Question

$\text{n}$ and $\ell $ for four electrons are given as $\left(\text{i}\right) \, \text{n} = 4 , \, \ell = 1 \, \left(\text{ii}\right) \, \text{n} = 4 , \, \ell = 0 \, \left(\text{iii}\right) \, \text{n} = 3 , \, \ell = 2 \, \left(\text{iv}\right) \, \text{n} = 3 , \, \ell = 1.$
Arrange them in order of increasing energy from the lowest to highest:

• A. (ii) < (iv) < (i) < (iii)
• B. (i) < (iii) < (ii) < (iv)
• C. (iii) < (i) < (iv) < (ii)
• D. (iv) < (ii) < (iii) < (i)

Answer 1

Answer: (D)

Lower the value of $\left(\text{n} + \ell \right)$ , lower is the energy. When $\left(\text{n} + \ell \right)$ value is the same for two orbitals, lower $\text{n}$ value implies lower energy of the electron. Thus, the order will be:
(iv) < (ii) < (iii) < (i).