron up to the ${{\text{d}}^{\text{5}}}$ configuration and then decreases after the middle.
If we follow this trend then $\text{Mn}$ should have a higher melting point than $\text{Fe}$.
However, there is a dip in the melting point$\text{Mn}$. This is an unexpected observation probably because $\text{Mn}$ has a stable electronic configuration(half-filled ${{\text{d}}^{\text{5}}}$and filled \[\text{4}{{\text{s}}^{\text{2}}}\] ).as a result the stable configuration decrease the localization of electrons resulting in the weaker metallic bonding than the other elements.
Thus $\text{Mn}$have a lower melting point than the $\text{Fe}$.this is because the $\text{Mn}$ has a higher number of unpaired e than Fe in the atomic state.
Therefore the assertion is incorrect but the reason is the correct statement.
Hence, (D) is the correct option.
Note: The same trend of the melting point and abnormal lower melting point is observed in the second and third transition metal series with \[\text{Tc}\] and $\text{ Re}$.
Complete step by step solution:
The transition metals have very high melting and boiling points. The melting point of the metals rises and reaches the maximum and then decreases with an increase in atomic number.
The high boiling or melting point of transition metal is due to the strong metallic bond between the atoms of these elements. This is also evident from the fact that these metals have high enthalpies of atomization. The metallic bond is formed due to the interaction of electrons in outermost orbitals. The strength of bonding is roughly related to the number of unpaired electrons. In general, the greater is the number of valence electrons, the higher is the melting point. Therefore, as we move along a 3d series, the metallic strength increases up to the middle with increasing the availability of unpaired electron up to the ${{\text{d}}^{\text{5}}}$ configuration and then decreases after the middle.
If we follow this trend then $\text{Mn}$ should have a higher melting point than $\text{Fe}$.
However, there is a dip in the melting point$\text{Mn}$. This is an unexpected observation probably because $\text{Mn}$ has a stable electronic configuration(half-filled ${{\text{d}}^{\text{5}}}$and filled \[\text{4}{{\text{s}}^{\text{2}}}\] ).as a result the stable configuration decrease the localization of electrons resulting in the weaker metallic bonding than the other elements.
Thus $\text{Mn}$have a lower melting point than the $\text{Fe}$.this is because the $\text{Mn}$ has a higher number of unpaired e than Fe in the atomic state.
Therefore the assertion is incorrect but the reason is the correct statement.
Hence, (D) is the correct option.
Note: The same trend of the melting point and abnormal lower melting point is observed in the second and third transition metal series with \[\text{Tc}\] and $\text{ Re}$.
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