Animated Solution for Chemistry - d and f-Block Elements: The set having ions which are coloured and paramagnetic both is

Color and Magnetism

When we dive into the fascinating world of d-block elements, two properties immediately stand out: their vibrant colors and their intriguing magnetic behaviors. But what is the hidden link between a brilliant blue copper solution and its ability to be weakly attracted to a magnet? The answer lies in a single, powerful concept: unpaired electrons.

For a transition metal ion to exhibit both color and paramagnetism, it must have at least one unpaired electron in its outermost orbitals (usually the d-orbitals). These lonely electrons act like tiny bar magnets, giving rise to paramagnetism. Simultaneously, they can absorb specific wavelengths of visible light to jump to higher energy levels within the d-subshell (a process known as a d-d transition). The light that isn't absorbed is transmitted or reflected, painting the compound with beautiful complementary colors.

Let's break down the first set of ions to see if they fit our criteria.

1. Copper(II) Ion (${\text{Cu}}^{2+}$) Copper is a classic exception to the Aufbau principle. A neutral copper atom (${}_{29}\text{Cu}$) has the configuration $[\text{Ar}]3d^{10}4s^1$. When it loses two electrons to form ${\text{Cu}}^{2+}$, it loses the $4s$ electron first, followed by one from the $3d$ subshell. This leaves us with $[\text{Ar}]3d^9$. If you visualize the five d-orbitals, four are completely filled with pairs, but one orbital holds a single, unpaired electron ($n=1$). Thus, ${\text{Cu}}^{2+}$ is both paramagnetic and colored (typically blue).

2. Chromium(III) Ion (${\text{Cr}}^{3+}$) Chromium is another famous exception. Neutral chromium (${}_{24}\text{Cr}$) is $[\text{Ar}]3d^{5}4s^1$. To form the ${\text{Cr}}^{3+}$ ion, we strip away three electrons: the single $4s$ electron and two from the $3d$ subshell. The resulting configuration is $[\text{Ar}]3d^3$. Here, we have three unpaired electrons ($n=3$) sitting in three separate d-orbitals. This guarantees strong paramagnetism and a distinct color (often green or violet depending on the ligands).

3. Scandium(I) Ion (${\text{Sc}}^{+}$) Scandium (${}_{21}\text{Sc}$) normally has the configuration $[\text{Ar}]3d^{1}4s^2$. The question specifically asks about the ${\text{Sc}}^{+}$ ion. To form a +1 ion, we remove just one electron from the outermost shell, which is the $4s$ orbital. This leaves the configuration as $[\text{Ar}]3d^{1}4s^1$. Notice that we now have one unpaired electron in the $3d$ orbital and another unpaired electron in the $4s$ orbital. With two unpaired electrons ($n=2$), ${\text{Sc}}^{+}$ is definitely paramagnetic and colored.

Since all three ions in option (a)—${\text{Cu}}^{2+}$, ${\text{Cr}}^{3+}$, and ${\text{Sc}}^{+}$—possess unpaired electrons, they all satisfy the condition of being both colored and paramagnetic.

If we quickly glance at the other options, we find ions like ${\text{Zn}}^{2+}$ ($3d^{10}$), ${\text{Ti}}^{4+}$ ($3d^0$), and ${\text{Mn}}^{7+}$ ($3d^0$). These ions have either completely full or completely empty d-subshells. With zero unpaired electrons ($n=0$), they are diamagnetic (repelled by magnetic fields) and colorless. This confirms that our initial analysis was spot on!