Question

The electrons, identified by quantum numbers $n$ and $l$, (i) $n=4,l=1$, (ii) $n=4,l=0$, (iii) $n=3,l=2$, and (iv) $n=3,l=1$ can be placed in order of increasing energy, from the lowest to highest, as:

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

Answer 1

Answer: (A)

(iv) < (ii) < (iii) < (i)

The greater is the value of $(n+l)$, the greater is the energy of orbitals.

(i) $n=4,l=1\Rightarrow 4porbital$

(ii) $n=4,l=0\Rightarrow 4sorbital$

(iii) $n=3,l=2\Rightarrow 3dorbital$

(iv) $n=3,l=1\Rightarrow 3porbital$

According to the Aufbau principle, energies of above-mentioned orbitals are in the order of-

The increasing order of energy $\left(iv\right)3p<\left(ii\right)4s<\left(iii\right)3d<\left(i\right)4p$