To place the electrons in order of increasing energy based on their quantum numbers, we consider the principles that dictate electron energy levels in atoms. The energy of an electron in an atom is primarily determined by its principal quantum number ($n$) and azimuthal (or angular momentum) quantum number ($l$). Generally, the energy of an electron increases with an increase in the $n$ value. However, for electrons within the same $n$ level, those with higher $l$ values have slightly higher energies due to electron penetration and shielding effects. In multi-electron atoms, due to electron-electron repulsion and orbital shapes, orbitals with the same $n$ but different $l$ values do not have the same energy.

Given the quantum numbers:

• (i) $n = 4, l = 1$
• (ii) $n = 4, l = 0$
• (iii) $n = 3, l = 2$
• (iv) $n = 3, l = 1$

We'll arrange them in order of increasing energy.

The principal quantum number, $n$, primarily determines the energy level. Hence, states with $n = 3$ would generally have lower energy compared to states with $n = 4$. Within the same $n$ level, the energy increases with $l$. Thus, a higher $l$ value indicates a slightly higher energy within the same $n$ level.

Based on these guidelines, we can place the given states in the following order:

1. (iv) $n = 3, l = 1$
2. (iii) $n = 3, l = 2$
3. (ii) $n = 4, l = 0$
4. (i) $n = 4, l = 1$

So, the electrons in order of increasing energy, from the lowest to highest, are Option A (iv)<(iii)<(ii)<(i). This ranking reflects that within the same principal quantum number, $n$, the electron energy increases with the azimuthal quantum number, $l$, and electrons with a lower principal quantum number typically have lower energy than those with a higher $n$.