The spin only magnetic moment of the complex present in Fehling's reagent is _____ B.M. (Round off your answer to the nearest integer)

Fehling’s reagent contains the copper(II) bis(tartrate) complex, $$[\text{Cu}(\text{C}_4\text{H}_4\text{O}_6)_2]^{2-}$$, where copper is in the +2 oxidation state.

Copper has atomic number 29 and its electronic configuration is $$[\text{Ar}]\,3d^{10}\,4s^1$$. Removing two electrons to form Cu$$^{2+}$$ gives $$[\text{Ar}]\,3d^9$$.

In the $$3d^9$$ configuration, five d-orbitals accommodate nine electrons, so four orbitals are fully occupied (8 electrons) and one orbital contains a single unpaired electron. Therefore, the number of unpaired electrons is $$n = 1$$.

The spin-only magnetic moment is given by $$\mu = \sqrt{n(n+2)} \text{ B.M.}$$. Substituting $$n = 1$$ yields $$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.732 \text{ B.M.}$$.

Rounding to the nearest integer gives the magnetic moment as $$\boxed{2}$$ B.M.