Question

Here, \[4d,5d,5f\] and $6p$ orbitals are arranged in the order of decreasing energy. The correct option is:
$
  (1)5f > 6p > 5p > 4d \\
  (2)6p > 5f > 5p > 4d \\
  (3)6p > 5f > 4d > 5p \\
  (4)5f > 6p > 4d > 5p \\
 $

Answer 1

Hint: The energy of an electron is defined by the principal quantum number. The azimuthal quantum number represents the shape of an atomic orbital and determines its orbital angular momentum.

Complete answer:
The energy of orbitals is determined by the principal quantum number $\left( n \right)$ as well as the azimuthal quantum number $\left( l \right)$. Therefore, the lower the sum of the Principal quantum number $\left( n \right)$ and the Azimuthal Quantum Number $\left( l \right)$ for an orbital, the lower the energy of the orbital.
When two orbitals have an equal value of $\left( {n + l} \right)$, the orbital with the lower value of $n$ (principal quantum number) has the lower energy.
For orbital $4d$:
The Principal quantum number $\left( n \right) = 4$
The Azimuthal Quantum Number $\left( l \right) = 2$
The total of the Principal quantum number $\left( n \right)$ and Azimuthal Quantum Number $\left( l \right)$ for $4d$ orbital $ \Rightarrow n + l = 4 + 2 = 6$
For orbital $5p$:
The Principal quantum number $\left( n \right) = 5$
The Azimuthal Quantum Number $\left( l \right) = 1$
The total of the Principal quantum number $\left( n \right)$ and Azimuthal Quantum Number $\left( l \right)$ for $5p$ orbital $ \Rightarrow n + l = 5 + 1 = 6$
For orbital $5f$:
The Principal quantum number $\left( n \right) = 5$
The Azimuthal Quantum Number $\left( l \right) = 3$
The total of the Principal quantum number $\left( n \right)$ and Azimuthal Quantum Number $\left( l \right)$ for $5f$ orbital $ \Rightarrow n + l = 5 + 3 = 8$
For orbital $6p$:
The Principal quantum number $\left( n \right) = 6$
The Azimuthal Quantum Number $\left( l \right) = 1$
The total of the Principal quantum number $\left( n \right)$ and Azimuthal Quantum Number $\left( l \right)$ for $6p$ orbital $ \Rightarrow n + l = 6 + 1 = 7$
Since $4d$ and $5p$ have the same $\left( {n + l} \right)$ value, among these two, $4d$ has a lower value of $n$, and therefore, $4d$ has lower energy as compared to $5p$.
Therefore, the decreasing order of energy of the given orbitals is $5f > 6p > 5p > 4d$

Hence, the correct option is $\left( 1 \right)$

Note:
Electrons first fill low-energy orbitals (those closest to the nucleus) before moving on to higher-energy orbitals. They fill the orbitals singly as much as possible when there is a preference between orbitals of equal energy. Hund's law states that orbitals should be filled singly wherever possible.