Identify the elements X and Y using the ionisation energy values given below.

Ionisation energy (Is

Q: Why is the first ionization energy of Magnesium higher than that of Sodium?

Magnesium has a fully filled $3s^2$ orbital, which is more stable than Sodium's half-filled $3s^1$ orbital. Additionally, Magnesium has a higher effective nuclear charge, holding its valence electrons more tightly.

Q: Why doesn't element Y show a huge jump in its second ionization energy?

Element Y (Magnesium) has two valence electrons ($3s^2$). The second electron is still being removed from the valence shell, not from the tightly bound inner noble gas core, so the energy required only roughly doubles.

Q: What would the third ionization energy of Magnesium look like?

It would be astronomically high! After losing two electrons, Magnesium achieves a stable Neon core ($1s^2 2s^2 2p^6$). Removing a third electron would break this core, resulting in a massive energy jump.

Q: Could element X be Fluorine?

No. Fluorine is a highly electronegative non-metal with a very high first ionization energy (around $1681 \text{ kJ/mol}$). It has 7 valence electrons, so a massive jump would only occur after the 7th ionization energy, not the 1st.

Q: How do we mathematically define a 'huge jump' in ionization energy?

While there's no strict mathematical rule, a jump of 5x to 10x or more usually indicates a change in the principal quantum number (breaking a noble gas core). A normal jump within the same shell is typically around 1.5x to 3x.

Question

Identify the elements $X$ and $Y$ using the ionisation energy values given below.

$\begin{array}{lcc}
& 	ext{Ionisation energy (Ist)} & 	ext{kJ/mol (IInd)} \
X & 495 & 4563 \
Y & 731 & 1450
\end{array}$

Options:

(A) $X = 	ext{Na}, Y = 	ext{Mg}$

(B) $X = 	ext{Mg}, Y = 	ext{F}$

(C) $X = 	ext{Mg}, Y = 	ext{Na}$

(D) $X = 	ext{F}, Y = 	ext{Mg}$

Sigma Percentile · Accepted Answer

### Cracking the Code of Ionization Energies

Imagine you are trying to take money from someone. Taking a few loose bills from their pocket might be relatively easy, but trying to break into their highly secure bank vault? That requires an astronomical amount of energy. This is exactly how atoms behave when we try to remove their electrons, a concept we measure as **Ionization Energy ($IE$)**.

In this problem, we are given two mystery elements, $X$ and $Y$, along with their first and second ionization energies. Our mission is to play atomic detectives and identify them based on how tightly they hold onto their electrons.

### Analyzing Element X: The Ten-Fold Leap

Let's focus our attention on element $X$. Its first ionization energy ($IE_1$) is a modest $495 	ext{ kJ/mol}$. This is the energy required to pluck the outermost, most loosely bound electron.

But look what happens when we try to take a second electron! The second ionization energy ($IE_2$) skyrockets to $4563 	ext{ kJ/mol}$. That is nearly a ten-fold increase!

What does this massive jump tell us? It screams that after losing just one electron, element $X$ achieved a state of ultimate stability—a **noble gas configuration**. The second electron is not sitting in a loose outer shell; it is locked away inside a deeply stable, fully-filled inner core. Therefore, element $X$ must have exactly **one valence electron**, placing it squarely in Group 1 (the alkali metals).

Looking at our options, Sodium ($	ext{Na}$) fits this profile perfectly. Its electronic configuration is $[	ext{Ne}] 3s^1$. Removing the $3s^1$ electron is easy, but trying to break into the $[	ext{Ne}]$ core for the second electron requires immense energy.

### Analyzing Element Y: The Steady Climb

Now, let's examine element $Y$. Its $IE_1$ is $731 	ext{ kJ/mol}$, and its $IE_2$ is $1450 	ext{ kJ/mol}$.

Notice the difference here. The energy roughly doubles. While it is always harder to remove a negative electron from an increasingly positive ion, there is no sudden, astronomical leap. This tells us a crucial fact: the second electron is being removed from the **same valence shell** as the first one. It is not coming from a stable noble gas core.

This implies that element $Y$ has more than one valence electron. Among our options, Magnesium ($	ext{Mg}$) is a Group 2 alkaline earth metal with the configuration $[	ext{Ne}] 3s^2$. The first electron comes from the $3s$ orbital, and the second electron also comes from the $3s$ orbital. Hence, the energy jump is normal and expected.

### The Final Verdict

By simply observing the ratios of successive ionization energies, we have decoded the atomic structures. A massive jump indicates the breaking of a noble gas core, while a steady increase indicates the continued emptying of a valence shell.

Therefore, we can confidently conclude that **$X$ is Sodium ($	ext{Na}$)** and **$Y$ is Magnesium ($	ext{Mg}$)**. The correct option is (a).

Animated Solution for Chemistry - Periodicity in Properties: Identify the elements $X$ and $Y$ using the ionisation energy values given below. $\begin{array}{lcc} & \text{Ionisation energy (Ist)} & \text{kJ/mol (IInd)} \\ X & 495 & 4563 \\ Y & 731 & 1450 \end{array}$

Select Answer:

Visualized Solution

$Given Data$

$Concept of Successive I E$

$Analyzing Element X$

$Identifying X$

$Analyzing Element Y$

$Identifying Y$

$Conclusion$

$The Way Forward$

The Sigma Insight: Periodic Table and Periodic Properties

Cracking the Code of Ionization Energies

Analyzing Element X

Analyzing Element Y

The Final Verdict

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