201. An octahedral complex of a metal ion with configuration \(d^4\) is prepared in two forms, P and Q. P has the ground-state arrangement \(t_{2g}^{3}e_g^{1}\), while Q has \(t_{2g}^{4}e_g^{0}\). Which statement correctly combines their valence-bond descriptions and magnetic classifications?
ⓐ. P is outer orbital with four unpaired electrons; Q is inner orbital with two unpaired electrons.
ⓑ. P is inner orbital with four unpaired electrons; Q is outer orbital with two unpaired electrons.
ⓒ. Both are inner orbital because each has four d electrons.
ⓓ. Both are outer orbital because neither configuration contains enough electrons to pair.
Correct Answer: P is outer orbital with four unpaired electrons; Q is inner orbital with two unpaired electrons.
Explanation: \(\textbf{Electronic arrangement of P:}\)
P contains four d electrons distributed as \(t_{2g}^{3}e_g^{1}\).
The three \(t_{2g}\) orbitals each contain one electron.
One additional electron occupies an \(e_g\) orbital.
Therefore, all four electrons are unpaired:
\[
n_P=4
\]
This is the high-spin arrangement.
In the corresponding standard valence-bond description, only one inner d orbital remains vacant.
Two vacant inner d orbitals are required for \(d^2sp^3\) hybridisation.
Consequently, P is described using outer-orbital \(sp^3d^2\) hybridisation.
\(\textbf{Electronic arrangement of Q:}\)
Q contains four electrons distributed as \(t_{2g}^{4}e_g^{0}\).
One \(t_{2g}\) orbital contains a pair of electrons.
The other two \(t_{2g}\) orbitals are singly occupied.
Therefore:
\[
n_Q=2
\]
This is the low-spin arrangement.
The corresponding valence-bond description has two vacant inner d orbitals.
These can participate in \(d^2sp^3\) hybridisation.
Thus, Q is an inner-orbital octahedral complex.
\(\textbf{Combined conclusion:}\)
Both forms have coordination number six and the same total d-electron count.
Their different electron-pairing patterns account for the difference in magnetic behaviour and inner-/outer-orbital descriptions.
The comparison demonstrates how an orbital-occupancy representation can be translated into a valence-bond hybridisation assignment.
\(\textbf{Final answer:}\)
\[
P:sp^3d^2,\ n=4;\qquad Q:d^2sp^3,\ n=2
\]
202. A coordination chemist compares two observations involving the same metal oxidation state:
Observation P: Two complexes have identical hybridisation assignments but different visible absorption spectra.
Observation Q: Two complexes have identical hybridisation assignments but different ligand-substitution rates.
Which statement correctly describes the explanatory limitations of the elementary valence-bond approach?
ⓐ. It quantitatively explains P but not Q because hybridisation uniquely determines orbital splitting.
ⓑ. It quantitatively explains Q but not P because hybridisation uniquely determines reaction activation energies.
ⓒ. Neither observation presents a limitation because hybridisation labels are unrelated to coordination compounds.
ⓓ. Neither colours nor ligand-substitution kinetics are reliably predicted from a shared valence-bond hybridisation assignment.
Correct Answer: Neither colours nor ligand-substitution kinetics are reliably predicted from a shared valence-bond hybridisation assignment.
Explanation: The elementary valence-bond approach successfully describes the formation of many coordination bonds using hybrid orbitals.
It is useful for relating suitable hybridisation schemes to common coordination geometries.
It also provides qualitative explanations for the magnetic behaviour of several standard complexes.
However, its explanatory scope is limited.
The theory does not satisfactorily account for the colours of coordination compounds.
It does not supply a quantitative description of the relevant d-orbital electronic excitation energies responsible for many optical absorptions.
Crystal-field theory provides a useful additional framework by describing the splitting of metal d orbitals in ligand fields.
Thus, two complexes may possess the same hybridisation assignment yet exhibit different absorption spectra.
The valence-bond approach also does not provide a quantitative explanation of the kinetic stabilities of coordination compounds.
Kinetic stability concerns reaction rates, including rates of ligand substitution.
A hybridisation label alone does not specify the reaction pathway or activation-energy barrier for ligand exchange.
Consequently, different ligand-substitution rates are compatible with the same qualitative hybridisation assignment.
The common hybridisation description therefore does not determine either observation.
Recognising these limitations is important because a model may remain useful for geometry and magnetic classification without quantitatively explaining every physical and chemical property.
The appropriate conclusion is not that the model has no value, but that additional concepts are required for the phenomena it does not adequately describe.
203. Carbon monoxide lies towards the strong-field end of the NCERT spectrochemical series, whereas chloride is a weak-field ligand. Yet carbon monoxide is neutral and chloride carries charge \(-1\). Which statement best reconciles this observation with the limitations of elementary crystal-field theory?
ⓐ. Neutral ligands must always produce greater splitting than anionic ligands because they have no electrostatic repulsion.
ⓑ. Formal ligand charge alone cannot determine field strength; the point-charge model omits effects such as CO back-donation.
ⓒ. Chloride cannot form coordinate bonds because its negative charge prevents donation of an electron pair.
ⓓ. Carbon monoxide must be negatively charged after coordination to produce a stronger field than chloride.
Correct Answer: Formal ligand charge alone cannot determine field strength; the point-charge model omits effects such as CO back-donation.
Explanation: The spectrochemical series is based on experimentally observed differences in ligand-field splitting.
Chloride is a negatively charged ligand and lies in the weak-field region of the standard series.
Carbon monoxide is neutral but lies towards the strong-field end.
Therefore, formal electrical charge alone cannot account for their relative ligand-field strengths.
Elementary crystal-field theory represents ligands as point charges or point dipoles.
This electrostatic treatment explains why a geometrically arranged ligand field splits the metal d orbitals.
However, it does not explicitly describe the full covalent character of metal–ligand bonding.
Carbon monoxide provides an important example because its bonding involves more than ordinary electron-pair sigma donation.
The carbon atom donates an electron pair into an appropriate vacant metal orbital.
