The calculated spin-only magnetic moment of Ti²⁺,3d², is:

We need the spin‑only magnetic moment (μ) for the ion Ti²⁺ which has a 3d² configuration.

1. Determine the number of unpaired electrons (n).
• A neutral Ti atom has the electron configuration [Ar] 3d² 4s².
• Removing two electrons to form Ti²⁺ takes away the two 4s electrons, leaving only the two 3d electrons.
• In a free ion (or in a weak crystal field) the two 3d electrons occupy separate orbitals, so both are unpaired.
• Therefore, n = 2.

2. Use the spin‑only formula.
The spin‑only magnetic moment is given by

$\mu \; =\; sqrt(n · (n + 2))$

where μ is expressed in Bohr magnetons (BM). This formula comes from μ = √[S(S+1)] · 2 · √3 · μ_B, which simplifies to the expression above when expressed in BM.

3. Insert n = 2 into the formula.
$\mu \; =\; sqrt(2 · (2 + 2))\; =\; sqrt(2 · 4)\; =\; sqrt(8)$

4. Evaluate the square root.
sqrt(8) ≈ 2.8284 BM.

5. Round to the usual reporting precision.
Rounding to two decimal places gives μ ≈ 2.84 BM.

Thus, the calculated spin‑only magnetic moment of Ti²⁺ (3d²) is 2.84 BM, which matches the provided correct option.