The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is ______ BM.

The element with atomic number 29 is copper (Cu), which has the electronic configuration $$[Ar]3d^{10}4s^1$$ (note the exceptional configuration due to the stability of a completely filled d-subshell).

The divalent ion $$Cu^{2+}$$ is formed by removing two electrons. The $$4s^1$$ electron is removed first, followed by one $$3d$$ electron, giving the configuration $$[Ar]3d^9$$.

With 9 electrons in the 3d subshell, there is 1 unpaired electron (since 4 orbitals are fully paired and 1 orbital has a single electron).

The spin-only magnetic moment is calculated using $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons. Substituting $$n = 1$$: $$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$$ BM.

Rounding to the nearest integer, the spin-only magnetic moment is $$\boxed{2}$$ BM.