301. A hypothetical platinum(II) coordination entity contains one ammonia ligand, one bromido ligand, one chlorido ligand and one nitrite ligand attached through oxygen. All four ligands are unidentate, and the entity is square planar. Which systematic name and formal charge are correct?
ⓐ. Amminebromidochloridonitrito-O-platinate(II), charge \(-1\)
ⓑ. Amminebromidochloridonitrito-N-platinate(II), charge \(-1\)
ⓒ. Bromidochloridoamminenitrito-O-platinum(II), charge \(+1\)
ⓓ. Amminebromidochloridonitrito-O-platinum(II), charge \(0\)
Correct Answer: Amminebromidochloridonitrito-O-platinate(II), charge \(-1\)
Explanation: \(\textbf{Formal-charge reconstruction:}\)
Platinum is in oxidation state \(+2\).
Ammonia is neutral.
Bromido and chlorido each contribute charge \(-1\).
The nitrite ligand also carries charge \(-1\), regardless of whether it coordinates through nitrogen or oxygen.
Therefore:
\[
q=(+2)+0+(-1)+(-1)+(-1)
\]
\[
q=-1
\]
The coordination entity is anionic.
Consequently, the metal takes the name platinate rather than platinum.
\(\textbf{Ligand-name ordering:}\)
Ammonia is named ammine.
The other ligand names are bromido, chlorido and nitrito-O.
The names are arranged alphabetically as ammine, bromido, chlorido and nitrito-O.
Each ligand occurs once, so no numerical multiplicative prefix is required.
\(\textbf{Donor-atom specification:}\)
The nitrite ligand is explicitly oxygen-bound.
Its name must therefore contain the donor-atom specification O.
Using nitrito-N would describe a different metal–ligand connectivity.
The specified square-planar geometry is consistent with four unidentate donor attachments.
\(\textbf{Complete nomenclature:}\)
The correct name is amminebromidochloridonitrito-O-platinate(II).
The Roman numeral denotes platinum's oxidation state \(+2\), not the coordination-entity charge.
The negative entity charge and the positive metal oxidation state are therefore compatible.
An external monovalent cation would be needed if this entity were isolated as an ordinary neutral salt.
302. Two formally related octahedral iron coordination entities are:
P: \(\mathrm{[Fe(NH_3)_6]^{2+}}\)
Q: \(\mathrm{[Fe(NH_3)_4(CN)_2]}\)
Both contain iron in oxidation state \(+2\). Q is obtained by replacing exactly two coordinated ammonia ligands in P with two cyanido ligands. Which statement correctly identifies the change in nomenclature and charge?
ⓐ. P is hexaammineferrate(II), whereas Q is tetraamminedicyanidoferrate(II); both entities are anionic.
ⓑ. P is hexaammineiron(II), whereas Q is tetraamminedicyanidoiron(II); the entity charge changes from \(+2\) to zero while coordination number remains six.
ⓒ. P is hexaammineiron(II), whereas Q is tetraamminedicyanidoiron(IV); the entity charge remains \(+2\).
ⓓ. P is hexaammineiron(II), whereas Q is tetraamminedicyanidoiron(II); the coordination number decreases from six to four.
Correct Answer: P is hexaammineiron(II), whereas Q is tetraamminedicyanidoiron(II); the entity charge changes from \(+2\) to zero while coordination number remains six.
Explanation: P contains six neutral ammonia ligands.
Iron has oxidation state \(+2\).
The coordination entity therefore carries charge \(+2\).
Since it is cationic, the metal is named iron rather than ferrate.
The name of P is hexaammineiron(II).
In Q, four ammonia ligands remain coordinated.
Two cyanido ligands replace the other two ammonia molecules.
Every ligand is unidentate, so Q still has coordination number six.
However, the two cyanido ligands contribute a combined charge of \(-2\).
The charge of Q is:
\[
q_Q=(+2)+4(0)+2(-1)
\]
\[
q_Q=0
\]
Q is therefore a neutral coordination entity.
The metal name remains iron because the -ate suffix is reserved for anionic coordination entities.
The ligand names are ammine and cyanido.
Alphabetically, ammine precedes cyanido.
Four ammonia ligands require tetraammine, while two cyanido ligands require dicyanido.
The resulting name is tetraamminedicyanidoiron(II).
The formal oxidation state remains \(+2\), even though the overall entity charge has changed.
The example demonstrates that replacing neutral ligands with anionic ligands can change the charge classification without changing metal oxidation state or coordination number.
