Animated Solution for Chemistry - d and f-Block Elements: Which of the following arrangements does not represent the correct order of the property stated against it?

Imagine you are a detective, and your suspects are four different chemical trends. Your job is to find the one that is lying—the one that does not represent the correct order. Let's start our investigation with the first suspect: paramagnetic behaviour.

The Role of Unpaired Electrons

Paramagnetism is like a magnetic personality; it depends entirely on the number of unpaired electrons ($n$) an ion possesses. The more unpaired electrons, the stronger the magnetic pull. To figure out if the given order is correct, we need to calculate $n$ for each ion in the lineup: $V^{2+}$, $C{r}^{2+}$, $M{n}^{2+}$, and $F{e}^{2+}$.

Let's break them down one by one:

1. Vanadium ($V^{2+}$): Vanadium has an atomic number of 23. When it loses two electrons to become $V^{2+}$, its electronic configuration becomes $[Ar]3d^3$. This gives it exactly 3 unpaired electrons ($n=3$).

2. Chromium ($C{r}^{2+}$): Chromium's atomic number is 24. As $C{r}^{2+}$, its configuration is $[Ar]3d^4$. That means it has 4 unpaired electrons ($n=4$).

3. Manganese ($M{n}^{2+}$): Manganese sits at atomic number 25. The $M{n}^{2+}$ ion has a configuration of $[Ar]3d^5$. It has a full house of 5 unpaired electrons ($n=5$), which is the maximum possible for the 3d subshell!

4. Iron ($F{e}^{2+}$): Here is where the plot twists. Iron has an atomic number of 26. Its $F{e}^{2+}$ ion has a configuration of $[Ar]3d^6$. Because there are only 5 orbitals, the 6th electron must pair up with one of the existing electrons. This pairing leaves only 4 unpaired electrons ($n=4$).

Now, let's look at the order given in option (a): $V^{2+}<C{r}^{2+}<M{n}^{2+}<F{e}^{2+}$.

Based on our calculations, the actual order of unpaired electrons is $3<4<5>4$. Iron ($F{e}^{2+}$) has fewer unpaired electrons than Manganese ($M{n}^{2+}$), so it should be less paramagnetic, not more! Therefore, option (a) is the incorrect arrangement and is our culprit.

Size, Stability, and Oxidation States

Just to be thorough detectives, let's quickly clear the other suspects:

- Ionic Size: As we move from left to right across a period, the effective nuclear charge ($Z_{eff}$) increases, pulling the electron cloud closer and decreasing the ionic size. Thus, $N{i}^{2+}<C{o}^{2+}<F{e}^{2+}<M{n}^{2+}$ is perfectly correct.

- Stability in Aqueous Solution: $S{c}^{3+}$ is incredibly stable because it achieves a noble gas configuration ($[Ar]$). On the other end, $C{o}^{3+}$ is a strong oxidizing agent and is quite unstable in water. The order $C{o}^{3+}<F{e}^{3+}<C{r}^{3+}<S{c}^{3+}$ holds true.

- Number of Oxidation States: From group 3 to group 7, the number of available electrons for bonding increases. Scandium only shows +3, while Manganese can show a wide range from +2 all the way to +7. The order $Sc<Ti<Cr<Mn$ is absolutely correct.

And there you have it! By understanding the fundamental principles of electronic configuration and periodic trends, we successfully identified the imposter.