The 'spin-only' magnetic moment ($\mu$) is calculated using the formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons.

1. Analyze the given value: A magnetic moment of 2.84 B.M. corresponds to $\sqrt{8}$. Solving $\sqrt{n(n+2)} \approx 2.84$ gives $n=2$. Thus, we are looking for the ion with 2 unpaired electrons.

2. Check Electronic Configurations:

• $Ni^{2+}$ (Z=28): Configuration $[Ar] 3d^8$. In $d$-orbitals (5 orbitals), 8 electrons are arranged as 3 pairs and 2 singles. $n = 2$.
• $Ti^{3+}$ (Z=22): Configuration $[Ar] 3d^1$. $n = 1$ ($\mu \approx 1.73$ B.M.).
• $Cr^{3+}$ (Z=24): Configuration $[Ar] 3d^3$. $n = 3$ ($\mu \approx 3.87$ B.M.).
• $Co^{2+}$ (Z=27): Configuration $[Ar] 3d^7$. In $d$-orbitals, 7 electrons are arranged as 2 pairs and 3 singles. $n = 3$ ($\mu \approx 3.87$ B.M.).

Therefore, $Ni^{2+}$ is the correct answer .