B has a smaller first ionisation enthalpy than Be. Consider the following statements :
(I) It is eas

Q: Why does Beryllium have a higher ionisation energy if Boron has more protons?

While Boron does have more protons (higher nuclear charge), the electron being removed in Boron is in a $2p$ orbital, whereas in Beryllium it is in a $2s$ orbital. The $2s$ orbital penetrates closer to the nucleus and is more tightly bound, overcoming the effect of Boron's extra proton.

Q: What exactly is 'penetration power'?

Penetration power describes how close an electron in a specific orbital can get to the nucleus. Because of their spherical shape, $s$-orbitals have a higher probability of being near the nucleus than $p$-orbitals. The order is always $s > p > d > f$.

Q: Does the fully filled $2s^2$ configuration in Beryllium make it extra stable?

Yes! A fully filled subshell (like $2s^2$) has a symmetrical electron distribution and extra stability. Breaking this stable configuration requires more energy, contributing to Beryllium's higher ionisation enthalpy.

Q: Why is Statement IV false? Doesn't adding an electron make the atom bigger?

When you move from Be to B, you add an electron, but you also add a proton to the nucleus. Because the new electron goes into the same principal shell ($n=2$), the increased positive charge of the nucleus pulls the entire shell inwards, making the atom smaller.

Q: How does shielding work in Boron compared to Beryllium?

In Beryllium, the outermost $2s$ electrons are shielded only by the inner $1s^2$ core. In Boron, the outermost $2p$ electron is shielded by both the $1s^2$ core AND the $2s^2$ electrons. This extra shielding makes the $2p$ electron easier to remove.

Question

B has a smaller first ionisation enthalpy than Be. Consider the following statements :
(I) It is easier to remove $2p$ electron than $2s$ electron
(II) $2p$ electron of B is more shielded from the nucleus by the inner core of electrons than the $2s$ electrons of Be
(III) $2s$ electron has more penetration power than $2p$ electron
(IV) atomic radius of B is more than Be
(atomic number B = 5, Be = 4)
The correct statements are

Options:

(A) (I), (II) and (III)

(B) (II), (III) and (IV)

(C) (I), (III) and (IV)

(D) (I), (II) and (IV)

Sigma Percentile · Accepted Answer

### The Anomaly of Beryllium and Boron

When we study the periodic table, we learn a general rule: **Ionisation enthalpy increases as we move from left to right across a period.** This happens because the effective nuclear charge ($Z_{	ext{eff}}$) increases, pulling the electrons closer and making them harder to remove.

However, chemistry is full of fascinating exceptions, and the pair of Beryllium (Be) and Boron (B) is a classic one. Even though Boron is to the right of Beryllium, its first ionisation enthalpy is actually *lower*. Let's dive into the quantum mechanics of why this happens by evaluating the given statements.

### Analyzing the Electronic Configurations

The secret lies in the electronic configurations of these two elements.

For Beryllium ($Z=4$), the configuration is $1s^2 2s^2$. The outermost electrons are in a fully filled, stable $2s$ subshell.

For Boron ($Z=5$), the configuration is $1s^2 2s^2 2p^1$. The outermost electron is a lone occupant in the $2p$ subshell.

### The Power of Penetration

Let's look at **Statement III**: "$2s$ electron has more penetration power than $2p$ electron."

Penetration power refers to the ability of an orbital to get close to the nucleus. Because of the shape of the orbitals, an $s$-orbital has a higher probability of being found near the nucleus compared to a $p$-orbital of the same principal quantum number. The order of penetration is $s > p > d > f$.

Because the $2s$ electrons in Beryllium penetrate closer to the nucleus, they are held more tightly by the positive nuclear charge. This makes Statement III **true**.

### The Ease of Removal

This brings us directly to **Statement I**: "It is easier to remove $2p$ electron than $2s$ electron."

Since the $2p$ electron in Boron is further away from the nucleus (less penetrating) than the $2s$ electrons in Beryllium, it experiences a weaker electrostatic pull. Consequently, it requires less energy to pluck that $2p$ electron away. This is the primary reason why Boron has a lower ionisation enthalpy. Statement I is **true**.

### The Shielding Effect

Now, let's examine **Statement II**: "$2p$ electron of B is more shielded from the nucleus by the inner core of electrons than the $2s$ electrons of Be."

Shielding (or screening) occurs when inner electrons repel outer electrons, effectively reducing the full nuclear charge to a smaller effective nuclear charge ($Z_{	ext{eff}}$).

In Beryllium, the $2s$ electrons are shielded only by the two electrons in the $1s$ shell.

In Boron, the single $2p$ electron is shielded by the two $1s$ electrons **and** the two $2s$ electrons. This increased shielding from the inner core makes the $2p$ electron even more loosely bound. Thus, Statement II is also **true**.

### The Atomic Radius Trend

Finally, let's check **Statement IV**: "atomic radius of B is more than Be."

As we move from left to right across a period (from Be to B), the number of protons in the nucleus increases. This increases the effective nuclear charge, which pulls the entire electron cloud closer to the nucleus. Therefore, the atomic radius generally decreases across a period.

Boron actually has a **smaller** atomic radius than Beryllium. This makes Statement IV **false**.

### Final Conclusion

By carefully analyzing the quantum mechanical properties of these atoms, we found that Statements I, II, and III are correct, while Statement IV is incorrect. This perfectly aligns with option (a).

Select Answer:

Visualized Solution

\text{Electronic Configurations}

\text{Penetration Effect}

\text{Evaluating Statements I \& III}

\text{Shielding Effect}

\text{Evaluating Statement II}

\text{Atomic Radius Trend}

\text{Conclusion}

The Sigma Insight: Periodic Table and Periodic Properties

The Anomaly of Beryllium and Boron

Analyzing the Electronic Configurations

The Power of Penetration

The Ease of Removal

The Shielding Effect

The Atomic Radius Trend

Final Conclusion

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