At the same time, occupied metal d orbitals can donate electron density into vacant CO antibonding orbitals through pi back-bonding.
These interactions contribute to the distinctive bonding behaviour of metal carbonyls.
A simple ranking based only on whether a ligand is neutral or anionic cannot capture these effects.
Consequently, the unexpectedly strong position of CO relative to chloride illustrates a limitation of a purely electrostatic account.
The observation does not require CO to acquire a formal negative charge.
Nor does it mean that chloride is incapable of acting as a donor ligand.
The correct interpretation distinguishes formal ligand charge from the electronic interactions governing ligand-field strength.
204. A hypothetical metal ion has a \(d^6\) configuration and forms two octahedral complexes P and Q. The ligand field of P is weak enough to give four unpaired electrons, while that of Q is strong enough to give no unpaired electrons. Both have coordination number six and the same metal oxidation state. Which combination of statements is correct?
Statement I: The valence-bond description assigns outer-orbital hybridisation to P and inner-orbital hybridisation to Q.
Statement II: The crystal-field description assigns \(t_{2g}^{4}e_g^{2}\) to P and \(t_{2g}^{6}e_g^{0}\) to Q.
Statement III: The different magnetic behaviour necessarily means that six metal–ligand coordinate bonds are broken when P changes into Q.
Statement IV: Both complexes have the same total number of metal d electrons despite their different magnetic properties.
ⓐ. Statements I and II only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, II and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because P has the high-spin \(d^6\) arrangement.
In the standard valence-bond description, it lacks the two vacant inner d orbitals required for \(d^2sp^3\) hybridisation.
P is consequently described as an outer-orbital \(sp^3d^2\) complex.
Q has the low-spin arrangement in which the six d electrons are paired in three inner d orbitals.
Two inner d orbitals remain vacant, permitting \(d^2sp^3\) hybridisation.
Q is therefore an inner-orbital complex.
Statement II is correct because the specified magnetic behaviours correspond to the standard high-spin and low-spin octahedral \(d^6\) configurations.
P contains four unpaired electrons in \(t_{2g}^{4}e_g^{2}\).
Q contains no unpaired electrons in \(t_{2g}^{6}e_g^{0}\).
Statement III is incorrect because changing the electron-pairing arrangement does not itself require the destruction of all six coordination bonds.
A change in ligand environment may change the preferred spin state while preserving octahedral coordination number six.
The problem does not describe the breaking of six bonds as a necessary consequence.
Statement IV is correct because both complexes contain the same metal ion in the same oxidation state.
Each therefore has six metal d electrons.
The different magnetic classifications arise from electron distribution rather than a change in total electron count.
This comparison integrates the crystal-field and valence-bond descriptions without treating them as identical physical models.
The crystal-field description emphasises orbital-energy splitting and occupancy.
The valence-bond description uses the corresponding electron-pairing picture to assign inner- or outer-orbital hybridisation.
Therefore, Statements I, II and IV are correct.
205. A chemist studies an octahedral chromium(III) coordination entity and records that three metal d electrons occupy separate \(t_{2g}\) orbitals in the ground state. The ligands are then replaced by ligands producing a substantially greater octahedral splitting, while chromium remains in oxidation state \(+3\) and the coordination number remains six. Which combination of predictions is correct?
ⓐ. The three electrons pair within \(t_{2g}\), leaving one unpaired electron; the hybridisation changes from \(d^2sp^3\) to \(sp^3d^2\).
ⓑ. The electron configuration becomes \(t_{2g}^{2}e_g^{1}\); the number of unpaired electrons remains three.
ⓒ. The three electrons remain singly occupied in \(t_{2g}\); the number of unpaired electrons remains three, although the splitting energy changes.
ⓓ. The electron configuration remains \(t_{2g}^{3}e_g^{0}\), but the complex becomes diamagnetic because the ligand field is stronger.
Correct Answer: The three electrons remain singly occupied in \(t_{2g}\); the number of unpaired electrons remains three, although the splitting energy changes.
Explanation: The chromium oxidation state is \(+3\), corresponding to the electronic configuration \(3d^3\).
In an octahedral field, the three \(t_{2g}\) orbitals lie below the two \(e_g\) orbitals.
The first three d electrons occupy the three lower-energy orbitals singly according to Hund's rule.
Therefore, the ground-state configuration is \(t_{2g}^{3}e_g^{0}\).
Increasing the splitting does not introduce any need to pair these electrons.
Each electron already occupies a separate orbital in the lower-energy set.
Pairing two electrons in one \(t_{2g}\) orbital would leave another degenerate \(t_{2g}\) orbital vacant and would not provide an energetic advantage.
Moving an electron into \(e_g\) would instead require additional excitation energy.
Consequently, the ordinary ground-state occupancy remains unchanged.
All three d electrons remain unpaired, and the complex continues to be paramagnetic.
The octahedral splitting energy can nevertheless increase because the ligand environment has changed.
The standard inner-orbital \(d^2sp^3\) description is also compatible with this \(d^3\) ion, since two inner d orbitals are already vacant.
Thus, magnetic behaviour need not change merely because the crystal-field splitting becomes larger.
The result distinguishes a change in orbital energy separation from a change in electron occupancy.
206. Decacarbonyldimanganese(0), \(\mathrm{Mn_2(CO)_{10}}\), contains two \(\mathrm{Mn(CO)_5}\) units joined by a manganese–manganese bond. Each CO is a neutral ligand and is attached to one manganese atom through carbon. Which statement correctly describes the structural and oxidation-state consequences?
ⓐ. Each manganese has oxidation state zero, and the Mn–Mn bond does not require either metal to acquire a nonzero formal oxidation state.
ⓑ. Each manganese has oxidation state \(+1\), because a metal–metal bond always contributes a positive oxidation-state unit.
ⓒ. Each manganese has oxidation state \(+5\), because five CO molecules are attached to each metal.
ⓓ. One manganese has oxidation state \(+1\) and the other \(-1\), because the metal–metal bond requires opposite formal charges.
Correct Answer: Each manganese has oxidation state zero, and the Mn–Mn bond does not require either metal to acquire a nonzero formal oxidation state.
