1. We check each statement one by one.

2. Option A: $[CoCl(NH_3)_5]^{2+}$ is violet as it absorbs yellow light.

• The observed color of a complex is complementary to the color absorbed.
• Yellow light and violet color are complementary.
• Hence, if the complex absorbs yellow light, it can appear violet.

So, Option A is correct.

3. Option B: the gemstone, ruby, has $Cr^{3+}$ ions occupying the octahedral sites of beryl.

• Ruby is not beryl.
• Ruby is a variety of corundum, which is $Al_2O_3$.
• Its red color is due to trace amounts of $Cr^{3+}$ replacing $Al^{3+}$ in the crystal lattice.
• Beryl is a different mineral: $Be_3Al_2Si_6O_{18}$.

Therefore this statement is incorrect.

4. Option C: the spin-only magnetic moment of $[Ni(NH_3)_4(H_2O)_2]^{2+}$ is $2.83\,\text{BM}$.

• For $Ni^{2+}$, electronic configuration is $3d^8$.
• In an octahedral complex with $NH_3$ and $H_2O$, the configuration remains with 2 unpaired electrons.
• Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ where $n$ = number of unpaired electrons.
• Here $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83\,\text{BM}$

So, Option C is correct.

5. Option D: the spin-only magnetic moments of $[Fe(H_2O)_6]^{2+}$ and $[Cr(H_2O)_6]^{2+}$ are nearly similar.

Let us calculate both.

For $[Fe(H_2O)_6]^{2+}$

• $Fe^{2+}$ is $3d^6$.
• $H_2O$ is a weak-field ligand, so this is high spin octahedral.
• Number of unpaired electrons $n=4$.
• Magnetic moment: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\,\text{BM}$

For $[Cr(H_2O)_6]^{2+}$

• $Cr^{2+}$ is $3d^4$.
• With weak-field ligand $H_2O$, this is high spin octahedral.
• Number of unpaired electrons $n=4$.
• Magnetic moment: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\,\text{BM}$

Thus, the two magnetic moments are indeed nearly the same.

So, Option D is correct.

6. Hence, the only incorrect statement is:

$\boxed{\text{B}}$