303. An octahedral cobalt(III) entity contains four ammonia molecules, one nitrite ligand and one thiocyanate ligand. The nitrite and thiocyanate ligands are each monodentate in every permitted structure. Nitrite can attach through N or O, while thiocyanate can attach through N or S. Suppose their relative positions are held fixed throughout and only donor-atom connectivity is allowed to change. How many linkage arrangements are possible, and what is the formal charge on each entity?
ⓐ. Two linkage arrangements, each carrying charge \(+2\)
ⓑ. Three linkage arrangements, each carrying charge \(+1\)
ⓒ. Four linkage arrangements, each carrying charge \(+1\)
ⓓ. Six linkage arrangements, each carrying charge \(+3\)
Correct Answer: Four linkage arrangements, each carrying charge \(+1\)
Explanation: \(\textbf{Formal-charge calculation:}\)
Cobalt is in oxidation state \(+3\).
The four ammonia molecules are neutral.
Nitrite contributes charge \(-1\).
Thiocyanate also contributes charge \(-1\).
Thus:
\[
q=(+3)+4(0)+(-1)+(-1)
\]
\[
q=+1
\]
Every allowed donor-atom attachment preserves these formal ligand charges.
Therefore, each linkage arrangement has charge \(+1\).
\(\textbf{Nitrite linkage choices:}\)
Nitrite can coordinate through nitrogen or oxygen.
Under the stated monodentate condition, these are alternative donor atoms.
Thus, nitrite has two linkage choices.
\(\textbf{Thiocyanate linkage choices:}\)
Thiocyanate can coordinate through nitrogen or sulphur.
It therefore provides two additional independent linkage choices.
\(\textbf{Combination count:}\)
The relative ligand positions are fixed, so geometrical arrangements are not being counted.
The two donor-atom choices are independent.
Hence:
\[
N_{\mathrm{linkage}}=2\times2
\]
\[
N_{\mathrm{linkage}}=4
\]
\(\textbf{Coordination check:}\)
Four ammonia ligands contribute four donor atoms.
Nitrite and thiocyanate each contribute one.
The coordination number remains six in every arrangement.
The four structures retain the same overall elemental composition and cobalt oxidation state.
They differ only in which atom of each ambidentate ligand is attached to cobalt.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{linkage}}=4,\quad q=+1}
\]
304. Two octahedral cobalt(III) salts have formulae:
P: \(\mathrm{[Co(NH_3)_4Cl_2]Cl}\)
Q: \(\mathrm{[Co(NH_3)_4Cl(H_2O)]Cl_2}\)
Which statement correctly compares their systematic ligand names, ionisation behaviour and empirical compositions?
ⓐ. They are hydrate isomers because Q contains coordinated water, and both release one chloride ion per formula unit.
ⓑ. They are geometrical isomers because each has four coordinated ammonia molecules and a six-coordinate cobalt centre.
ⓒ. They are coordination isomers because chloride exchanges between two different complex metal centres.
ⓓ. They are not isomers: P has two coordinated chlorides, Q one aqua and one chloride; counter-ion counts also differ.
Correct Answer: They are not isomers: P has two coordinated chlorides, Q one aqua and one chloride; counter-ion counts also differ.
Explanation: \(\textbf{Coordination structure of P:}\)
P contains four ammonia ligands and two coordinated chloride ligands.
The cobalt oxidation state is \(+3\).
Its complex cation therefore has charge \(+1\).
One chloride counter ion balances this charge.
The corresponding name is tetraamminedichloridocobalt(III) chloride.
\(\textbf{Coordination structure of Q:}\)
Q contains four ammonia ligands, one aqua ligand and one coordinated chlorido ligand.
All six ligands are unidentate.
The cobalt coordination number consequently remains six.
The coordinated chloride contributes charge \(-1\).
The complex cation therefore has charge \(+2\), requiring two external chloride ions.
Alphabetical ligand naming gives ammine, aqua and chlorido.
The name is tetraammineaquachloridocobalt(III) chloride.
\(\textbf{Isomerism test:}\)
The two compounds do not have identical complete elemental compositions.
Q contains an additional oxygen and hydrogen contribution from coordinated water.
It also has a different total ammonia–chloride–water composition from P.
Therefore, they cannot be isomers of each other.
A difference in coordination-sphere composition alone is not sufficient to establish hydrate isomerism.
Hydrate isomers must preserve the same overall composition while redistributing water between coordinated and lattice positions.
\(\textbf{Ionisation consequence:}\)
P produces one external chloride ion per formula unit.
