The spin-only magnetic moment value of B₂ + species is _____________ × 10⁻² BM. (Nearest integer) [Given : √3 = 1.73]

Solution

Chemistry ma’am · 4 steps

Fill the 9 electrons of B2+ into MOs, remembering the π orbitals come before σ2p for boron.

1. Count electrons. Each boron atom has 5 electrons, so B₂ has 10. Removing one electron for the positive charge leaves 9 in B₂⁺.
2. Use the right order. For B₂, C₂ and N₂, s-p mixing pushes σ 2p_z above the two π 2p orbitals. So after σ 2s and σ^* 2s, the next electrons go into π 2pₓ and π 2p_y.

   B₂⁺: σ 1s² σ^* 1s² σ 2s² σ^* 2s² π 2pₓ¹ π 2p_y⁰

3. Unpaired electrons. Only one electron is left after the σ^* 2s level, and it sits alone in a π orbital. So n = 1.
4. So. μ = √n(n+2) = √(1 × 3) = √3 = 1.73 BM = 173 × 10⁻² BM, so the answer is 173.

The common slip: 283. Counting the 2 unpaired electrons of neutral B₂ instead of 1 gives √8 = 2.83 BM, that is 283, not 173.

The idea to keep. Up to N₂, the π 2p orbitals fill before σ 2p_z. Get the order right, count the lone electrons, then use μ = √n(n+2) BM.

Answer from the NTA final answer key. Solution by PrepWiser.