Match List-I with List-II.
List-I (Property) | List-II (Example)
A. Diamagnetism | 1. MnO
B. Ferrima

Q: Why is $O_2$ paramagnetic when it has 16 electrons (an even number)?

While it seems intuitive that an even number of electrons would all pair up, Molecular Orbital Theory (MOT) reveals the truth. The last two electrons in $O_2$ enter two degenerate $\pi^*$ antibonding orbitals. According to Hund's Rule, they occupy these orbitals singly with parallel spins, making the molecule paramagnetic.

Q: What is the difference between Ferromagnetism and Ferrimagnetism?

In Ferromagnetism (like pure Iron), all magnetic domains align in the same direction, creating a very strong magnet. In Ferrimagnetism (like $Fe_3O_4$), domains align in opposite directions but in unequal numbers, resulting in a weaker net magnetic moment.

Q: How can I easily remember that $MnO$ is antiferromagnetic?

Think of the "Anti" in Antiferromagnetic as "Anti-Manganese". $MnO$ is the standard NCERT example where the $Mn^{2+}$ ions have spins that perfectly alternate and cancel each other out to zero.

Q: Are diamagnetic substances affected by magnets at all?

Yes, but they are weakly repelled by magnetic fields. This happens because the applied magnetic field induces a small opposing magnetic moment in the paired electrons (Lenz's Law at the atomic level).

Q: Do I need to calculate the magnetic moments for these questions?

For this specific type of question, no. These are standard, factual examples from the Solid State chapter. Memorizing the classic examples for each magnetic property is highly recommended for JEE/NEET.

Question

Match List-I with List-II.

\begin{array}{ll} 	ext{List-I (Property)} & 	ext{List-II (Example)} \ 	ext{A. Diamagnetism} & 	ext{1. MnO} \ 	ext{B. Ferrimagnetism} & 	ext{2. } 	ext{O}_2 \ 	ext{C. Paramagnetism} & 	ext{3. NaCl} \ 	ext{D. Antiferromagnetism} & 	ext{4. } 	ext{Fe}_3	ext{O}_4 \end{array}

Choose the most appropriate answer from the options given below.

Options:

(A) A $ightarrow$ 2, B $ightarrow$ 1, C $ightarrow$ 3, D $ightarrow$ 4

(B) A $ightarrow$ 1, B $ightarrow$ 3, C $ightarrow$ 4, D $ightarrow$ 2

(C) A $ightarrow$ 3, B $ightarrow$ 4, C $ightarrow$ 2, D $ightarrow$ 1

(D) A $ightarrow$ 4, B $ightarrow$ 2, C $ightarrow$ 1, D $ightarrow$ 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to a classic challenge from the Solid State chapter! We are presented with a matching problem that tests our memory and understanding of the magnetic properties of solids.

On one side, we have four distinct magnetic behaviors: Diamagnetism, Ferrimagnetism, Paramagnetism, and Antiferromagnetism. On the other side, we have four chemical compounds: $MnO$, $O_2$, $NaCl$, and $Fe_3O_4$.

Our goal is to logically pair each property with its textbook example. Let's break them down one by one.

### Diamagnetism: The Power of Pairs

Let's start with **Diamagnetism**. A substance is diamagnetic when all of its electrons are perfectly paired up. Because there are no unpaired electrons, these substances are weakly repelled by external magnetic fields.

Looking at our list, **Sodium Chloride** ($NaCl$) is the perfect fit. In $NaCl$, the sodium ion ($Na^+$) has lost its valence electron to achieve a stable neon core, and the chloride ion ($Cl^-$) has gained an electron to achieve an argon core.

Both ions have completely filled, paired electron shells. Therefore, $NaCl$ is diamagnetic, and we can confidently match **A with 3**.

### Paramagnetism: The Unpaired Attraction

Next, we move to **Paramagnetism**. Paramagnetic substances contain one or more unpaired electrons. These unpaired spins create tiny magnetic dipoles that align with an external magnetic field, causing a weak attraction.

If you recall the Molecular Orbital Theory (MOT) from Chemical Bonding, the oxygen molecule ($O_2$) is a famous exception to the Lewis dot structure. Despite having an even number of total electrons, $O_2$ has two unpaired electrons residing in its degenerate $\pi^*$ antibonding orbitals.

This makes $O_2$ paramagnetic. Thus, we match **C with 2**.

### Ferrimagnetism: The Unequal Alignment

Now, let's tackle **Ferrimagnetism**. In ferrimagnetic substances, the magnetic domains (or moments) are aligned in both parallel and anti-parallel directions, but in **unequal numbers**.

Because the opposing moments do not perfectly cancel each other out, the substance retains a net magnetic moment, though it is weaker than true ferromagnetism.

The most iconic textbook example of a ferrimagnetic substance is magnetite, or iron(II,III) oxide ($Fe_3O_4$). Therefore, we match **B with 4**.

### Antiferromagnetism: The Perfect Cancellation

Finally, we have **Antiferromagnetism**. In these materials, the magnetic moments are aligned in a perfectly alternating parallel and anti-parallel pattern.

Because there are an **equal number** of moments pointing in opposite directions, they completely cancel each other out, resulting in a net magnetic moment of exactly zero.

Manganese(II) oxide ($MnO$) is the classic example of this behavior. The spins of the $Mn^{2+}$ ions perfectly oppose one another in the crystal lattice. So, we match **D with 1**.

### Final Calculation

Let's compile our logical deductions into a final sequence:
- **A** (Diamagnetism) $\rightarrow$ **3** ($NaCl$)
- **B** (Ferrimagnetism) $\rightarrow$ **4** ($Fe_3O_4$)
- **C** (Paramagnetism) $\rightarrow$ **2** ($O_2$)
- **D** (Antiferromagnetism) $\rightarrow$ **1** ($MnO$)

Our final matching sequence is **A-3, B-4, C-2, D-1**. Scanning through the given options, we find that this exact combination corresponds to **Option (c)**.

Mastering these standard examples is a surefire way to secure quick marks in your exams!

Select Answer:

Visualized Solution

Magnetic Properties

Diamagnetism: $N a C l$

Paramagnetism: $O_{2}$

Ferrimagnetism: $F e_{3} O_{4}$

Antiferromagnetism: $M n O$

Final Matching Sequence

The Sigma Insight: Solid State

Analyzing the Setup

Diamagnetism

Paramagnetism

Ferrimagnetism

Antiferromagnetism

Final Calculation

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