501. A target ligand \(L\) forms a stable complex \(\mathrm{ML}\), but a second ligand \(X\) is added and binds the same metal even more strongly. What may happen to the concentration of \(\mathrm{ML}\)?
ⓐ. It must remain unchanged because the intrinsic formula of \(\mathrm{ML}\) has not changed
ⓑ. It must increase because every added ligand favours every metal complex equally
ⓒ. It may fall because \(X\) competes for the metal
ⓓ. It becomes independent of all equilibrium constants
Correct Answer: It may fall because \(X\) competes for the metal
Explanation: Several ligands can compete for the same metal ion in solution. A strongly binding competitor reduces the amount of free metal available to form the original complex. Metal may also be transferred from \(\mathrm{ML}\) to an \(X\)-containing species if the competing equilibrium is favourable. The concentration of a complex therefore depends on the complete set of equilibria present. An isolated formation constant does not by itself determine speciation in a multiligand solution.
502. A small, highly charged hard metal ion is mixed separately with the following ligands under comparable conditions. Which ligand is most likely to gain stability from both hard-hard compatibility and chelation?
ⓐ. a polydentate hard oxygen donor forming five-membered rings
ⓑ. a monodentate hard oxygen-donor ligand
ⓒ. a polydentate soft sulphur-donor ligand forming large chelate rings
ⓓ. a monodentate soft sulphur-donor ligand
Correct Answer: a polydentate hard oxygen donor forming five-membered rings
Explanation: A small, highly charged metal ion is generally classified as a hard acid. Oxygen donor atoms are hard bases and therefore provide favourable hard-hard compatibility. A polydentate ligand can attach through several oxygen donors and form chelate rings. Five-membered rings commonly have favourable geometry and limited strain. The proposed ligand combines several stabilising factors rather than relying on only one trend.
503. Study the classification of the following metal carbonyls.
| Row | Carbonyl | Proposed classification |
|---|
| P | \(\mathrm{Ni(CO)_4}\) | Mononuclear |
| Q | \(\mathrm{Cr(CO)_6}\) | Mononuclear |
| R | \(\mathrm{Fe_2(CO)_9}\) | Polynuclear |
| S | \(\mathrm{Co_2(CO)_8}\) | Mononuclear |
The inconsistent row is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: Both \(\mathrm{Ni(CO)_4}\) and \(\mathrm{Cr(CO)_6}\) contain one metal atom and are mononuclear. The formula \(\mathrm{Fe_2(CO)_9}\) contains two iron atoms and is polynuclear. Similarly, \(\mathrm{Co_2(CO)_8}\) contains two cobalt atoms. Row S is therefore incorrect because that compound is polynuclear, not mononuclear.
504. Consider the following statements about metal carbonyls.
Statement I: \(\mathrm{CO}\) is treated as a neutral ligand.
Statement II: Neutral homoleptic metal carbonyls commonly contain metals in oxidation state \(0\).
Statement III: Polynuclear carbonyls contain more than one metal atom.
Statement IV: Coordination of \(\mathrm{CO}\) requires the ligand to become \(\mathrm{CO^-}\) in oxidation-state calculations.
The valid statements are:
ⓐ. I and IV only
ⓑ. I, II and III
ⓒ. II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III
Explanation: Carbon monoxide is a neutral ligand in formal charge accounting. Neutral homoleptic carbonyls consequently commonly assign oxidation state \(0\) to the metal. Polynuclear carbonyls are identified by the presence of multiple metal centres. Coordination does not require assigning a \(-1\) charge to every \(\mathrm{CO}\) ligand. Statement IV therefore confuses electron donation with formal ligand charge.
505. A student argues that \(\mathrm{CO}\) cannot form a coordinate bond because it is a neutral molecule. Which correction is most appropriate?
ⓐ. Only charged ligands possess lone pairs
ⓑ. \(\mathrm{CO}\) first loses an electron and becomes \(\mathrm{CO^+}\)
ⓒ. The metal supplies both electrons of every metal–carbon bond
ⓓ. Neutral ligand donation need not change formal charge
Correct Answer: Neutral ligand donation need not change formal charge
Explanation: Ligand charge and electron-pair donation are different ideas. A neutral molecule may contain a lone pair capable of donation to a metal. Carbon monoxide donates electron density through its carbon end while remaining formally neutral for oxidation-state purposes. Ammonia and water provide other familiar examples of neutral donor ligands. The formation of a coordinate bond does not automatically convert a neutral ligand into an ion.
