Question bank — Ligand Field Theory (LFT) and MO description (overview)

False — they give the same measurable ${\Delta }_o$; LFT just explains it as an MO gap while CFT calls it electrostatic repulsion. Same number, different story.

True — the $t_{2g}$ lobes point between the ligands, so their σ-overlap integral is zero by symmetry; with no σ-partner they sit unchanged at pure metal energy.

False — that is only the σ-only special case. Once the ligand has π-orbitals of $t_{2g}$ symmetry, the $t_{2g}$ set bonds (or antibonds) and shifts up or down. (See Back-bonding and π-Acceptor Ligands.)

False — $e_g^{\ast }$ is antibonding and mostly metal; the bonding $e_g$ (unstarred, sitting far below, as defined above) is the mostly-ligand one. The starred set is where the metal $d$-electrons live.

False — $e_g^{\ast }$ is only weakly antibonding (mostly metal). Populating it slightly lengthens and weakens the M–L bonds (as in high-spin cases) but the complex stays intact.

False — this is the CFT point-charge intuition. Splitting tracks orbital overlap and π-effects, not raw charge, which is why neutral CO beats ${\text{F}}^{-}$. (See Spectrochemical Series.)

True — filled ligand π orbitals push up the metal $t_{2g}$, which shrinks ${\Delta }_o$; a smaller gap is the definition of a weaker field.

True — back-bonding donates metal $t_{2g}$ electrons into empty ligand π*; a $d^0$ metal has no $t_{2g}$ electrons to give, so back-bonding is negligible.

False — it is a covalency signature: the $d$-electron cloud spreads onto the ligands, so electron–electron repulsion drops. Pure point charges could never cause cloud-expansion.

Signs reversed. Since $e_g^{\ast }$ sits above $t_{2g}$, the correct order is ${\Delta }_o=E(e_g^{\ast })-E(t_{2g})>0$. Writing it backwards gives a negative gap, which is meaningless.

They point between the axes, not at the ligands. The set that points at the ligands is $e_g$ ($d_{z^2},d_{x^2-y^2}$), which is why those become antibonding.

Wrong symmetry set. The 6 σ-LGOs (ligand group orbitals, defined above) span $a_{1g}+e_g+t_{1u}$ — there is no $t_{2g}$ σ-LGO, which is exactly why metal $t_{2g}$ finds no σ-partner.

Incomplete — σ-donation alone only raises $e_g^{\ast }$. CO's real power is π-acceptance: back-bonding lowers $t_{2g}$, so the gap widens from both sides at once.

That describes a π-donor. A π-acceptor uses empty π* orbitals to accept electrons from the metal, which lowers $t_{2g}$.

The negativity is a red herring; ${\text{F}}^{-}$ raises $t_{2g}$ because it is a π-donor (filled lone pairs of $t_{2g}$ symmetry), not because of its charge.

The d–d transition is between $t_{2g}$ and $e_g^{\ast }$ — the frontier $d$-based sets — with energy exactly ${\Delta }_o=h\nu$. The deep bonding $e_g$ is not involved.

If symmetries differ, the net overlap integral is zero (positive and negative overlap regions cancel), so no bonding/antibonding interaction can occur. Symmetry is the matchmaker.

Better overlap and closer energy match push the antibonding $e_g^{\ast }$ higher; since ${\Delta }_o=E(e_g^{\ast })-E(t_{2g})$, raising $e_g^{\ast }$ directly widens the gap.

Among simple σ/π-donors, more polarizable donors happen to correlate loosely with charge, so CFT limps along. But neutral π-acceptors like CO sit at the top, which a charge-only model cannot place — only the MO/π picture can.

$e_g^{\ast }$ is antibonding along the M–L axes, so its electrons partially cancel the bonding electrons' glue, reducing net bond order and stretching the bond.

${\text{NH}}_3$ is a σ-only donor keeping a big clean ${\Delta }_o>P$ (pairing energy, defined above), forcing electrons to pair ($t_{2g}^6$). ${\text{F}}^{-}$ is a π-donor that raises $t_{2g}$, shrinking ${\Delta }_o<P$, so electrons spread into $e_g^{\ast }$.

The absorbed photon energy equals the gap: ${\Delta }_o=h\nu =hc/\lambda$. A bigger ${\Delta }_o$ demands a higher-frequency, shorter-$\lambda$ photon.

MO theory lets metal and ligand orbitals mix into real covalent bonds, which is the only framework that explains π-bonding, back-bonding, the true spectrochemical order, and the nephelauxetic effect.

The gap still exists (it's a property of the orbitals, not the electrons), but with no $d$-electrons there is no d–d transition, so such ions are typically colourless.

Essentially no — with an empty $t_{2g}$ there are no metal electrons to donate into ligand π*, so the $t_{2g}$-lowering back-bond effect vanishes.

Its σ-donation raises $e_g^{\ast }$ (grows gap) but its π-donation raises $t_{2g}$ (shrinks gap); the π-donor effect usually wins for halides, netting a small ${\Delta }_o$ — hence their low position in the series.

It collapses to zero: $e_g^{\ast }$ falls back onto the non-bonding $t_{2g}$ energy, all five $d$-orbitals become degenerate again, exactly the free-ion limit.

None — symmetry mismatch forces zero overlap, so the ligand behaves as effectively σ-only regardless of whether it carries π orbitals.

No — ${\Delta }_o$ is fixed by the ligand field. Spin state only decides whether the five electrons pair up ($t_{2g}^5$, low-spin) or spread out ($t_{2g}^3{e}_g^{\ast 2}$, high-spin), depending on whether ${\Delta }_o$ beats the pairing energy $P$.

Not a hard cap, but practically CO/CN${}^{-}$ sit at the series' top because both effects add: raising $e_g^{\ast }$ and lowering $t_{2g}$ simultaneously pushes ${\Delta }_o$ to its largest common values.

It undergoes a Jahn–Teller distortion — the octahedron stretches (or squashes) along one axis to split the degenerate $e_g^{\ast }$ level, lowering the energy of the occupied orbital. Any orbitally degenerate ground state is unstable and distorts.

No — by lowering the symmetry from perfect $O_h$, it splits both the $t_{2g}$ and $e_g^{\ast }$ sets into sub-levels, so a single clean ${\Delta }_o$ no longer fully describes the spectrum; the absorption band typically broadens or splits (a fingerprint of the distortion).