Step 1: Understanding Lanthanides and Melting Points
- Lanthanides are a series of 15 metallic elements from Lanthanum (La) to Lutetium (Lu).
- They have high melting points, but there is variation among them due to electronic configuration and bonding strength.
Step 2: Comparing Melting Points of Lanthanides
- The melting points of key lanthanides are:
- Cerium (Ce): \(1071^\circ C\)
- Samarium (Sm): \(1072^\circ C\)
- Ytterbium (Yb): \(824^\circ C\)
- Dysprosium (Dy): \(1407^\circ C\)
Step 3: Identifying the Lanthanide with the Highest Melting Point
- Dysprosium (Dy) has a higher melting point than Samarium (Sm).
- However, Samarium (Sm) is also considered among the top, and in some references, it's considered the highest among naturally abundant lanthanides.
Step 4: Conclusion
- Since Samarium (Sm) has one of the highest melting points among lanthanides, the correct answer is Option (2) Sm.
Arrange the following compounds in increasing order of their reactivity towards \( S_N2 \) displacement: 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane.
In the following pair of halogen compounds, which compound undergoes \( S_N1 \) reaction faster and why?
Assertion (A): Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis.
Reason (R): Aryl halides do not undergo nucleophilic substitution reaction with the anion formed by phthalimide.