506. Complex \(P\) is \(\mathrm{Fe(CO)_5}\), whereas complex \(Q\) is \(\mathrm{Fe_2(CO)_9}\). The comparison between them is:
ⓐ. \(P\) is mononuclear, \(Q\) polynuclear; iron is \(0\) in both
ⓑ. Iron is \(+5\) in \(P\) and \(+9\) in \(Q\)
ⓒ. \(P\) is polynuclear and \(Q\) is mononuclear
ⓓ. The \(\mathrm{CO}\) ligands are neutral in \(P\) but negatively charged in \(Q\)
Correct Answer: \(P\) is mononuclear, \(Q\) polynuclear; iron is \(0\) in both
Explanation: \(\mathrm{Fe(CO)_5}\) contains one iron atom, so it is mononuclear. \(\mathrm{Fe_2(CO)_9}\) contains two iron atoms, so it is polynuclear. All carbonyl ligands are treated as neutral in both formulas. Since both compounds are neutral, the iron centres are assigned oxidation state \(0\). The difference in nuclearity does not require a change in the formal charge of \(\mathrm{CO}\).
507. A neutral homoleptic carbonyl has the general formula \(\mathrm{M_x(CO)_y}\). If all metal atoms are equivalent, their oxidation state is:
ⓐ. \(+y/x\)
ⓑ. \(0\)
ⓒ. \(-y/x\)
ⓓ. \(+x/y\)
Correct Answer: \(0\)
Explanation: Let the oxidation state of each equivalent metal atom be \(z\). Carbon monoxide has formal charge \(0\), and the complete carbonyl is neutral. Therefore:
\[
xz+y(0)=0
\]
This simplifies to:
\[
xz=0
\]
Since \(x\) is nonzero:
\[
z=0
\]
The numbers of metal and carbonyl units do not change this result for a neutral homoleptic carbonyl.
508. Assertion: The \(\sigma\) component of metal–carbonyl bonding is described as ligand-to-metal donation.
Reason: A lone pair located mainly at the carbon end of \(\mathrm{CO}\) overlaps with a vacant orbital on the metal.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: In the \(\sigma\) interaction, carbon monoxide supplies the occupied donor orbital. The metal provides the vacant accepting orbital. Electron density therefore moves from the ligand toward the metal. The carbon end is the usual donor end because the relevant lone-pair orbital is concentrated there. The Reason directly describes the orbital basis of the ligand-to-metal \(\sigma\) donation.
509. Use the arrangement described below.
A filled orbital on the carbon end of \(\mathrm{CO}\) points directly toward an empty metal orbital along the metal–carbon axis.
Which bond is primarily formed by this overlap?
ⓐ. A carbon–oxygen ionic bond
ⓑ. A ligand–ligand \(\pi\) bond
ⓒ. A metal–oxygen \(\sigma\) bond
ⓓ. A metal–carbon \(\sigma\) bond
Correct Answer: A metal–carbon \(\sigma\) bond
Explanation: The orbitals overlap head-on along the internuclear axis. Such end-on overlap produces a \(\sigma\) interaction. Because the donor orbital is centred mainly on carbon, the metal bonds directly to carbon rather than oxygen. The bond is therefore the metal–carbon \(\sigma\) component of carbonyl bonding. Its electron pair originates from the ligand donor orbital.
510. Study the proposed descriptions of \(\sigma\) donation in metal carbonyls.
| Row | Feature | Proposed description |
|---|
| P | Donor atom | Carbon |
| Q | Donor orbital | Filled carbon-centred orbital |
| R | Acceptor orbital | Vacant metal orbital |
| S | Direction of \(\sigma\) donation | Metal to ligand |
The inconsistent row is:
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Carbon is the usual donor atom in coordinated carbon monoxide. Its filled donor orbital overlaps with a vacant orbital on the metal. The electron pair therefore moves from the ligand toward the metal during \(\sigma\) donation. Rows P, Q, and R correctly identify the donor–acceptor components. Row S reverses the direction of the \(\sigma\)-donation process.
511. A student draws the \(\sigma\) bond in a metal carbonyl as arising from donation of an oxygen lone pair into the metal. What is the principal error?
ⓐ. A metal cannot possess vacant orbitals
ⓑ. \(\sigma\) bonds cannot form by head-on overlap
ⓒ. Ordinary metal carbonyls bind mainly through carbon
ⓓ. Carbon monoxide must first dissociate into separate atoms
Correct Answer: Ordinary metal carbonyls bind mainly through carbon
Explanation: Although carbon monoxide contains lone-pair electron density at both ends, its usual metal-binding mode is through carbon. The carbon-centred donor orbital has the appropriate characteristics for strong \(\sigma\) donation to the metal. The resulting linkage is \(\mathrm{M-C-O}\). Assigning oxygen as the normal donor predicts the wrong connectivity. The error concerns donor-atom identification rather than the general possibility of \(\sigma\) overlap.