Explanation: Decacarbonyldimanganese is a neutral metal carbonyl containing two manganese atoms.
All ten carbon monoxide ligands are electrically neutral.
Each manganese atom is bonded to five CO ligands in the stated structure.
The two manganese-containing units are also joined by a direct metal–metal bond.
In oxidation-state bookkeeping, a bond between two identical atoms does not assign a positive oxidation state to one atom and a negative state to the other.
The bonding electrons are assigned equally for this formal purpose.
Since the carbonyl ligands contribute zero charge and the molecule is neutral, the manganese atoms retain oxidation state zero.
The existence of five metal–carbon bonds around each manganese does not imply oxidation state \(+5\).
Coordination attachments and metal oxidation states describe different chemical properties.
The Mn–Mn bond is an additional structural feature that must not be confused with a bond to an external ligand.
This example also shows that not every metal carbonyl is mononuclear.
The standard structural description consists of two square-pyramidal \(\mathrm{Mn(CO)_5}\) units connected through a metal–metal bond.
The correct oxidation-state assignment remains consistent with the compound's name, decacarbonyldimanganese(0).
207. Octacarbonyldicobalt(0), \(\mathrm{Co_2(CO)_8}\), has a structure containing a cobalt–cobalt bond and two bridging carbonyl groups. A student concludes that each bridging CO must have formal charge \(-1\), because it interacts with more than one cobalt atom. Which response is correct?
ⓐ. The conclusion is correct because every ligand that bridges two metal centres becomes an anion.
ⓑ. The conclusion is correct because bridging removes an electron from each carbonyl ligand.
ⓒ. The conclusion is incorrect only because cobalt must have oxidation state \(+2\) in every carbonyl compound.
ⓓ. The conclusion is incorrect because bridging coordination does not by itself change the formal neutral-ligand classification of CO.
Correct Answer: The conclusion is incorrect because bridging coordination does not by itself change the formal neutral-ligand classification of CO.
Explanation: Carbon monoxide is conventionally treated as a neutral ligand in oxidation-state calculations for metal carbonyls.
Its neutral classification does not depend on whether its bonding arrangement is terminal or bridging.
A terminal carbonyl interacts directly with one metal centre.
A bridging carbonyl participates in bonding involving more than one metal centre.
These bonding arrangements differ structurally but do not automatically change the ligand's assigned formal charge.
The compound \(\mathrm{Co_2(CO)_8}\) contains two cobalt atoms and eight neutral carbonyl ligands.
Its overall molecular charge is zero.
The standard oxidation-state assignment is therefore cobalt(0).
The direct cobalt–cobalt bond does not independently require either identical cobalt atom to acquire a formal positive charge.
Assigning a charge of \(-1\) to each bridging CO would confuse the number of metal–ligand interactions with formal ionic charge.
Metal carbonyl bonding can also contain synergic sigma-donation and pi-back-donation components.
These interactions redistribute electron density but do not necessarily imply a corresponding change in the formal oxidation-state convention.
Consequently, the presence of bridging carbonyls must be considered when describing structure, but it does not justify treating neutral CO as an anionic ligand.
This illustrates the distinction between actual bonding interactions and formal oxidation-state bookkeeping.
208. An aqueous nickel(II) complex undergoes the stepwise substitution shown below:
\[
\mathrm{[Ni(H_2O)_6]^{2+}\rightarrow[Ni(H_2O)_4(en)]^{2+}\rightarrow[Ni(H_2O)_2(en)_2]^{2+}\rightarrow[Ni(en)_3]^{2+}}
\]
The spectrochemical series places en above water in ligand-field strength. Assume the coordination number remains six and all four complexes retain the same ordinary octahedral \(d^8\) ground-state occupancy. Which inference is most appropriate?
ⓐ. The absorption energies must remain unchanged because the nickel oxidation state is constant.
ⓑ. The coordination number increases at every substitution, so the orbital splitting cannot be compared.
ⓒ. The ligand-field splitting may increase while the number of unpaired d electrons remains two.
ⓓ. Every addition of en must remove one unpaired electron until the final complex becomes diamagnetic.
Correct Answer: The ligand-field splitting may increase while the number of unpaired d electrons remains two.
Explanation: The initial coordination entity contains six unidentate water ligands.
Each en molecule replaces two coordinated water molecules because en is didentate.
Consequently, every specified complex retains six directly coordinated donor atoms.
The coordination number therefore remains six throughout the sequence.
Both water and en are neutral ligands, so nickel retains oxidation state \(+2\).
Nickel(II) has the electronic configuration \(3d^8\).
In an ordinary octahedral crystal field, its occupancy is \(t_{2g}^{6}e_g^{2}\).
The three \(t_{2g}\) orbitals are completely filled, while the two \(e_g\) orbitals each contain one electron.
Thus, the ordinary octahedral \(d^8\) arrangement contains two unpaired electrons.
Increasing the splitting cannot place more than six electrons in the \(t_{2g}\) set.
The ordinary spin count therefore remains unchanged under the stated conditions.
However, en is higher than water in the spectrochemical series.
Replacing water by en can alter the ligand field and hence the energy separation between the two orbital sets.
Changes in electronic transition energies may consequently accompany changes in colour.
The full absorption spectrum cannot be predicted from ligand identity alone without additional information.
This example demonstrates that ligand substitution, chelation, magnetic behaviour and optical properties need not change together.
209. A hypothetical octahedral \(d^4\) complex can adopt a high-spin configuration \(t_{2g}^{3}e_g^{1}\) or a low-spin configuration \(t_{2g}^{4}e_g^{0}\). The octahedral splitting is \(18\,000\ \mathrm{cm^{-1}}\), and the pairing energy is \(12\,000\ \mathrm{cm^{-1}}\). Treat the energy of each \(t_{2g}\) electron as \(-0.4\Delta_o\), each \(e_g\) electron as \(+0.6\Delta_o\), and add \(P_{\mathrm{pair}}\) for every electron pair. Ignoring other contributions, which conclusion is correct?