Q produces two.
Their ideal ionisation behaviours therefore differ.
This difference follows directly from the coordination-entity charges and must not be interpreted as proof that the compounds are structural isomers.
305. A coordination compound is named potassium diamminebis(oxalato)ferrate(III). Every oxalato ligand is didentate. A student writes the formula \(\mathrm{K[Fe(NH_3)_2(C_2O_4)_2]}\). Which correction is required?
ⓐ. The formula is correct because two ammonia ligands neutralise two oxalato ligands.
ⓑ. Three potassium ions are required because the central metal is \(\mathrm{Fe^{3+}}\).
ⓒ. Two potassium ions balance the \(-2\) complex charge, whose coordination number is six.
ⓓ. One potassium ion is correct, but the coordination number is four rather than six.
Correct Answer: Two potassium ions balance the \(-2\) complex charge, whose coordination number is six.
Explanation: \(\textbf{Identifying formal ligand charges:}\)
Iron is in oxidation state \(+3\).
The two ammonia ligands are neutral.
Each oxalato ligand carries charge \(-2\).
The two oxalato ligands therefore contribute total charge \(-4\).
\(\textbf{Coordination-entity charge:}\)
\[
q=(+3)+2(0)+2(-2)
\]
\[
q=+3-4
\]
\[
q=-1
\]
The coordination anion therefore carries charge \(-1\).
One potassium ion is sufficient to balance it.
The student's formula is consequently already charge-balanced.
\(\textbf{Coordination-number check:}\)
Each ammonia ligand contributes one nitrogen donor atom.
Each didentate oxalato ligand contributes two oxygen donor atoms.
Thus:
\[
CN=2(1)+2(2)
\]
\[
CN=6
\]
\(\textbf{Nomenclature check:}\)
Because the entity is anionic, the metal name ferrate is appropriate.
The two ammonia ligands are represented by diammine.
The two oxalato ligands are represented by bis(oxalato) if the provided name explicitly uses that multiplicative form.
The Roman numeral (III) correctly describes the iron oxidation state.
The number of potassium counter ions is one, not two or three.
The distinction is that coordination number six does not imply an entity charge of \(-2\) or \(+6\).
The formula and the stated ligand inventory are consistent.
\(\textbf{Final answer:}\)
\[
\mathrm{K[Fe(NH_3)_2(C_2O_4)_2]},\quad CN=6
\]
306. Two proposed names are assigned to the coordination entity \(\mathrm{[Pt(NH_3)_2Cl_2]}\):
Name P: diamminedichloridoplatinum(II)
Name Q: diamminedichloridoplatinate(II)
The entity is neutral and can have cis and trans square-planar geometrical forms. Which statement correctly evaluates the names and the stereochemical information?
ⓐ. P is the appropriate base name; Q misuses the -ate suffix, and a specific cis/trans form needs identification.
ⓑ. Q is appropriate because any compound containing anionic chloride ligands must use the suffix -ate.
ⓒ. Both P and Q are equally correct because platinum is in oxidation state \(+2\).
ⓓ. P is incorrect because a neutral coordination entity cannot display geometrical isomerism.
Correct Answer: P is the appropriate base name; Q misuses the -ate suffix, and a specific cis/trans form needs identification.
Explanation: The coordination entity contains platinum in oxidation state \(+2\).
Two ammonia ligands are neutral.
Two chlorido ligands contribute combined charge \(-2\).
The complete coordination entity therefore has charge zero.
In systematic coordination nomenclature, a neutral complex is named using the ordinary metal name.
The anionic suffix -ate is not required.
Thus, platinum is correct, while platinate would incorrectly imply an anionic coordination entity.
The ligand names ammine and chlorido are arranged alphabetically.
Two of each ligand give diammine and dichlorido.
The base name is consequently diamminedichloridoplatinum(II).
The complex has four unidentate donor attachments.
Its square-planar arrangement permits the two chloride ligands to occupy adjacent or opposite positions.
These arrangements are respectively cis and trans.
Both geometrical isomers have the same formula, ligand names and platinum oxidation state.
The positional descriptor must therefore be supplied when the name is intended to identify one particular geometrical isomer.
The presence of cis/trans isomerism does not require the entity to be electrically charged.
The case distinguishes coordination-entity charge, metal-name suffix and stereochemical specification.
307. In a hypothetical synthesis, two coordinated chlorido ligands in \(\mathrm{[Pt(NH_3)_2Cl_2]}\) are replaced by two neutral aqua ligands without changing platinum's oxidation state. The product is isolated as a nitrate salt. Which formula and systematic name are correct?
