Final answer:
In a low-spin [Ru(NH3)5H2O]2+ complex, the Ru2+ ion has 6 d electrons, all paired in the lower energy t2g orbitals, resulting in 6 electrons in the lower state and 0 electrons in the upper state. The correct answer is (b) 6, 0.
Step-by-step explanation:
When considering the complex [Ru(NH3)5H2O]2+ in a low-spin configuration, we need to determine the number of d electrons in the lower (t2g) and upper (eg) states. Ruthenium (Ru) has an atomic number of 44 and therefore has 8 d electrons in its neutral state. As the complex is in a 2+ oxidation state, the Ru ion has 6 d electrons (8 - 2 = 6). According to crystal field theory, a low-spin complex has a larger crystal field splitting energy (Δo) than the spin-pairing energy (P), favoring the pairing of electrons in the lower energy t2g orbitals before occupying the higher energy eg orbitals.
Given that Ru2+ is in a d6 configuration and the complex is low spin, all 6 electrons will be paired in the t2g orbitals, meaning there are 6 electrons in the lower (t2g) state and 0 electrons in the upper (eg) state.