Question

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

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

Answer 1

The correct option is A (iv)< (ii)< (iii)< (i)

(i) $4p$ (ii) $4s$ (iii) $3d$ (iv) $3p$

According to Aufbau rule, order of increasing energy is $3p<4s<3d<4p$