ⓐ. \(\mathrm{[Pt(NH_3)_2(H_2O)_2](NO_3)_2}\); diamminediaquaplatinum(II) nitrate
ⓑ. \(\mathrm{[Pt(NH_3)_2(H_2O)_2]NO_3}\); diamminediaquaplatinum(I) nitrate
ⓒ. \(\mathrm{[Pt(NH_3)_2(H_2O)_2](NO_3)_4}\); diamminediaquaplatinum(IV) nitrate
ⓓ. \(\mathrm{[Pt(NH_3)_2(H_2O)_2](NO_3)_2}\); diaquadiammineplatinate(II) nitrate
Correct Answer: \(\mathrm{[Pt(NH_3)_2(H_2O)_2](NO_3)_2}\); diamminediaquaplatinum(II) nitrate
Explanation: \(\textbf{Original coordination entity:}\)
Platinum is in oxidation state \(+2\).
Two chlorido ligands contribute a total charge of \(-2\).
The two ammonia ligands are neutral.
The starting entity is therefore neutral.
\(\textbf{Ligand-substitution change:}\)
Both coordinated chlorido ligands are replaced by neutral aqua ligands.
The product contains two ammonia and two water ligands.
All four ligands are unidentate.
The coordination number remains four.
\(\textbf{Product charge:}\)
No anionic ligand remains coordinated.
Platinum retains oxidation state \(+2\).
Therefore:
\[
q_{\mathrm{product}}=(+2)+2(0)+2(0)
\]
\[
q_{\mathrm{product}}=+2
\]
The product is a complex cation.
\(\textbf{Nitrate requirement:}\)
Each nitrate counter ion has charge \(-1\).
Two nitrate ions are needed to balance the \(+2\) complex cation.
The complete salt formula is:
\[
\mathrm{[Pt(NH_3)_2(H_2O)_2](NO_3)_2}
\]
\(\textbf{Naming the coordination cation:}\)
Ammonia is named ammine.
Water is named aqua.
Alphabetically, ammine precedes aqua.
Two ligands of each type give diammine and diaqua.
The metal is named platinum because its coordination entity is cationic.
The oxidation state is indicated as (II).
The complete name is diamminediaquaplatinum(II) nitrate.
The substitution changes formal entity charge because two anionic ligands have been replaced by neutral ligands.
It does not, by itself, change the platinum oxidation state.
308. A salt is reported to have the systematic name tris(ethane-1,2-diamine)chromium(III) phosphate. A student writes \(\mathrm{[Cr(en)_3]_3(PO_4)}\). Which statement identifies the necessary correction and the underlying reasoning?
ⓐ. The student's formula is correct because the prefix tris requires three coordination cations.
ⓑ. The formula should be \(\mathrm{[Cr(en)_3]PO_4}\), because the complex cation and phosphate ion each carry charge of magnitude three.
ⓒ. The formula should be \(\mathrm{[Cr(en)_3](PO_4)_3}\), because chromium is trivalent.
ⓓ. The formula should be \(\mathrm{Cr(en)_3PO_4}\), because the coordination brackets are unnecessary when every ligand is neutral.
Correct Answer: The formula should be \(\mathrm{[Cr(en)_3]PO_4}\), because the complex cation and phosphate ion each carry charge of magnitude three.
Explanation: The name specifies one coordination entity containing chromium and three en ligands.
The prefix tris applies to the number of en ligands within that entity.
It does not indicate three separate complex cations in the complete salt.
Each en ligand is neutral and didentate.
Chromium has oxidation state \(+3\).
The charge on the coordination entity is therefore \(+3\).
The corresponding coordination cation is \(\mathrm{[Cr(en)_3]^{3+}}\).
Phosphate carries charge \(-3\).
One phosphate ion is sufficient to balance one complex cation.
The neutral salt formula is consequently \(\mathrm{[Cr(en)_3]PO_4}\).
The student's proposed formula contains three complex cations but only one phosphate ion.
Its formal charges would not balance.
The three en ligands contribute six nitrogen donor atoms, giving coordination number six.
This coordination-number calculation is independent of the number of phosphate counter ions.
The brackets must be retained because they identify the ligands belonging to the coordination sphere.
The case illustrates why a multiplicative ligand prefix must not be transferred to the stoichiometric ratio of complex ions and counter ions.
The correct cation-to-anion ratio follows from electrical neutrality, not from the prefix used inside the coordination name.
