For the compound BeN:

Question

a. Construct a fully labeled MO diagram.

b. Determine the bond order.

c. Is the compound paramagnetic or diamagnetic?

Answer 1

Short answers: bond order 1.5, and BeN is paramagnetic (one unpaired electron). Here is the reasoning.

Step 1 — count valence electrons

Be is [He]2s² → 2 valence electrons. N is [He]2s²2p³ → 5 valence electrons. Total = 7. Core 1s electrons are not placed in the diagram.

Step 2 — pick the right orbital ordering (this is the whole problem)

Second-row diatomics come in two orderings. For Li₂ through N₂, s–p mixing pushes σ_2p above the π_2p pair. For O₂, F₂, Ne₂ the mixing is weak and σ_2p drops below π_2p. Be and N both sit in the first group, so BeN uses the mixed ordering:

σ_2s < σ*_2s < π_2p (two degenerate) < σ_2p < π*_2p (two degenerate) < σ*_2p

Getting this backwards is the single most common error on these problems — it changes which orbital the last electrons enter and therefore changes your answer to part (c).

Step 3 — fill, bottom up (a)

σ*_2p  ——  (empty)
π*_2p  —— ——  (empty)
σ_2p  ——  (empty)
π_2p  ⇅  ↑  ← 3 electrons, Hund's rule leaves one unpaired
σ*_2s  ⇅
σ_2s  ⇅

Running total: 2 + 2 + 3 = 7. ✓

One labelling note for a "fully labeled" diagram: because Be and N differ in electronegativity, the atomic 2s and 2p levels on the nitrogen side sit lower than Be's. Draw N's atomic orbitals below Be's on the flanks. The consequence is that the bonding MOs have more nitrogen character and the antibonding MOs more beryllium character — the electron density is polarised toward N.

Step 4 — bond order (b)

Bonding electrons: 2 (σ_2s) + 3 (π_2p) = 5. Antibonding: 2 (σ*_2s).

BO = (5 − 2) / 2 = 1.5

A fractional bond order is perfectly legitimate — it is one of the things MO theory gives you that Lewis structures cannot.

Step 5 — magnetism (c)

The π_2p level holds three electrons across two degenerate orbitals. By Hund's rule they occupy both orbitals before pairing, so one electron is left unpaired → paramagnetic. Part (c) is never a separate question; it is read directly off the diagram you drew in (a). Any odd total electron count guarantees at least one unpaired electron, so 7 valence electrons tells you the answer before you fill anything.

Worth knowing: BeN is a textbook exercise rather than a stable species you will meet in the lab. That does not affect the method — MO theory handles heteronuclear diatomics exactly this way, and the same procedure works on CN, NO, and CO.