Oxygen Atom and Molecule Properties Analysis

Let's analyze each statement regarding oxygen properties to determine which ones are correct.

Statement A: The ground state for the O atom is \(^3P_1\).

• The electronic configuration of an oxygen atom (O) is \(1s^22s^22p^4\).
• The ground state term symbol is determined by the electrons in the outermost subshell, which is \(2p^4\).
• For a \(p^4\) configuration, the total orbital angular momentum (L) is calculated from the individual \(m_l\) values. The possible arrangements giving the maximum M_L (and thus L) is when the two paired electrons are in one orbital (say \(m_l = +1\)) and the two unpaired electrons are in the other two (\(m_l = 0, -1\)). Sum of \(m_l\) = (+1) + (+1) + (0) + (-1) = +1. So, the maximum M_L is 1, meaning L=1. This corresponds to a P state.
• The total spin angular momentum (S) is the sum of individual spins. With two unpaired electrons, \(S = +1/2 + +1/2 = 1\). So, the spin multiplicity is \(2S+1 = 2(1)+1 = 3\). This corresponds to a Triplet state.
• The possible values for the total angular momentum (J) for a \(^3P\) term (L=1, S=1) are \(L+S, L+S-1, \dots, |L-S|\). So, \(J\) can be \(1+1=2\), \(1+1-1=1\), \(|1-1|=0\). The possible states are \(^3P_2\), \(^3P_1\), and \(^3P_0\).
• For subshells that are more than half-filled (like \(p^4\)), the lowest energy state (ground state) corresponds to the smallest J value.
• The smallest J value among 2, 1, 0 is 0. Therefore, the ground state term symbol for the oxygen atom is \(^3P_0\).
• Statement A says the ground state is \(^3P_1\), which is incorrect.

Statement B: Both the atom and the diatomic molecule are paramagnetic with two unpaired electrons.

• For the O atom, as discussed above (2p⁴ configuration: ↑↓ ↑ ↑), there are two unpaired electrons. Atoms with unpaired electrons are paramagnetic. So, the O atom is paramagnetic with two unpaired electrons. This part is correct.
• For the diatomic molecule \(O_2\), we use the Molecular Orbital (MO) theory. The valence electrons are 12 (6 from each O). The MO configuration is: \(\sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}\)
• The highest occupied molecular orbitals (HOMOs) are the degenerate \(\pi^*\) antibonding orbitals. According to Hund's rule, the two electrons in the \(\pi^*\) orbitals occupy them individually with parallel spins (\(\uparrow \uparrow\)).
• Thus, the \(O_2\) molecule has two unpaired electrons in its ground state. Molecules with unpaired electrons are paramagnetic. So, the \(O_2\) molecule is paramagnetic with two unpaired electrons. This part is also correct.
• Since both the atom and the diatomic molecule are paramagnetic with two unpaired electrons, statement B is correct.

Statement C: The most readily accessible singlet excited state for dioxygen has an empty \(\pi^*\) orbital.

• The ground state of \(O_2\) is a triplet state (\(^3\Sigma_g^-\)) with two unpaired electrons in the \(\pi^*\) orbitals (\(\pi_{2p_x}^{*1} \pi_{2p_y}^{*1}\)).
• A singlet state has a total spin of S=0 (spin multiplicity 2S+1=1). To achieve a singlet state from the \(\pi^*\) configuration, the two electrons must be paired with opposite spins (e.g., \(\uparrow\downarrow\)).
• The most readily accessible excited singlet states arise from promoting one of the unpaired electrons to pair with the other in the same \(\pi^*\) orbital. This results in configurations like \(\pi_{2p_x}^{*2} \pi_{2p_y}^{*0}\) or \(\pi_{2p_x}^{*0} \pi_{2p_y}^{*2}\).
• These configurations correspond to the \(^1\Delta_g\) state, which is the lowest energy singlet excited state of \(O_2\). In this state, one of the \(\pi^*\) orbitals contains two paired electrons, while the other \(\pi^*\) orbital is indeed empty.
• Therefore, statement C is correct.

Based on the analysis:

• Statement A is incorrect.
• Statement B is correct.
• Statement C is correct.

The correct statements are B and C.

The final answer is the option that includes only B and C.