309. An octahedral complex contains cobalt(III), four ammonia ligands and two different anionic ligands, chlorido and bromido. Each ligand is unidentate. A student writes the name tetraamminebromidochloridocobalt(III) and argues that the compound must have exactly two geometrical forms because the two halido ligands can be cis or trans. Which evaluation is correct?
ⓐ. The name is alphabetically ordered; adjacent and opposite halido arrangements follow from the four identical ammine ligands.
ⓑ. The name is alphabetically incorrect because chlorido must precede bromido, and the entity cannot be octahedral.
ⓒ. The name is alphabetically ordered, and exactly three geometrical forms arise because the two halido ligands are different.
ⓓ. The entity must be anionic because it contains bromido and chlorido ligands, so its metal name must be cobaltate.
Correct Answer: The name is alphabetically ordered; adjacent and opposite halido arrangements follow from the four identical ammine ligands.
Explanation: The complex contains four ammonia ligands, one bromido ligand and one chlorido ligand.
These six unidentate ligands give octahedral coordination number six.
The ligand names are ammine, bromido and chlorido.
Their alphabetical order is therefore ammine, bromido and chlorido.
The naming sequence tetraamminebromidochlorido is correct.
Cobalt has oxidation state \(+3\).
The bromido and chlorido ligands each contribute charge \(-1\).
The coordination entity consequently carries charge \(+1\).
The cationic entity must retain the metal name cobalt, not cobaltate.
The geometrical arrangement requires more careful interpretation.
The four ammonia ligands are identical.
Only the relative positions of bromido and chlorido distinguish the geometrical alternatives.
They may occupy adjacent positions or opposite positions in the octahedron.
These are distinct arrangements.
However, the mere existence of two different halido ligands is not by itself the complete argument; the equivalence of the four ammonia positions and overall rotations must also be considered.
The resulting octahedral \(\mathrm{MA_4BC}\) pattern has two geometrical classes under the stated assumptions.
The key structural distinction is whether the B and C donor positions are cis or trans.
The full reasoning therefore combines ligand alphabetisation, charge balance and symmetry of the octahedral arrangement.
310. A cobalt(III) coordination entity has formula \(\mathrm{[Co(NH_3)_3Cl_2(H_2O)]^+}\). All ligands are unidentate. Which statement correctly identifies the number of coordinated donor atoms, formal ligand charge and an appropriate alphabetical order for the ligand names?
ⓐ. Six donor atoms; total ligand charge \(-2\); ammine, aqua, chlorido
ⓑ. Five donor atoms; total ligand charge \(-2\); ammine, chlorido, aqua
ⓒ. Six donor atoms; total ligand charge \(-1\); chlorido, ammine, aqua
ⓓ. Six donor atoms; total ligand charge \(-2\); aqua, ammine, chlorido
Correct Answer: Six donor atoms; total ligand charge \(-2\); ammine, aqua, chlorido
Explanation: \(\textbf{Coordination-number determination:}\)
The complex contains three ammonia ligands.
It also contains two chlorido ligands and one aqua ligand.
All six ligands are unidentate.
Therefore, six donor atoms are directly attached to cobalt.
The coordination number is six.
\(\textbf{Formal ligand charges:}\)
Ammonia is neutral.
Water is neutral.
Each chlorido ligand contributes charge \(-1\).
The total ligand charge is consequently \(-2\).
Cobalt(III) contributes charge \(+3\).
The complete coordination entity therefore has charge \(+1\), consistent with the stated formula.
\(\textbf{Alphabetical ligand-name order:}\)
Ammonia is named ammine.
Water is named aqua.
Coordinated chloride is named chlorido.
Alphabetisation is based on these ligand names rather than on their molecular formulae.
Ammine precedes aqua because the third letters distinguish their alphabetical positions.
Chlorido follows both.
The appropriate name sequence is therefore ammine, aqua, chlorido.
With multiplicative prefixes, the coordination-entity name is triamminediaquadichloridocobalt(III).
The cationic charge requires the ordinary metal name cobalt.
The example separates alphabetical naming order from the ligand order displayed in a chemical formula.
It also shows that total coordinated ligand charge, metal oxidation state and entity charge are different quantities.
311. A coordination compound has the formula \(\mathrm{[Co(NH_3)_2(en)_2](NO_3)_3}\). Each en ligand uses both nitrogen atoms. Two claims are made:
Claim P: The correct name must use the expression bis(ethane-1,2-diamine), because the ligand name contains a numerical descriptor.
