The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is:
Show Hint
- Fe < Mn, Ru < Tc and Re < Os
- Mn < Fe, Tc < Ru and Os < Re
- Mn < Fe, Tc < Ru and Re < Os
- Fe < Mn, Ru < Tc and Os < Re
The Correct Option is C
Approach Solution - 1
The question involves determining the order of melting points for pairs of elements (Mn, Fe), (Tc, Ru), and (Re, Os). We have four options to consider. Let's analyze each pair:
- Mn (Manganese) and Fe (Iron):
Manganese has a lower melting point compared to Iron. Therefore, Mn < Fe. - Tc (Technetium) and Ru (Ruthenium):
Technetium, being in the same column group earlier than Ruthenium, has a lower melting point. Thus, Tc < Ru. - Re (Rhenium) and Os (Osmium):
Rhenium has a lower melting point compared to Osmium, so Re < Os.
Based on the analysis, the correct order of the melting points is Mn < Fe, Tc < Ru and Re < Os.
This corresponds to the correct option given: Mn < Fe, Tc < Ru and Re < Os
Approach Solution -2
The melting point of transition metals depends mainly on the strength of metallic bonding, which arises due to the participation of (n−1)d and ns electrons in bond formation. The number of unpaired d-electrons significantly affects metallic bonding strength — more unpaired electrons mean stronger metallic bonds and thus higher melting points.
Step 2: Analyze each pair.
(A) (Mn, Fe)
- Manganese (Mn) has the electronic configuration [Ar] 3d⁵ 4s², with a half-filled d-subshell. Due to its stable configuration, the metallic bonding is weaker, resulting in a relatively lower melting point.
- Iron (Fe) has [Ar] 3d⁶ 4s², allowing more delocalized electrons to participate in bonding, leading to stronger metallic bonds.
Thus, Mn < Fe.
(B) (Tc, Ru)
- Technetium (Tc) and Ruthenium (Ru) belong to the same group (Group 7 and 8 respectively) in the second transition series.
- Ruthenium (Ru) has a higher number of unpaired d-electrons available for metallic bonding than Tc, resulting in a stronger metallic bond.
Hence, Tc < Ru.
(C) (Re, Os)
- Rhenium (Re) and Osmium (Os) belong to the third transition series (5d block).
- Osmium (Os) has more unpaired d-electrons and forms stronger metallic bonds than Rhenium (Re).
Therefore, Re < Os.
Step 3: Conclusion.
The correct order of melting points based on metallic bonding strength is:
\[ \boxed{\text{Mn < Fe, Tc < Ru, and Re < Os}} \]
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Periodic properties Questions
- The element having the highest first ionization enthalpy is
- Given below are two statements:
Statement I: Fluorine has most negative electron gain enthalpy in its group.
Statement II: Oxygen has least negative electron gain enthalpy in its group.
In the light of the above statements, choose the most appropriate from the options given below. - Anomalous behaviour of oxygen is due to its
- In case of isoelectronic species the size of F– , Ne and Na+ is affected by:
- Lowest oxidation number of an atom in a compound \( A_2B \) is -2. The number of electrons in its valence shell is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.