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.

Given $\mu = 2.84$ B.M., we can calculate $n$: $\sqrt{n(n+2)} \approx 2.84$ $n(n+2) \approx 8$ For $n = 2$, $\sqrt{2(4)} = \sqrt{8} \approx 2.83$ B.M. Thus, the ion must have 2 unpaired electrons.

Now, let's analyze the electronic configurations of the given ions:

1. $Ni^{2+}$ (Z=28): Configuration is $[Ar] 3d^8$. The d-orbitals have the arrangement $\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow$. This gives 2 unpaired electrons. (Calculated $\mu \approx 2.84$ B.M.)
2. $Ti^{3+}$ (Z=22): Configuration is $[Ar] 3d^1$. This gives 1 unpaired electron. (Calculated $\mu = 1.73$ B.M.)
3. $Cr^{2+}$ (Z=24): Configuration is $[Ar] 3d^4$. This gives 4 unpaired electrons. (Calculated $\mu = 4.90$ B.M.)
4. $Co^{2+}$ (Z=27): Configuration is $[Ar] 3d^7$. This gives 3 unpaired electrons. (Calculated $\mu = 3.87$ B.M.)

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