Claim Q: The name must include the prefix tri before nitrate, because three nitrate counter ions occur outside the coordination sphere.
Which evaluation is correct?
ⓐ. Both P and Q are correct.
ⓑ. P is incorrect, but Q is correct.
ⓒ. Both P and Q are incorrect.
ⓓ. P is correct, but Q is incorrect.
Correct Answer: P is correct, but Q is incorrect.
Explanation: The coordination cation contains two ammonia ligands and two en ligands.
Each ammonia ligand is neutral and unidentate.
Each en ligand is neutral and didentate.
The total coordination number is six.
The nitrate ions occur outside the coordination sphere.
They do not contribute to the cobalt coordination number.
Three nitrate ions each carry charge \(-1\).
The complex cation must therefore carry charge \(+3\).
Because all its ligands are neutral, cobalt has oxidation state \(+3\).
Claim P is correct.
The systematic ligand name ethane-1,2-diamine contains numerical descriptors.
When two such ligand units are present, the multiplicative form bis(ethane-1,2-diamine) is appropriate.
The two ammonia ligands are indicated by diammine.
The ligand names are ordered alphabetically as ammine followed by ethane-1,2-diamine.
Claim Q is incorrect.
The number of separate cations and anions in an ionic coordination compound is not normally expressed by adding a numerical prefix to the counter-ion name.
The three nitrate ions are required by charge balance, but the salt is simply named nitrate at the end.
The complete name is diamminobis(ethane-1,2-diamine)cobalt(III) nitrate.
The distinction illustrates how ligand multiplicative prefixes differ from the stoichiometric numbers used when writing an electrically neutral salt formula.
312. The neutral ligand tris(2-aminoethyl)amine, represented by \(\mathrm{N(CH_2CH_2NH_2)_3}\) and abbreviated as tren, contains four nitrogen donor atoms. In a hypothetical cobalt(III) complex, tren coordinates through all four nitrogen atoms, while two chloride ions occupy the remaining positions of an octahedral coordination sphere. Which statement correctly describes the resulting entity?
ⓐ. It has coordination number four and charge \(+1\), because tren is one ligand molecule.
ⓑ. It has coordination number six and charge \(+3\), because all four nitrogen donors are neutral.
ⓒ. It has coordination number six and charge \(+1\), with one tetradentate ligand and two unidentate ligands.
ⓓ. It has coordination number eight and charge \(+1\), because the four nitrogen atoms and four carbon atoms of tren are donor atoms.
Correct Answer: It has coordination number six and charge \(+1\), with one tetradentate ligand and two unidentate ligands.
Explanation: \(\textbf{Actual donor-atom count:}\)
The ligand tren contains one central nitrogen and three terminal amino nitrogen atoms.
All four nitrogen atoms are specified to coordinate to the same cobalt centre.
Therefore, tren acts as a tetradentate ligand in this structure.
The two coordinated chloride ions each contribute one additional donor atom.
The coordination number is:
\[
CN=4+2(1)
\]
\[
CN=6
\]
This is consistent with the stated octahedral coordination environment.
\(\textbf{Formal-charge calculation:}\)
Cobalt has oxidation state \(+3\).
Tren is neutral, whereas each chlorido ligand carries charge \(-1\).
Thus:
\[
q=(+3)+0+2(-1)
\]
\[
q=+1
\]
The resulting coordination entity is \(\mathrm{[Co(tren)Cl_2]^+}\).
\(\textbf{Structural interpretation:}\)
The entity contains only three ligand molecules or ions, but these provide six donor attachments.
The four nitrogen atoms of tren are part of one ligand rather than four independent ammonia molecules.
The carbon atoms forming the ligand framework do not become additional donor atoms merely because they are present in the molecule.
The ligand is polydentate and forms a chelating attachment around cobalt.
The result distinguishes ligand-molecule count, actual denticity, coordination number and coordination-entity charge.
\(\textbf{Final answer:}\)
\[
CN=6,\quad q=+1
\]
313. A hypothetical dinuclear coordination cation contains two cobalt(III) centres, eight terminal ammonia ligands and two chloride ligands that bridge the cobalt centres. Each cobalt is attached to four ammonia nitrogen atoms and to both bridging chloride ions. Every bridging chloride forms one donor attachment to each cobalt, and there are no other metal–ligand attachments. Which statement is correct?
ⓐ. Each cobalt has coordination number six, the complete entity has charge \(+4\), and the two bridging chlorides provide four metal–chloride donor attachments in total.
ⓑ. Each cobalt has coordination number five, the complete entity has charge \(+4\), and the two bridging chlorides provide two donor attachments in total.