ⓐ. The low-spin state is lower by \(6\,000\ \mathrm{cm^{-1}}\).
ⓑ. The high-spin state is lower by \(6\,000\ \mathrm{cm^{-1}}\).
ⓒ. The low-spin state is lower by \(18\,000\ \mathrm{cm^{-1}}\).
ⓓ. The two states have equal energy because they contain four d electrons each.
Correct Answer: The low-spin state is lower by \(6\,000\ \mathrm{cm^{-1}}\).
Explanation: \(\textbf{High-spin orbital contribution:}\)
The high-spin \(d^4\) configuration is:
\[
t_{2g}^{3}e_g^{1}
\]
Its crystal-field energy is:
\[
E_{\mathrm{CF,HS}}=3(-0.4\Delta_o)+1(+0.6\Delta_o)
\]
\[
E_{\mathrm{CF,HS}}=-0.6\Delta_o
\]
Substituting the given splitting:
\[
E_{\mathrm{CF,HS}}=-0.6(18\,000)
\]
\[
E_{\mathrm{CF,HS}}=-10\,800\ \mathrm{cm^{-1}}
\]
The four electrons occupy separate orbitals, so no electron-pairing cost is added.
\[
E_{\mathrm{HS}}=-10\,800\ \mathrm{cm^{-1}}
\]
\(\textbf{Low-spin orbital contribution:}\)
The low-spin configuration is:
\[
t_{2g}^{4}e_g^{0}
\]
Its crystal-field energy is:
\[
E_{\mathrm{CF,LS}}=4(-0.4\Delta_o)
\]
\[
E_{\mathrm{CF,LS}}=-1.6(18\,000)
\]
\[
E_{\mathrm{CF,LS}}=-28\,800\ \mathrm{cm^{-1}}
\]
One \(t_{2g}\) orbital contains an electron pair.
Therefore, the total energy includes one pairing-energy contribution:
\[
E_{\mathrm{LS}}=-28\,800+12\,000
\]
\[
E_{\mathrm{LS}}=-16\,800\ \mathrm{cm^{-1}}
\]
\(\textbf{Energy comparison:}\)
\[
E_{\mathrm{LS}}-E_{\mathrm{HS}}=-16\,800-(-10\,800)
\]
\[
E_{\mathrm{LS}}-E_{\mathrm{HS}}=-6\,000\ \mathrm{cm^{-1}}
\]
The negative sign indicates that the low-spin arrangement is energetically preferred.
This is consistent with \(\Delta_o\gt P_{\mathrm{pair}}\).
\(\textbf{Final answer:}\)
\[
\boxed{E_{\mathrm{LS}}-E_{\mathrm{HS}}=-6\,000\ \mathrm{cm^{-1}}}
\]
210. A first-row transition-metal ion M forms two octahedral complexes with the same oxidation state. The weak-field complex has five unpaired electrons, while the strong-field complex has one unpaired electron. The neutral metal atom has the electronic configuration \(\mathrm{[Ar]\,3d^5 4s^2}\). Which combination correctly identifies the metal oxidation state and the two crystal-field configurations?
ⓐ. \(+2\); weak field \(t_{2g}^{3}e_g^{2}\); strong field \(t_{2g}^{5}e_g^{0}\)
ⓑ. \(+3\); weak field \(t_{2g}^{4}e_g^{2}\); strong field \(t_{2g}^{6}e_g^{0}\)
ⓒ. \(+2\); weak field \(t_{2g}^{4}e_g^{2}\); strong field \(t_{2g}^{6}e_g^{0}\)
ⓓ. \(+3\); weak field \(t_{2g}^{3}e_g^{2}\); strong field \(t_{2g}^{5}e_g^{0}\)
Correct Answer: \(+2\); weak field \(t_{2g}^{3}e_g^{2}\); strong field \(t_{2g}^{5}e_g^{0}\)
Explanation: \(\textbf{Inference from the magnetic evidence:}\)
An octahedral complex with five unpaired d electrons requires at least five d electrons.
A strong-field arrangement retaining exactly one unpaired electron is consistent with a low-spin \(d^5\) ion.
The weak-field \(d^5\) arrangement is \(t_{2g}^{3}e_g^{2}\), containing five singly occupied orbitals.
The corresponding low-spin arrangement is \(t_{2g}^{5}e_g^{0}\).
It contains two electron pairs and one unpaired electron.
\(\textbf{Metal-ion oxidation state:}\)
The neutral atom is:
\[
\mathrm{M:[Ar]\,3d^5 4s^2}
\]
Removal of the two \(4s\) electrons gives:
\[
\mathrm{M^{2+}:[Ar]\,3d^5}
\]
Therefore, oxidation state \(+2\) is consistent with the inferred \(d^5\) configuration.
Removal of an additional \(3d\) electron would instead produce a \(d^4\) ion.
\(\textbf{Checking the alternatives:}\)
The \(d^6\) configurations \(t_{2g}^{4}e_g^{2}\) and \(t_{2g}^{6}e_g^{0}\) contain four and zero unpaired electrons, respectively.
They cannot reproduce the observed five-to-one change.
The \(d^5\) configurations satisfy both magnetic measurements.
The neutral-atom configuration then fixes the metal oxidation state as \(+2\).
\(\textbf{Final answer:}\)
\[
\mathrm{M^{2+}:3d^5};\quad \mathrm{HS}:t_{2g}^{3}e_g^{2};\quad \mathrm{LS}:t_{2g}^{5}e_g^{0}
\]
211. Two octahedral metal ions P and Q have different d-electron configurations but exhibit the same spin-only magnetic moment corresponding to two unpaired electrons. P has the configuration \(t_{2g}^{2}e_g^{0}\), while Q has \(t_{2g}^{6}e_g^{2}\). If their octahedral splitting energies are identical and positive, which statement correctly compares their crystal-field stabilisation energies, excluding pairing costs?
ⓐ. P and Q have identical crystal-field energies because their unpaired-electron counts are the same.
ⓑ. P has greater stabilisation because both of its electrons occupy \(t_{2g}\).
