The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is :

<p>Melting point of the pairs $\left(M_n, F_e\right),\left(T_c, R_u\right)$, and $\left(\mathrm{Re}_{\mathrm{e}}, \mathrm{O}_{\mathrm{s}}\right)$ :</p> <p>$M n, T_c, \mathrm{Re}$ are same group elements.</p> <p>Also, $\mathrm{Fe}, \mathrm{Bu}, \mathrm{Os}$, are same group elements.</p> <p>$\mathrm{Mn}, \mathrm{Fe}$</p> <p>Fe has a higher melting point than $M_n, M_n< F_e$</p> <p>$\left[\begin{array}{cc}M_n & F_e \\ T_c & R_u \\ R_e & O_s\end{array}\right]$</p> <p>"Melting point property is directly proportional to the strength of metallic bond, which proportional to the delocalize electrons. The more electrons are free to move around, the stronger will be the metallic bond."</p> <p>For Mn and $\mathrm{Fe}, \quad\left(\mathrm{Mn}-d^5 s^2, \mathrm{Fe}-d^6 s^2\right)$.</p> <p>$M_n$ has a weaker metallic bond due to its electronic configuration (half filled configuration). The half filled configuration (stable) results in less delocalization of electrons and weaker attraction between atoms, requiring less energy to break apart and met. But, Fe has localized electrons and hence stronger metallic bonds, leading to higher melting point.</p> <p>$T_c, R u$</p> <p>Tc has lower melting point than $\mathrm{Ru}, \mathrm{Tc}_c<\mathrm{Ru}$ This is because of the electronic configuration of its $d$-orbitals.</p> <p>$T c-d^5 s^2, R u-d^7 s^1$</p> <p>Tc has halt filled d-orbitals, which makes its electronic configuration stable. This configuration causes the nucleus to hold the electrons tightly, which weakens the metallic bond and it results in lower melting point.</p> <p>$\mathrm{Re}, \mathrm{Os}_{\mathrm{s}}$</p> <p>$\operatorname{Re}$ has a higher melting point than $\mathrm{Os}_{\mathrm{s}}, \mathrm{Os}<\operatorname{Re}$ $\left(\mathrm{Re}-d^5 s^2, \mathrm{O} s-d s_s^6\right)$</p> <p>The higher melting point of $R e$ is due to the strength of atomic bonding.</p> <p>Re trams strong metallic bonds than Us. Due to the langer size of Re , its valence elections are not tightly hooded by the nucleus and hence, the valence elections causes stronger metallic bond. As a result, melting point increases.</p> <p>Osmium tills weaker metallic bonds than $\operatorname{Re}$ due to its crystal arrangement and electronic Structure.</p> <p>Comparing Re, and OS,</p> <p>$R_e$ is slightly smaller in atomic radius which allows atoms to pack more efficiently than is and has stronger bonds and higher meeting point.</p> <p>So,</p> <p>$$\begin{aligned} & M_n<\mathrm{Fe} \\ & T_c<\mathrm{Ru} \\ & \mathrm{O}_s<\mathrm{Re} \end{aligned}$$</p> <p>Correct answer:</p> <p>Option 4) $\mathrm{Mn}<\mathrm{Fe}, \mathrm{TC}<\mathrm{Ru}$ and $\mathrm{Os}<\mathrm{Re}$.</p>