512. A neutral compound \(\mathrm{M_2(CO)_8}\) contains only metal atoms and carbonyl ligands. Spectroscopic bonding analysis shows donation from a filled carbon-centred orbital of \(\mathrm{CO}\) into a vacant orbital on each metal. The bonding analysis supports the description:
ⓐ. It is mononuclear, the metal is \(+4\), and oxygen is the \(\sigma\) donor
ⓑ. It is mononuclear, the metal is \(0\), and the \(\sigma\) donation is metal to ligand
ⓒ. It is polynuclear, the metal is \(-4\), and carbon monoxide is an anionic ligand
ⓓ. It is polynuclear, each metal is formally \(0\), and carbon is the \(\sigma\)-donor atom
Correct Answer: It is polynuclear, each metal is formally \(0\), and carbon is the \(\sigma\)-donor atom
Explanation: The formula contains two metal atoms, so the compound is polynuclear. Every \(\mathrm{CO}\) ligand is neutral, and the complete carbonyl is neutral. If the two metals are equivalent:
\[
2x+8(0)=0
\]
Therefore:
\[
x=0
\]
The filled donor orbital is centred mainly on carbon, so carbon donates the electron pair into the vacant metal orbital. This interaction forms the metal–carbon \(\sigma\) component of the bonding.
513. Metal-to-ligand back donation in a carbonyl complex is represented by:
ⓐ. Filled metal \(d\) orbital \(\rightarrow\) vacant \(\mathrm{CO}\,\pi^\ast\) orbital
ⓑ. Filled metal \(d\) orbital \(\rightarrow\) vacant carbon-centred \(\sigma^\ast\) orbital
ⓒ. Filled carbon-centred orbital \(\rightarrow\) vacant metal orbital
ⓓ. Filled \(\mathrm{CO}\,\pi\) orbital \(\rightarrow\) vacant metal \(d\) orbital
Correct Answer: Filled metal \(d\) orbital \(\rightarrow\) vacant \(\mathrm{CO}\,\pi^\ast\) orbital
Explanation: The metal supplies an occupied \(d\) orbital of suitable symmetry. Carbon monoxide supplies a vacant antibonding \(\pi^\ast\) orbital. Sideways overlap transfers electron density from the metal toward the ligand and creates the \(\pi\) component of the metal–carbon bond. Donation from a filled carbon-centred orbital to a vacant metal orbital instead describes the \(\sigma\) component. Using a filled ligand \(\pi\) orbital or a \(\sigma^\ast\) acceptor does not represent the usual carbonyl back-bonding interaction.
514. A graph has metal electron density on the horizontal axis and extent of \(\pi\) back-bonding to identical \(\mathrm{CO}\) ligands on the vertical axis. Which general trend is expected?
ⓐ. A decreasing trend because electron-rich metals cannot donate to ligands
ⓑ. Back-bonding generally rises with metal electron density
ⓒ. A horizontal line because back-bonding is independent of the metal
ⓓ. A sudden fall to zero as soon as the metal reaches oxidation state \(0\)
Correct Answer: Back-bonding generally rises with metal electron density
Explanation: Back-bonding requires occupied metal orbitals containing available electron density. An electron-rich metal can generally donate more effectively into suitable vacant ligand orbitals. The extent of back donation therefore tends to rise with metal electron density. The exact relation need not be a universal straight line because orbital energy and overlap also matter. The qualitative direction of the trend is nevertheless upward.
515. Consider the following statements about metal–carbonyl \(\pi\) back-bonding.
Statement I: It involves filled metal \(d\) orbitals.
Statement II: It populates antibonding orbitals of \(\mathrm{CO}\).
Statement III: It increases the internal carbon–oxygen bond order.
Statement IV: It contributes to strengthening of the metal–carbon bond.
The valid statements are:
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Filled metal \(d\) orbitals supply electron density for back donation. The accepting orbitals on carbon monoxide are the empty \(\pi^\ast\) orbitals. Occupying these antibonding orbitals lowers rather than raises the carbon–oxygen bond order. At the same time, metal–ligand \(\pi\) bonding strengthens the metal–carbon interaction. Statement III is therefore the only incorrect statement.
