Animated Solution for Chemistry - Periodicity in Properties: Match List-I with List-II. $\begin{array}{clcl}& \text{List-I (Electronic configuration of elements)}& & \text{List-II (}{\Delta }_{i}H{\text{ in kJ mol}}^{-1}\text{)}\\ \text{A.}& 1s^{2}2s^2& \text{(i)}& 801\\ \text{B.}& 1s^{2}2s^{2}2p^4& \text{(ii)}& 899\\ \text{C.}& 1s^{2}2s^{2}2p^3& \text{(iii)}& 1314\\ \text{D.}& 1s^{2}2s^{2}2p^1& \text{(iv)}& 1402\end{array}$ Choose the most appropriate answer from the options given below.

Before we can even think about matching values, we need to know exactly which elements we are dealing with. The question provides us with the outermost electronic configurations of four elements from the second period of the periodic table.

Let's break them down: - A: $1s^{2}2s^2$ corresponds to an atomic number of 4, which is Beryllium (Be). - B: $1s^{2}2s^{2}2p^4$ corresponds to an atomic number of 8, which is Oxygen (O). - C: $1s^{2}2s^{2}2p^3$ corresponds to an atomic number of 7, which is Nitrogen (N). - D: $1s^{2}2s^{2}2p^1$ corresponds to an atomic number of 5, which is Boron (B).

A Game of Nuclear Pull

Ionization enthalpy ($ \Delta_i H $) is the energy required to rip an electron away from an isolated gaseous atom. What is the general rule of thumb here?

As we march from left to right across a period, the atomic number increases. This means more protons are added to the nucleus, increasing the effective nuclear charge ($Z_{\text{eff}}$). The nucleus pulls the outermost electrons more tightly, shrinking the atomic radius and making it progressively harder to remove an electron.

Based purely on this general trend, we would expect the ionization energy to increase smoothly: $\mathrm{B}<\mathrm{Be}<\mathrm{N}<\mathrm{O}$. But nature loves exceptions, and this is where the real chemistry begins!

Beryllium vs. Boron

Let's zoom in on Beryllium and Boron. Beryllium has a perfectly stable, fully filled $2s$ orbital ($2s^2$). Boron, however, has one lonely electron sitting in the $2p$ orbital ($2p^1$).

Because the $2p$ orbital is slightly higher in energy and less penetrating (further from the nucleus) than the $2s$ orbital, it is actually easier to remove that single $2p$ electron from Boron than it is to break the stable $2s^2$ pair in Beryllium.

Therefore, we see our first anomaly: ${\Delta }_{i}H(\mathrm{Be})>{\Delta }_{i}H(\mathrm{B})$.

Nitrogen vs. Oxygen

Now, let's look at Nitrogen and Oxygen. Nitrogen has an exactly half-filled $2p$ subshell ($2p^3$). According to Hund's Rule, this means it has three unpaired electrons in three separate, degenerate orbitals. This highly symmetrical arrangement provides extra quantum mechanical stability.

Oxygen, on the other hand, has four electrons in the $2p$ subshell ($2p^4$). This forces two electrons to pair up in one of the orbitals. These paired electrons repel each other strongly (inter-electronic repulsion). Losing one electron actually relieves this repulsion and allows Oxygen to achieve that highly stable half-filled state.

Because of this, it is easier to remove an electron from Oxygen than from Nitrogen. Hence, our second anomaly: ${\Delta }_{i}H(\mathrm{N})>{\Delta }_{i}H(\mathrm{O})$.

Combining the general trend with our two crucial exceptions, the correct increasing order of first ionization enthalpy is:

Now, we simply match this order with the numerical values provided in List-II ($801<899<1314<1402{\text{ kJ mol}}^{-1}$): - Boron (D) gets the lowest value: $801$ (i) - Beryllium (A) gets the next value: $899$ (ii) - Oxygen (B) gets the next value: $1314$ (iii) - Nitrogen (C) gets the highest value: $1402$ (iv)

This perfectly aligns with option (a): A-(ii), B-(iii), C-(iv), D-(i).