ⓒ. Each cobalt has coordination number six, the complete entity has charge \(+2\), and each bridging chloride is counted as a neutral ligand.
ⓓ. Each cobalt has coordination number ten, the complete entity has charge \(+4\), and every ammonia ligand coordinates to both metals.
Correct Answer: Each cobalt has coordination number six, the complete entity has charge \(+4\), and the two bridging chlorides provide four metal–chloride donor attachments in total.
Explanation: \(\textbf{Coordination around one cobalt centre:}\)
Each cobalt is directly attached to four ammonia nitrogen atoms.
Both bridging chloride ions also attach to that same cobalt.
Therefore, the coordination number of either metal is:
\[
CN_{\mathrm{Co}}=4+2
\]
\[
CN_{\mathrm{Co}}=6
\]
The specified local coordination is consequently compatible with octahedral geometry.
\(\textbf{Counting attachments across the complete entity:}\)
Eight terminal ammonia ligands provide eight metal–nitrogen donor attachments.
Each bridging chloride attaches to two different cobalt atoms.
The two bridges therefore provide:
\[
N_{\mathrm{Co-Cl}}=2(2)=4
\]
The complete dinuclear entity contains twelve metal–ligand donor attachments.
This agrees with two cobalt centres, each having coordination number six.
\(\textbf{Entity-charge calculation:}\)
The two cobalt(III) ions contribute:
\[
q_{\mathrm{metals}}=2(+3)=+6
\]
Ammonia is neutral.
The two chloride ligands contribute charge \(-2\) altogether.
Hence:
\[
q_{\mathrm{entity}}=+6-2
\]
\[
q_{\mathrm{entity}}=+4
\]
\(\textbf{Counting distinction:}\)
Each bridging chloride is one ligand ion in the complete formula.
However, it contributes one coordination attachment to each of two metals.
This does not double its formal electrical charge.
The ligand count, metal–ligand attachment count and formal-charge count are therefore different quantities.
\(\textbf{Final answer:}\)
\[
CN_{\mathrm{each\ Co}}=6,\quad q_{\mathrm{entity}}=+4,\quad N_{\mathrm{Co-Cl}}=4
\]
314. A hypothetical ligand L contains four suitable donor atoms. Two donor atoms of L coordinate simultaneously to metal centre M₁, while the other two coordinate simultaneously to a different metal centre M₂. The two attachments to each metal form a metal-containing ring. M₁ also has four terminal ammonia ligands, while M₂ has two didentate en ligands. Which interpretation is correct?
ⓐ. L is only bridging and cannot be chelating, because its donor atoms coordinate to different metals.
ⓑ. L is chelating but not bridging, because every pair of donor atoms forms a ring.
ⓒ. M₁ has coordination number six, whereas M₂ has coordination number eight because en is didentate.
ⓓ. L acts as both a bridging ligand and a chelating ligand in the specified arrangement, and each metal has coordination number six.
Correct Answer: L acts as both a bridging ligand and a chelating ligand in the specified arrangement, and each metal has coordination number six.
Explanation: The classification of a ligand depends on its actual metal–donor connectivity.
A bridging ligand connects two or more metal centres within a coordination structure.
L connects M₁ and M₂ because donor atoms belonging to the same ligand are attached to both metals.
It therefore has a bridging role.
Chelation requires multiple donor atoms of one ligand to coordinate simultaneously to the same metal centre and form a metal-containing ring.
The question explicitly states that two donor atoms of L attach to M₁ and form such a ring.
The other two donor atoms similarly attach to M₂ and form another ring.
Therefore, L also acts as a chelating ligand at both metal centres.
These classifications are not mutually exclusive when the connectivity permits both.
For M₁, four ammonia ligands contribute four donor atoms.
The two donor atoms supplied by L raise its coordination number to six.
For M₂, two en ligands contribute four nitrogen donor atoms altogether.
L provides two additional donor attachments to M₂.
Its coordination number is consequently also six.
The complete structure contains one ligand L that connects two metals, but its four attachments must be distributed between them.
Counting all four L donor atoms towards the coordination number of either one metal would be incorrect.
The example extends the distinction between bridging and chelation by showing that both behaviours can coexist in one explicitly defined attachment arrangement.
