In comparison to boron, beryllium has
(a) lesser nuclear charge and lesser first ionisation enthalpy

Q: Doesn't ionization energy always increase as we move from left to right across a period?

Generally, yes. As nuclear charge increases, electrons are held more tightly. However, half-filled and fully-filled subshells provide extra quantum mechanical stability. This causes exceptions in the trend, such as Beryllium ($2s^2$) having a higher ionization energy than Boron ($2p^1$), and Nitrogen ($2p^3$) having a higher ionization energy than Oxygen ($2p^4$).

Q: Why is the 2p electron easier to remove than the 2s electron?

The $2p$ subshell is at a slightly higher energy level than the $2s$ subshell. Additionally, the $2s$ electrons shield the $2p$ electron from the nucleus, reducing the effective nuclear charge it experiences. This makes the $2p$ electron more loosely bound and easier to remove.

Q: What exactly is nuclear charge?

Nuclear charge is simply the total positive charge of the protons in the nucleus. It is directly equal to the atomic number ($Z$). Effective nuclear charge ($Z_{eff}$), on the other hand, is the net positive charge experienced by an electron after accounting for shielding by inner electrons.

Q: Does Beryllium have a higher second ionization energy too?

No! After losing one electron, Beryllium becomes $1s^2 2s^1$, while Boron becomes $1s^2 2s^2$. Now, Boron has the highly stable fully filled $2s$ subshell. Therefore, Boron will have a much higher second ionization energy than Beryllium.

Q: How does the penetration effect play a role here?

The penetration effect describes how close an electron in a specific orbital can get to the nucleus. The $s$-orbital is more penetrating than the $p$-orbital. Thus, an electron in the $2s$ orbital is held more tightly by the nucleus than an electron in the $2p$ orbital, increasing the ionization energy.

Question

In comparison to boron, beryllium has

Options:

(A) lesser nuclear charge and lesser first ionisation enthalpy

(B) greater nuclear charge and lesser first ionisation enthalpy

(C) greater nuclear charge and greater first ionisation enthalpy

(D) lesser nuclear charge and greater first ionisation enthalpy

Sigma Percentile · Accepted Answer

### The Setup: Meet the Neighbors

When we look at the periodic table, Beryllium (Be) and Boron (B) sit right next to each other in the second period. Beryllium is the alkaline earth metal at group 2, while Boron is the metalloid kicking off group 13. To compare their properties, we first need to look at their most fundamental characteristic: the atomic number ($Z$).

Beryllium has an atomic number of $Z = 4$, meaning it has exactly 4 protons in its nucleus. Boron, being one step to the right, has an atomic number of $Z = 5$, giving it 5 protons.

### The Nuclear Charge: A Simple Count

The nuclear charge of an atom is simply the total positive charge residing in its nucleus, which is directly proportional to the number of protons. Since Boron has 5 protons and Beryllium has only 4, it is a straightforward conclusion that **Beryllium has a lesser nuclear charge than Boron**.

Normally, as nuclear charge increases across a period, the atomic radius decreases, and the electrons are held more tightly. This usually means that it becomes harder to remove an electron. But nature loves exceptions, and this is where the story gets interesting.

### The Plot Twist: Electronic Configuration

To understand how tightly the outermost electrons are held, we must look at the electronic configuration. Let's write them down:

For Beryllium ($Z=4$): 
$$1s^2, 2s^2$$

For Boron ($Z=5$): 
$$1s^2, 2s^2, 2p^1$$

Notice something special about Beryllium? Its outermost subshell, the $2s$ orbital, is **completely filled** with two electrons. In quantum mechanics, fully filled and half-filled subshells possess extra stability due to symmetry and exchange energy. It is like a perfectly packed suitcase; it is very hard to pull something out of it.

Boron, however, has a lone electron sitting in the $2p$ subshell. This $2p$ electron is at a slightly higher energy level than the $2s$ electrons. Furthermore, the $s$-orbitals are more penetrating than $p$-orbitals. This means the $2s$ electrons spend more time closer to the nucleus and are shielded less effectively than the $2p$ electron.

### The Final Verdict

First ionization enthalpy ($
\Delta_i H_1
$) is the energy required to remove the most loosely bound electron from an isolated gaseous atom.

Because Beryllium's outermost electrons are in a highly stable, fully filled, and deeply penetrating $2s$ orbital, it requires a significantly **greater amount of energy** to pluck one away compared to Boron's loosely bound $2p$ electron.

Therefore, despite having a lesser nuclear charge, **Beryllium has a greater first ionization enthalpy than Boron**. This beautiful interplay of quantum mechanics and electrostatic forces makes option (d) the undisputed correct answer.

Animated Solution for Chemistry - Periodicity in Properties: In comparison to boron, beryllium has

Select Answer:

Visualized Solution

$Z_{Be} vs Z_{B}$

$Nuclear Charge$

$Electronic Configuration$

$Configuration of Be$

$Configuration of B$

$First Ionization Enthalpy (Δ_{i} H_{1})$

$Stability & Penetration$

$Comparing Δ_{i} H_{1}$

$Conclusion$

$The Way Forward$

The Sigma Insight: Periodic Table and Periodic Properties

The Setup

The Nuclear Charge

The Plot Twist

The Final Verdict

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