ⓒ. Q has greater stabilisation by \(0.4\Delta_o\), despite having the same number of unpaired electrons as P.
ⓓ. Q has less stabilisation by \(0.4\Delta_o\), because its two \(e_g\) electrons are destabilised.
Correct Answer: Q has greater stabilisation by \(0.4\Delta_o\), despite having the same number of unpaired electrons as P.
Explanation: \(\textbf{Crystal-field energy of P:}\)
P has two electrons in the lower-energy \(t_{2g}\) set.
Its crystal-field energy is:
\[
E_P=2(-0.4\Delta_o)
\]
\[
E_P=-0.8\Delta_o
\]
Both electrons occupy different \(t_{2g}\) orbitals and remain unpaired.
\(\textbf{Crystal-field energy of Q:}\)
Q has six electrons in \(t_{2g}\) and two in \(e_g\).
The lower-energy contribution is:
\[
E_{\mathrm{lower}}=6(-0.4\Delta_o)
\]
\[
E_{\mathrm{lower}}=-2.4\Delta_o
\]
The higher-energy contribution is:
\[
E_{\mathrm{upper}}=2(+0.6\Delta_o)
\]
\[
E_{\mathrm{upper}}=+1.2\Delta_o
\]
Therefore:
\[
E_Q=-2.4\Delta_o+1.2\Delta_o
\]
\[
E_Q=-1.2\Delta_o
\]
\(\textbf{Comparison of stabilisation:}\)
\[
E_Q-E_P=-1.2\Delta_o-(-0.8\Delta_o)
\]
\[
E_Q-E_P=-0.4\Delta_o
\]
Thus, Q has the more negative crystal-field energy.
Its stabilisation magnitude exceeds that of P by \(0.4\Delta_o\).
\(\textbf{Interpretation:}\)
Both configurations contain two unpaired electrons and therefore have the same spin-only magnetic moment.
However, their total d-electron counts and orbital populations differ.
Equal magnetic moments do not imply equal crystal-field stabilisation.
The conclusion concerns the crystal-field contribution alone, not the complete thermodynamic stability of the complexes.
\(\textbf{Final answer:}\)
\[
\boxed{|E_Q|-|E_P|=0.4\Delta_o}
\]
212. A high-spin tetrahedral complex contains a metal ion with configuration \(d^4\). A hypothetical low-spin arrangement is proposed in which all four electrons pair within the two lower-energy \(e\) orbitals. The tetrahedral splitting is \(8\,000\ \mathrm{cm^{-1}}\), while the pairing energy is \(13\,000\ \mathrm{cm^{-1}}\). Use \(E(e)=-0.6\Delta_t\), \(E(t_2)=+0.4\Delta_t\), and add \(P_{\mathrm{pair}}\) for each electron pair. Which conclusion correctly compares the proposed arrangements?
ⓐ. The low-spin arrangement is lower by \(10\,000\ \mathrm{cm^{-1}}\).
ⓑ. The low-spin arrangement is lower by \(5\,000\ \mathrm{cm^{-1}}\).
ⓒ. The high-spin arrangement is lower by \(5\,000\ \mathrm{cm^{-1}}\).
ⓓ. The high-spin arrangement is lower by \(10\,000\ \mathrm{cm^{-1}}\).
Correct Answer: The high-spin arrangement is lower by \(10\,000\ \mathrm{cm^{-1}}\).
Explanation: \(\textbf{High-spin orbital occupancy:}\)
For tetrahedral \(d^4\), the ordinary high-spin configuration is:
\[
e^2t_2^2
\]
The two \(e\) electrons occupy separate orbitals.
The two \(t_2\) electrons also occupy separate orbitals.
Thus, the high-spin arrangement contains no electron pairs.
\(\textbf{High-spin energy:}\)
\[
E_{\mathrm{HS}}=2(-0.6\Delta_t)+2(+0.4\Delta_t)
\]
\[
E_{\mathrm{HS}}=-0.4\Delta_t
\]
Substituting:
\[
E_{\mathrm{HS}}=-0.4(8\,000)
\]
\[
E_{\mathrm{HS}}=-3\,200\ \mathrm{cm^{-1}}
\]
\(\textbf{Proposed low-spin occupancy:}\)
The proposed arrangement is:
\[
e^4t_2^0
\]
Four electrons completely fill the two \(e\) orbitals.
This arrangement contains two electron pairs.
Its orbital contribution is:
\[
E_{\mathrm{CF,LS}}=4(-0.6\Delta_t)
\]
\[
E_{\mathrm{CF,LS}}=-2.4(8\,000)
\]
\[
E_{\mathrm{CF,LS}}=-19\,200\ \mathrm{cm^{-1}}
\]
Including two pairing-energy contributions:
\[
E_{\mathrm{LS}}=-19\,200+2(13\,000)
\]
\[
E_{\mathrm{LS}}=6\,800\ \mathrm{cm^{-1}}
\]
\(\textbf{Energy comparison:}\)
\[
E_{\mathrm{LS}}-E_{\mathrm{HS}}=6\,800-(-3\,200)
\]
\[
E_{\mathrm{LS}}-E_{\mathrm{HS}}=10\,000\ \mathrm{cm^{-1}}
\]
The positive result means the proposed low-spin arrangement is higher in energy.
Therefore, the high-spin arrangement is preferred under the stated conditions.
The result is consistent with the usual tendency of tetrahedral complexes to remain high spin because tetrahedral splitting is comparatively small.
\(\textbf{Final answer:}\)
\[
\boxed{E_{\mathrm{LS}}-E_{\mathrm{HS}}=10\,000\ \mathrm{cm^{-1}}}
\]
213. A chemist compares two ligand environments around the same octahedral \(d^4\) metal ion. The ligand-field splittings are:
| Environment | \(\Delta_o\) | Observed number of unpaired electrons |
|---|
| P | \(11\,000\ \mathrm{cm^{-1}}\) | 4 |
| Q | \(19\,000\ \mathrm{cm^{-1}}\) | 2 |
Assuming the elementary crystal-field model applies and that the relevant pairing energy is unchanged between P and Q, which interval must contain the pairing energy?
