Animated Solution for Chemistry - Coordination Compounds: The incorrect statement is

This question is a beautiful blend of solid-state chemistry and coordination chemistry. It challenges us to not only understand the abstract theories of d-orbitals but also to connect them to the tangible, colorful world of gemstones and chemical solutions. Let's embark on a journey to dissect each statement and uncover the truth.

Ruby vs. Emerald

Let's start with the first statement, which takes us into the fascinating realm of mineralogy. The statement claims that ruby has $C{r}^{3+}$ ions occupying the octahedral sites of beryl.

To verify this, we need to know what ruby and beryl actually are. Ruby is a precious gemstone that is fundamentally a variety of the mineral corundum. Corundum is crystalline aluminum oxide, with the chemical formula $A{l}_2{O}_3$. In its pure form, corundum is completely colorless. However, nature is an artist. When a small fraction of the $A{l}^{3+}$ ions in the corundum lattice are replaced by chromium ions ($C{r}^{3+}$), the crystal absorbs certain wavelengths of light and transmits a deep, mesmerizing red.

On the other hand, beryl is a completely different beast. It is a beryllium aluminum cyclosilicate with the formula $B{e}_{3}A{l}_{2}S{i}_6{O}_{18}$. If $C{r}^{3+}$ ions happen to occupy the octahedral sites in a beryl lattice instead of corundum, the resulting gemstone is not red, but a brilliant green. We call this green gemstone emerald!

Therefore, the statement that ruby is formed in a beryl lattice is fundamentally flawed. Ruby belongs to the corundum family. This makes statement (a) the incorrect one.

Moving on to statement (b), we encounter the complex $[CoCl(N{H}_3{)}_5{]}^{2+}$. The statement says it is violet because it absorbs yellow light.

This is a classic application of the complementary color theory. When white light passes through a solution of a transition metal complex, the complex absorbs specific wavelengths of light to excite electrons from lower energy d-orbitals to higher energy d-orbitals (a process known as a d-d transition). The color that our eyes perceive is the light that is not absorbed—the transmitted light.

According to the color wheel, the complementary color of yellow is violet. If a complex absorbs photons in the yellow region of the visible spectrum, the remaining light that reaches our eyes will appear violet. This statement is perfectly accurate.

Crystal Field Theory

Statements (c) and (d) dive deep into Crystal Field Theory (CFT) and magnetic moments.

Statement (c) compares the magnetic moments of $[Fe(H_{2}O{)}_6{]}^{2+}$ and $[Cr(H_{2}O{)}_6{]}^{2+}$. Both complexes feature water ($H_{2}O$) as the ligand. Water is generally considered a weak field ligand, meaning it produces a relatively small crystal field splitting energy (${\Delta }_o$). Because the splitting is small, it is energetically more favorable for electrons to occupy the higher energy $e_g$ orbitals rather than pairing up in the lower $t_{2g}$ orbitals. This results in high-spin complexes.

Let's look at the electron configurations: - $F{e}^{2+}$ has a $3d^6$ configuration. In a high-spin octahedral field, the electrons arrange as $t_{2g}^4{e}_g^2$. This leaves us with $n=4$ unpaired electrons. - $C{r}^{2+}$ has a $3d^4$ configuration. In a high-spin octahedral field, the electrons arrange as $t_{2g}^3{e}_g^1$. This also leaves us with $n=4$ unpaired electrons.

The spin-only magnetic moment is calculated using the formula $\mu =\sqrt{n(n+2)}$ Bohr Magnetons (BM). Since both ions have 4 unpaired electrons, their magnetic moments will be identical: $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90$ BM. Thus, statement (c) is correct.

Finally, statement (d) evaluates the magnetic moment of $[Ni(N{H}_3{)}_4(H_{2}O{)}_2{]}^{2+}$. Here, the central metal is $N{i}^{2+}$, which has a $3d^8$ configuration.

There is a beautiful simplicity to $d^8$ configurations in an octahedral field. Whether the ligands are strong or weak, the first 6 electrons will completely fill the lower $t_{2g}$ level. The remaining 2 electrons must go into the higher $e_g$ level. According to Hund's rule, they will occupy separate orbitals and remain unpaired. Therefore, an octahedral $N{i}^{2+}$ complex will always have exactly 2 unpaired electrons ($n=2$).

Plugging this into our formula: $\mu =\sqrt{2(2+2)}=\sqrt{8}\approx 2.83$ BM. Statement (d) is absolutely correct.

After a thorough chemical investigation, we have confirmed that the physics of color and magnetism hold true for statements (b), (c), and (d). The only imposter is statement (a), which confuses the mineralogical origins of ruby and emerald. Thus, (a) is our final answer.