Question

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

• A. $\left(iv\right)<\left(ii\right)<\left(iii\right)<\left(i\right)$
• B. $\left(ii\right)<\left(iv\right)<\left(i\right)<\left(iii\right)$
• C. $\left(i\right)<\left(iii\right)<\left(ii\right)<\left(iv\right)$
• D. $\left(iii\right)<\left(i\right)<\left(iv\right)<\left(ii\right)$

Answer 1

The correct option is A $\left(iv\right)<\left(ii\right)<\left(iii\right)<\left(i\right)$

According to Aufbau's principle,
orbitals with lower n+l have lower energy so as valuse of n+l increases, its energy increases.
Also, if (n+l) is same, lower is n, lower is energy of orbit.
here,
(i)n=4,l=1 n+l=5
(ii)n=4,l=0 n+l=4
(iiii)n=3,l=2 n+l=5
(iv)n=3,l=1 n+l=4
so order is
(iv)<(ii)<(iii)<(i)