ⓐ. \(P_{\mathrm{pair}}\lt 11\,000\ \mathrm{cm^{-1}}\)
ⓑ. \(P_{\mathrm{pair}}\gt 19\,000\ \mathrm{cm^{-1}}\)
ⓒ. \(P_{\mathrm{pair}}=15\,000\ \mathrm{cm^{-1}}\) exactly
ⓓ. \(11\,000\lt P_{\mathrm{pair}}\lt 19\,000\ \mathrm{cm^{-1}}\)
Correct Answer: \(11\,000\lt P_{\mathrm{pair}}\lt 19\,000\ \mathrm{cm^{-1}}\)
Explanation: \(\textbf{Interpretation of P:}\)
A \(d^4\) ion with four unpaired electrons must have the high-spin octahedral arrangement:
\[
t_{2g}^{3}e_g^{1}
\]
The fourth electron occupies the higher-energy \(e_g\) orbital instead of pairing within \(t_{2g}\).
In the elementary model, this occurs when:
\[
\Delta_o\lt P_{\mathrm{pair}}
\]
Substituting the splitting for P:
\[
11\,000\lt P_{\mathrm{pair}}
\]
\(\textbf{Interpretation of Q:}\)
The same \(d^4\) ion has two unpaired electrons in Q.
This corresponds to the low-spin arrangement:
\[
t_{2g}^{4}e_g^{0}
\]
The fourth electron pairs within the lower-energy \(t_{2g}\) set.
This occurs when:
\[
\Delta_o\gt P_{\mathrm{pair}}
\]
Using the splitting for Q:
\[
19\,000\gt P_{\mathrm{pair}}
\]
\(\textbf{Combining the constraints:}\)
Both observations must be satisfied by the same pairing-energy value.
Therefore:
\[
11\,000\lt P_{\mathrm{pair}}\lt 19\,000\ \mathrm{cm^{-1}}
\]
The experiments establish an interval rather than one exact value.
Although \(15\,000\ \mathrm{cm^{-1}}\) lies inside the interval, the observations do not uniquely determine that value.
\(\textbf{Final answer:}\)
\[
\boxed{11\,000\lt P_{\mathrm{pair}}\lt 19\,000\ \mathrm{cm^{-1}}}
\]
214. Two octahedral coordination entities contain chromium(III), which has a \(d^3\) configuration. In P, the ligands produce a relatively weak field; in Q, they produce a much stronger field. Assume both remain octahedral and their metal oxidation states are unchanged. Which statement correctly identifies a property that must remain unchanged and one that may change?
ⓐ. The \(t_{2g}\) electron count must remain three, while the energy required for a suitable crystal-field electronic transition may change.
ⓑ. The number of unpaired electrons must decrease, while the energy required for a crystal-field transition remains unchanged.
ⓒ. The coordination number must increase because strong-field ligands contribute more donor atoms.
ⓓ. The central-metal \(d\)-electron count must change because strong-field ligands transfer electrons into the metal.
Correct Answer: The \(t_{2g}\) electron count must remain three, while the energy required for a suitable crystal-field electronic transition may change.
Explanation: Chromium(III) has the electronic configuration \(3d^3\).
The ligand substitutions are stated not to change chromium's oxidation state.
Therefore, its d-electron count remains three.
Both complexes retain octahedral geometry, so the lower orbital set is \(t_{2g}\).
The three d electrons occupy its three orbitals singly in the ordinary ground state.
This arrangement does not depend on whether the field is relatively weak or strong.
Accordingly, both complexes have configuration \(t_{2g}^{3}e_g^{0}\).
Their numbers of unpaired electrons remain three.
However, different ligands can produce different octahedral splitting energies.
A stronger ligand field may increase the separation between the \(t_{2g}\) and \(e_g\) sets.
The energy of a suitable electronic excitation can consequently change.
This may alter the positions of absorption bands even though the metal oxidation state, coordination number and ground-state spin count remain fixed.
The change in splitting should not be confused with a change in the total number of metal d electrons.
Nor does a stronger field inherently mean a greater coordination number.
The question illustrates why magnetic and optical properties must be interpreted through related but distinct electronic considerations.
215. A student compares four octahedral complexes and attempts to arrange their splitting energies using only their ligand identities:
| Complex | Central metal | Ligands |
|---|
| P | \(\mathrm{Fe^{2+}}\) | \(\mathrm{H_2O}\) |
| Q | \(\mathrm{Fe^{2+}}\) | \(\mathrm{CN^-}\) |
| R | \(\mathrm{Co^{3+}}\) | \(\mathrm{H_2O}\) |
| S | \(\mathrm{Co^{3+}}\) | \(\mathrm{CN^-}\) |
Which conclusion is justified by the spectrochemical series and the stated information?
ⓐ. \(\Delta_o(P)\lt \Delta_o(R)\lt \Delta_o(Q)\lt \Delta_o(S)\) must hold universally.
ⓑ. \(\Delta_o(P)\gt \Delta_o(Q)\), because water is neutral and cyanide is anionic.
ⓒ. For each metal, cyanide gives more splitting than water; the cross-metal ranking is undetermined.
ⓓ. All four splitting energies must be equal because every complex is octahedral.
Correct Answer: For each metal, cyanide gives more splitting than water; the cross-metal ranking is undetermined.
Explanation: The spectrochemical series ranks ligands according to their relative field strengths in suitable comparable coordination environments.
Cyanide lies above water in the standard series.
Therefore, for a fixed central metal and oxidation state, the cyanido complex is expected to show greater crystal-field splitting than the corresponding aqua complex.
For iron(II), this supports the qualitative comparison \(\Delta_o(Q)\gt \Delta_o(P)\).
For cobalt(III), it similarly supports \(\Delta_o(S)\gt \Delta_o(R)\).
However, the absolute splitting energy does not depend on ligand identity alone.
The central metal and its oxidation state also influence the energy separation between d-orbital sets.
P and R contain different metal ions.
Likewise, Q and S contain different metal ions.
