To determine the order of decreasing energy for the given orbitals, we use the $\\$(n + l) \text{ rule, where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.} \\ The values of $l$ for different orbitals are:$\\$• $s: l = 0 \\$• $p: l = 1 \\$• $d: l = 2 \\$• $f: l = 3 \\$Let's calculate $(n + l)$ for each orbital:$\\$• For $4d: n = 4, l = 2 \Rightarrow n + l = 4 + 2 = 6 \\$• For $5p: n = 5, l = 1 \Rightarrow n + l = 5 + 1 = 6 \\$• For $5f: n = 5, l = 3 \Rightarrow n + l = 5 + 3 = 8 \\$• For $6p: n = 6, l = 1 \Rightarrow n + l = 6 + 1 = 7 \\$According to the $(n + l)$ rule, orbitals with higher $(n + l)$ values have higher energy.$\\$If two orbitals have the same $(n + l)$ value, the one with the higher $n$ value has higher energy.$\\$Now, let's arrange the orbitals in order of decreasing energy:$\\$• $5f$ has the highest $(n + l)$ value of 8.$\\$• Next is $6p$ with $(n + l) = 7. \\$• Both $4d$ and $5p$ have $(n + l) = 6,$ but $5p$ has a higher $n$ value than $4d. \\$Therefore, the correct order is: $5f > 6p > 5p > 4d. \\$This corresponds to Option 1.