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

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

The two guiding rules to arrange the various orbitals in the increasing energy are:
(i) Energy of an orbital increases with increase in the value of $n+l$.
(ii) If orbitals having the same value of $n+l$, the orbital with lower value of $n$ has lower energy.
Thus for the given orbitals, we have
(i) $n+l=4+1=5$
(ii) $n+l=4+0=4$
(iii) $n+l=3+2=5$
(iv) $n+l=3+1=4$
Hence, the order of increasing energy is
(iv) < (ii) < (iii) < (i)