The spectrochemical ordering cannot, by itself, determine every cross-metal comparison.
In particular, it does not specify whether the splitting in R exceeds that in Q.
A complete numerical or qualitative four-complex ranking would require additional information about the relative metal-ion effects.
Thus, the two within-metal ligand comparisons are supported, but a universal order of all four complexes is not established.
The question illustrates why a valid spectrochemical-series argument must keep relevant comparison conditions under control.
216. An octahedral \(d^5\) metal ion forms two complexes P and Q with different ligands. P has configuration \(t_{2g}^{3}e_g^{2}\), while Q has configuration \(t_{2g}^{5}e_g^{0}\). Which statement correctly compares their crystal-field energies, electron-pairing contributions and magnetic behaviour?
ⓐ. P: \(E_{\mathrm{CF}}=0\), five unpaired electrons; Q: \(E_{\mathrm{CF}}=-2\Delta_o\), two electron pairs, one unpaired electron.
ⓑ. P: \(E_{\mathrm{CF}}=-2\Delta_o\), one unpaired electron; Q: \(E_{\mathrm{CF}}=0\), no electron pairs, five unpaired electrons.
ⓒ. P: \(E_{\mathrm{CF}}=0\), three unpaired electrons; Q: \(E_{\mathrm{CF}}=-2\Delta_o\), one electron pair, three unpaired electrons.
ⓓ. P: \(E_{\mathrm{CF}}=0\), five unpaired electrons; Q: \(E_{\mathrm{CF}}=-2\Delta_o\), no electron pairs, zero unpaired electrons.
Correct Answer: P: \(E_{\mathrm{CF}}=0\), five unpaired electrons; Q: \(E_{\mathrm{CF}}=-2\Delta_o\), two electron pairs, one unpaired electron.
Explanation: \(\textbf{High-spin configuration P:}\)
The occupancy is:
\[
t_{2g}^{3}e_g^{2}
\]
Three electrons occupy the lower-energy \(t_{2g}\) orbitals.
Two occupy the higher-energy \(e_g\) orbitals.
All five electrons are unpaired.
The crystal-field energy is:
\[
E_{\mathrm{CF},P}=3(-0.4\Delta_o)+2(+0.6\Delta_o)
\]
\[
E_{\mathrm{CF},P}=-1.2\Delta_o+1.2\Delta_o
\]
\[
E_{\mathrm{CF},P}=0
\]
\(\textbf{Low-spin configuration Q:}\)
The occupancy is:
\[
t_{2g}^{5}e_g^{0}
\]
Two lower-energy orbitals contain electron pairs, while the third contains one unpaired electron.
The crystal-field energy is:
\[
E_{\mathrm{CF},Q}=5(-0.4\Delta_o)
\]
\[
E_{\mathrm{CF},Q}=-2\Delta_o
\]
The number of unpaired electrons is one.
\(\textbf{Pairing contribution:}\)
P contains no paired d electrons in the stated arrangement.
Q contains two electron pairs.
Therefore, Q incurs two additional pairing-energy contributions relative to P.
The simplified total-energy difference is:
\[
E_Q-E_P=-2\Delta_o+2P_{\mathrm{pair}}
\]
When \(\Delta_o\gt P_{\mathrm{pair}}\), this difference is negative and the low-spin arrangement is favoured.
\(\textbf{Combined inference:}\)
The greater crystal-field stabilisation of Q must be considered together with its pairing-energy cost.
Its remaining unpaired electron means it is paramagnetic rather than diamagnetic.
A low-spin classification does not necessarily imply complete electron pairing for a \(d^5\) ion.
217. In a spectroscopic study, the metal-to-CO pi back-donation is experimentally found to be stronger in carbonyl environment Q than in environment P. The CO ligands are otherwise compared under suitable similar conditions. Which simultaneous trend is consistent with the molecular-orbital description of metal carbonyl bonding?
ⓐ. The internal C–O bond becomes stronger, while the metal–carbon pi contribution decreases.
ⓑ. The internal C–O bond becomes weaker, while the metal–carbon pi-bonding contribution increases.
ⓒ. Both the internal C–O bond and the metal–carbon pi contribution must weaken.
ⓓ. Both the internal C–O bond and the metal–carbon pi contribution must strengthen equally.
Correct Answer: The internal C–O bond becomes weaker, while the metal–carbon pi-bonding contribution increases.
Explanation: Metal carbonyl bonding involves two complementary electron-donation interactions.
The ligand-to-metal sigma component arises from donation of an electron pair on CO to an appropriate vacant metal orbital.
The metal-to-ligand pi component involves donation from an occupied metal d orbital into a vacant antibonding orbital of CO.
These interactions act synergistically.
The question states that metal-to-CO pi back-donation is stronger in Q.
This means that the antibonding orbital of CO receives greater electron density through the pi interaction.
Greater occupancy of an antibonding orbital tends to weaken the internal bond associated with that orbital.
Therefore, the internal C–O bond is weakened relative to the situation with less back-donation.
At the same time, the metal-to-ligand pi interaction contributes to bonding between the metal and carbon.
Increasing this contribution tends to reinforce the metal–carbon bond through the pi-bonding component.
The two effects are not contradictory because they concern different bonds.
Electron density that weakens the internal C–O bond can participate in strengthening the metal–carbon interaction.
The overall bond strengths in a real compound may also depend on additional structural and electronic factors.
Nevertheless, the stated directional trends follow from the orbital interactions.
The interpretation is consistent with the synergic bonding model of metal carbonyls.
218. A transition-metal ion with configuration \(d^6\) forms an octahedral complex X. Experimental evidence establishes that X has four unpaired electrons. Another complex Y of the same ion is also octahedral and has no unpaired electrons. A student claims that Y must have a greater total number of d electrons because its orbitals contain more electron pairs. Which statement correctly evaluates the claim?
ⓐ. The claim is correct because every newly formed electron pair adds two electrons to the d subshell.
ⓑ. The claim is correct because high-spin and low-spin complexes of the same metal cannot have the same oxidation state.
