The first and second ionisation enthalpies of a metal are 496 and 4560 kJ mol^-1, respectively. How

Q: Why is the second ionization energy so much higher than the first?

Because after losing the first electron, the alkali metal achieves a stable noble gas electron configuration. Removing an electron from this highly stable, fully-filled inner shell requires an immense amount of energy.

Q: How do we know the hydroxide is a monoacidic base?

Since the metal belongs to Group 1, it forms an $M^+$ ion. Therefore, its hydroxide formula is $MOH$. When dissolved in water, one molecule of $MOH$ yields exactly one $OH^-$ ion, making it a monoacidic base.

Q: What if the huge jump was after the second ionization energy?

If the jump occurred between $IE_2$ and $IE_3$, it would mean the metal easily loses two electrons to achieve a noble gas configuration. This would make it a Group 2 alkaline earth metal, forming a diacidic base like $M(OH)_2$.

Q: Why do we need only 0.5 moles of $H_2SO_4$ for 1 mole of $MOH$?

Sulfuric acid ($H_2SO_4$) is dibasic, meaning one molecule provides two $H^+$ ions. These two $H^+$ ions can neutralize two $OH^-$ ions. So, 1 mole of $H_2SO_4$ neutralizes 2 moles of $MOH$. Consequently, 1 mole of $MOH$ only needs 0.5 moles of $H_2SO_4$.

Q: Can we identify the exact metal from the given data?

Yes, the values $496 \text{ kJ mol}^{-1}$ and $4560 \text{ kJ mol}^{-1}$ perfectly match the first and second ionization enthalpies of Sodium (Na). However, identifying the exact metal is not strictly necessary to solve the stoichiometry part of the problem.

Question

The first and second ionisation enthalpies of a metal are $496$ and $4560	ext{ kJ mol}^{-1}$, respectively. How many moles of $	ext{HCl}$ and $	ext{H}_2	ext{SO}_4$, respectively, will be needed to react completely with $1$ mole of the metal hydroxide?

Options:

(A) $1$ and $2$

(B) $1$ and $1$

(C) $2$ and $0.5$

(D) $1$ and $0.5$

Sigma Percentile · Accepted Answer

### Decoding the Ionization Energy Jump

Imagine you are trying to pull an electron away from an atom. The energy you spend is the ionization enthalpy. In this problem, we are given a metal with a first ionization enthalpy ($IE_1$) of $496 	ext{ kJ mol}^{-1}$ and a second ionization enthalpy ($IE_2$) of $4560 	ext{ kJ mol}^{-1}$.

Look closely at these numbers. The first electron comes off relatively easily. But when you try to remove the second electron, the energy required skyrockets! The difference is massive: $\Delta IE = 4064 	ext{ kJ mol}^{-1}$.

What does this physical reality tell us? It means that after losing just one electron, the metal atom achieves a highly stable, noble gas electron configuration. Trying to break into this stable core requires an immense amount of energy. This behavior is the classic signature of **Group 1 elements**, the Alkali Metals (like Sodium or Potassium).

### The Nature of the Hydroxide

Since our mystery metal belongs to Group 1, it forms a stable $+1$ cation, which we can denote as $M^+$. When this metal forms a hydroxide, the chemical formula will simply be $MOH$.

Because it releases exactly one hydroxide ion ($OH^-$) per molecule in an aqueous solution, $MOH$ is classified as a **monoacidic base**.

### The Stoichiometry of Neutralization

Now, let's tackle the chemical reactions. We need to find out how many moles of Hydrochloric acid ($HCl$) and Sulfuric acid ($H_2SO_4$) are required to completely neutralize exactly $1$ mole of our metal hydroxide, $MOH$.

**Case 1: Reaction with $HCl$**
Hydrochloric acid is a monobasic acid; it provides one $H^+$ ion per molecule. The neutralization reaction is straightforward:

$$MOH + HCl \longrightarrow MCl + H_2O$$

From the balanced equation, it is clear that $1$ mole of $MOH$ requires exactly **$1$ mole of $HCl$** for complete neutralization.

**Case 2: Reaction with $H_2SO_4$**
Here is where you need to be careful. Sulfuric acid is a dibasic acid; it provides two $H^+$ ions per molecule. Let's write the balanced chemical equation:

$$2MOH + H_2SO_4 \longrightarrow M_2SO_4 + 2H_2O$$

This equation tells us that $2$ moles of $MOH$ are neutralized by $1$ mole of $H_2SO_4$. Therefore, by simple unitary method, $1$ mole of $MOH$ will require exactly half of that amount, which is **$0.5$ moles of $H_2SO_4$**.

### Final Answer

Bringing it all together, to neutralize $1$ mole of the metal hydroxide, we need $1$ mole of $HCl$ and $0.5$ moles of $H_2SO_4$. This perfectly matches option (d).

Animated Solution for Chemistry - Periodicity in Properties: The first and second ionisation enthalpies of a metal are $496$ and $4560 kJ mol^{- 1}$ , respectively. How many moles of $HCl$ and $H_{2} SO_{4}$ , respectively, will be needed to react completely with $1$ mole of the metal hydroxide?

Select Answer:

Visualized Solution

$Analyzing Ionisation Enthalpies$

$Identifying the Metal Group$

$Formula of the Hydroxide$

$Neutralization with HCl$

$Neutralization with H_{2} SO_{4}$

$Final Conclusion$

The Sigma Insight: Periodic Table and Periodic Properties

Decoding the Ionization Energy Jump

The Nature of the Hydroxide

The Stoichiometry of Neutralization

Final Answer

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