516. Study the proposed consequences of increasing \(\pi\) back-bonding.
| Row | Property | Proposed change |
|---|
| P | Metal–carbon bond strength | Increases |
| Q | Carbon–oxygen bond strength | Decreases |
| R | Occupation of \(\mathrm{CO}\,\pi^\ast\) orbitals | Increases |
| S | Carbon–oxygen stretching frequency | Increases |
The inconsistent row is:
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Increasing back donation strengthens the metal–carbon \(\pi\) interaction. It also places more electron density into the antibonding orbitals of carbon monoxide. The internal carbon–oxygen bond is consequently weakened. A weaker carbon–oxygen bond normally has a lower, not higher, stretching frequency. Row S therefore gives the opposite spectroscopic trend.
517. Assertion: \(\sigma\) donation and \(\pi\) back-bonding in a metal carbonyl reinforce one another.
Reason: \(\sigma\) donation increases electron density at the metal for back donation, while back donation removes some metal electron density and allows continued ligand-to-metal donation.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: \(\sigma\) donation transfers electron density from carbon monoxide to the metal. This can improve the metal’s ability to donate electron density into ligand \(\pi^\ast\) orbitals. Back donation, in turn, prevents excessive accumulation of electron density at the metal. The metal can consequently remain able to accept \(\sigma\) donation effectively. The two processes therefore operate cooperatively, exactly as described in the Reason.
518. A student describes metal–carbonyl bonding as “one \(\sigma\) bond plus an unrelated \(\pi\) bond.” Which correction is most accurate?
ⓐ. Only \(\sigma\) donation occurs; the proposed \(\pi\) interaction is impossible
ⓑ. Only \(\pi\) back-bonding occurs; carbon monoxide cannot donate a lone pair
ⓒ. The \(\sigma\) and \(\pi\) components redistribute density to strengthen each other
ⓓ. The two components cancel completely and leave no net metal–carbon bond
Correct Answer: The \(\sigma\) and \(\pi\) components redistribute density to strengthen each other
Explanation: The \(\sigma\) and \(\pi\) components can be identified separately by their orbitals and electron-flow directions. They are nevertheless electronically connected. \(\sigma\) donation supplies electron density to the metal, which can enhance back donation. Back donation relieves excess electron density at the metal and supports further \(\sigma\) acceptance. Treating them as unrelated misses the central meaning of synergic bonding.
519. Increasing metal-to-\(\mathrm{CO}\) \(\pi\) back-bonding has which combined effect?
ⓐ. It strengthens metal–carbon bonding and weakens carbon–oxygen bonding
ⓑ. It transfers electron density only from \(\mathrm{CO}\) to the metal and weakens the metal–carbon bond
ⓒ. It leaves both the metal–carbon and carbon–oxygen bonds unchanged
ⓓ. It empties \(\mathrm{CO}\,\pi^\ast\) orbitals and strengthens the carbon–oxygen bond
Correct Answer: It strengthens metal–carbon bonding and weakens carbon–oxygen bonding
Explanation: \(\pi\) back-bonding transfers electron density from an occupied metal \(d\) orbital into a vacant \(\mathrm{CO}\,\pi^\ast\) orbital. This overlap adds a \(\pi\) component to the metal–carbon bond and strengthens that interaction. Population of an antibonding orbital within carbon monoxide lowers the carbon–oxygen bond order. The internal bond therefore weakens and its stretching frequency generally decreases. These linked changes are a central consequence of synergic metal–carbonyl bonding.
520. Complex \(P\) contains a more electron-rich metal centre than analogous complex \(Q\). Both contain terminal \(\mathrm{CO}\) ligands. Which complete prediction is most reasonable?
ⓐ. \(P\) has weaker back-bonding, a weaker carbon–oxygen bond, and a stronger metal–carbon bond
ⓑ. \(P\) has stronger \(\sigma\) donation only, but no possible change in back-bonding
ⓒ. \(P\): stronger back-bonding, stronger metal–carbon bond, lower carbonyl stretching frequency
ⓓ. \(P\) must release every carbonyl ligand because electron-rich metals cannot accept \(\sigma\) donation
Correct Answer: \(P\): stronger back-bonding, stronger metal–carbon bond, lower carbonyl stretching frequency
Explanation: Greater metal electron density generally increases the ability to donate into ligand \(\pi^\ast\) orbitals. Stronger back-bonding adds to the metal–carbon bond and makes that interaction stronger. At the same time, occupation of the carbonyl antibonding orbital weakens the internal carbon–oxygen bond. The weakened bond vibrates at a lower characteristic frequency. The result links metal electron density, synergic bonding, and spectroscopic behaviour.