ⓒ. The claim is incorrect only if the coordination numbers of X and Y are different.
ⓓ. The claim is incorrect because electron pairing redistributes the same six d electrons without changing their total number.
Correct Answer: The claim is incorrect because electron pairing redistributes the same six d electrons without changing their total number.
Explanation: Both complexes contain the same metal ion in the same oxidation state.
Therefore, both possess six metal d electrons.
Complex X has four unpaired electrons, which corresponds to the high-spin octahedral \(d^6\) configuration.
Its occupancy is \(t_{2g}^{4}e_g^{2}\).
The \(t_{2g}\) set contains one electron pair and two unpaired electrons.
The two \(e_g\) orbitals each contain one additional unpaired electron.
Complex Y has no unpaired electrons.
Its low-spin configuration is \(t_{2g}^{6}e_g^{0}\).
All six electrons occupy the three lower-energy orbitals as three pairs.
The change from X to Y involves redistribution and pairing of the existing electrons.
No additional electrons are required.
The energy balance between crystal-field splitting and pairing energy determines which arrangement is favoured.
A stronger ligand field can make additional pairing within \(t_{2g}\) energetically favourable.
The total d-electron count is fixed by the metal's identity and oxidation state, not by the number of electron pairs present.
Thus, the difference in magnetic behaviour does not establish a change in oxidation state or total electron count.
The student's argument confuses electron pairing with electron acquisition.
219. Four observations are reported for coordination compounds:
I. A neutral nickel carbonyl is tetrahedral and diamagnetic.
II. An octahedral chromium(III) complex has three unpaired electrons despite being an inner-orbital complex.
III. A low-spin octahedral iron(III) complex contains one unpaired electron.
IV. An octahedral \(d^{10}\) complex exhibits no net crystal-field stabilisation relative to the unsplit d-orbital barycentre.
Which statement correctly evaluates these observations?
ⓐ. Only I and II are consistent with the elementary bonding models.
ⓑ. Only I, II and III are consistent; IV is impossible because d orbitals cannot split when filled.
ⓒ. All four observations are consistent, but for different electronic reasons that must be evaluated separately.
ⓓ. Only II, III and IV are consistent; I is impossible because tetrahedral complexes must be paramagnetic.
Correct Answer: All four observations are consistent, but for different electronic reasons that must be evaluated separately.
Explanation: Observation I is consistent with tetracarbonylnickel(0), \(\mathrm{[Ni(CO)_4]}\).
Its neutral carbonyl ligands give nickel oxidation state zero.
The standard valence-bond description has a filled \(3d^{10}\) subshell and tetrahedral \(sp^3\) hybridisation.
All d electrons are paired, so diamagnetism is expected.
Observation II is consistent with an octahedral chromium(III) complex having configuration \(3d^3\).
Three electrons occupy separate d orbitals, while two inner d orbitals remain vacant.
These vacant orbitals can participate in \(d^2sp^3\) hybridisation without requiring the three electrons to pair.
The complex can therefore be inner orbital and paramagnetic simultaneously.
Observation III is consistent with a low-spin \(d^5\) iron(III) ion.
Its octahedral occupancy is \(t_{2g}^{5}e_g^{0}\).
Two pairs and one unpaired electron occupy the lower-energy orbital set.
Thus, low spin does not necessarily mean diamagnetic.
Observation IV is also consistent with elementary crystal-field theory.
For an octahedral \(d^{10}\) ion, the occupancy is \(t_{2g}^{6}e_g^{4}\).
The six lower-level electrons contribute \(-2.4\Delta_o\), while the four upper-level electrons contribute \(+2.4\Delta_o\).
These energy contributions cancel, even though the ligand field splits the orbitals.
The four observations therefore concern distinct ideas: geometry, inner-orbital availability, spin-state occupancy and net crystal-field energy.
No single rule based only on magnetism, hybridisation or d-electron count can replace the separate reasoning required in each case.
All four observations are scientifically consistent.
220. The coordination entity \(\mathrm{[Ti(H_2O)_6]^{3+}}\) is violet in aqueous solution. Its absorption spectrum contains a band in the blue-green region, near \(498\,\mathrm{nm}\). Which explanation correctly relates its electronic configuration, absorbed radiation and observed colour?
ⓐ. Titanium(III) has a \(d^0\) configuration, and the violet colour arises because violet light is absorbed.
ⓑ. Titanium(III) has a \(d^1\) configuration, and absorption promotes an electron from \(e_g\) to \(t_{2g}\).
ⓒ. Titanium(III) is \(d^1\); a \(t_{2g}\)-to-\(e_g\) excitation removes light complementary to the observed colour.
ⓓ. Titanium(III) has a \(d^2\) configuration, and the violet colour results from complete pairing of its d electrons.
Correct Answer: Titanium(III) is \(d^1\); a \(t_{2g}\)-to-\(e_g\) excitation removes light complementary to the observed colour.
Explanation: The six water molecules are neutral ligands, so titanium has oxidation state \(+3\).
Neutral titanium has the electronic configuration \(\mathrm{[Ar]\,3d^2 4s^2}\).
Removal of two \(4s\) electrons and one \(3d\) electron gives \(\mathrm{Ti^{3+}:[Ar]\,3d^1}\).
The six oxygen donor atoms produce an octahedral coordination environment.
In an octahedral crystal field, the \(t_{2g}\) orbitals lie below the \(e_g\) orbitals.
The single d electron occupies a \(t_{2g}\) orbital in the ground state.
When suitable radiation is absorbed, this electron can be promoted to the higher-energy \(e_g\) set.
The excitation requires energy corresponding to the orbital separation.
The specified blue-green radiation supplies the energy for this electronic transition.
White light passing through the solution consequently loses a portion of its blue-green components.
The transmitted light is perceived as the complementary colour, violet.
The observed violet colour must therefore not be confused with the colour of the radiation absorbed.
The example connects oxidation state, d-electron configuration, crystal-field excitation and complementary-colour interpretation.
It also demonstrates why colour is an electronic property that the elementary valence-bond approach does not satisfactorily explain.
