Which orbital in each of the following pairs is lower in energy in a many electron atom?
`(i) 2s , 2p " " (ii) 3p, 3d`
`(iii) 3s, 4s " " (iv) 4d , 5f`

Explanation

To determine which orbital in each of the given pairs is lower in energy in a many-electron atom, we will apply the n + l rule. According to this rule, the energy of an orbital is determined by the sum of the principal quantum number (n) and the azimuthal quantum number (l). The orbital with the lower n + l value has lower energy. If two orbitals have the same n + l value, the orbital with the higher principal quantum number (n) has higher energy.

Let's analyze each pair step by step:

### Step 1: Analyze the pair (i) 2s, 2p
- For **2s**:
- n = 2, l = 0 (s orbital)
- n + l = 2 + 0 = 2
...