315. A solution initially contains \(0.080\,\mathrm{mol}\) of \(\mathrm{[M(H_2O)_6]^{2+}}\). Exactly \(0.120\,\mathrm{mol}\) of en is incorporated into its coordination spheres. The only final metal-containing species are \(\mathrm{[M(en)(H_2O)_4]^{2+}}\) and \(\mathrm{[M(en)_2(H_2O)_2]^{2+}}\), and all initial metal centres are converted into one of these two species. Every en ligand replaces two coordinated water molecules. What are the amounts of the two product species and the total amount of coordinated water released?
ⓐ. \(0.060,\ 0.020,\ 0.20\,\mathrm{mol}\), respectively
ⓑ. \(0.040,\ 0.040,\ 0.24\,\mathrm{mol}\), respectively
ⓒ. \(0.020,\ 0.060,\ 0.28\,\mathrm{mol}\), respectively
ⓓ. \(0.040,\ 0.040,\ 0.16\,\mathrm{mol}\), respectively
Correct Answer: \(0.040,\ 0.040,\ 0.24\,\mathrm{mol}\), respectively
Explanation: \(\textbf{Metal-centre conservation:}\)
Let \(x\) be the amount of \(\mathrm{[M(en)(H_2O)_4]^{2+}}\).
Let \(y\) be the amount of \(\mathrm{[M(en)_2(H_2O)_2]^{2+}}\).
Each product entity contains one metal centre.
Therefore:
\[
x+y=0.080
\]
\(\textbf{Incorporated-en balance:}\)
The first product contains one en ligand per metal.
The second contains two en ligands per metal.
All \(0.120\,\mathrm{mol}\) of en is incorporated.
Thus:
\[
x+2y=0.120
\]
Subtracting the metal balance:
\[
(x+2y)-(x+y)=0.120-0.080
\]
\[
y=0.040\,\mathrm{mol}
\]
Substitution gives:
\[
x=0.080-0.040
\]
\[
x=0.040\,\mathrm{mol}
\]
\(\textbf{Water-release balance:}\)
Each mole of the first product is formed by replacing two aqua ligands.
Each mole of the second product is formed by replacing four aqua ligands.
Therefore:
\[
n_{\mathrm{H_2O,released}}=2x+4y
\]
\[
n_{\mathrm{H_2O,released}}=2(0.040)+4(0.040)
\]
\[
n_{\mathrm{H_2O,released}}=0.080+0.160
\]
\[
n_{\mathrm{H_2O,released}}=0.240\,\mathrm{mol}
\]
\(\textbf{Independent donor-site check:}\)
Each mole of en supplies two donor attachments.
The \(0.120\,\mathrm{mol}\) of incorporated en therefore occupies \(0.240\,\mathrm{mol}\) of donor sites.
The same amount of coordinated water must have been displaced.
The resulting species retain coordination number six and charge \(+2\).
The calculation therefore satisfies metal conservation, en conservation and donor-site conservation simultaneously.
\(\textbf{Final answer:}\)
\[
\boxed{x=0.040\,\mathrm{mol},\quad y=0.040\,\mathrm{mol},\quad n_{\mathrm{H_2O}}=0.240\,\mathrm{mol}}
\]
316. The coordination cation \(\mathrm{[Co(NH_3)_6]^{3+}}\) undergoes three hypothetical changes, each considered separately:
Change I: Its three external chloride counter ions are replaced by three nitrate counter ions, without disturbing the coordination sphere.
Change II: One coordinated ammonia ligand is replaced by one water molecule.
Change III: The cobalt oxidation state changes from \(+3\) to \(+2\), while all six coordinated ammonia ligands remain attached.
Which statement correctly identifies the changes that alter the homoleptic or heteroleptic classification of the coordination entity?
ⓐ. Only Change I alters the classification, because the complete salt contains different counter ions.
ⓑ. Changes I and III alter the classification, because they change charge balance.
ⓒ. Only Change II alters the classification, because it introduces a second type of coordinated ligand.
ⓓ. All three changes alter the classification, because each changes some aspect of the coordination compound.
Correct Answer: Only Change II alters the classification, because it introduces a second type of coordinated ligand.
Explanation: Homoleptic and heteroleptic classifications refer to the ligand groups coordinated to the central metal.
They are not assigned merely from the complete elemental composition of the salt.
Initially, all six cobalt donor attachments arise from ammonia ligands.
The coordination entity is therefore homoleptic.
In Change I, only external counter ions are replaced.
Neither the coordinated ligand identity nor the number of ammonia attachments changes.
The coordination cation consequently remains homoleptic.
In Change II, one coordinated ammonia molecule is replaced by water.
The new entity contains five ammonia ligands and one aqua ligand.
Two different coordinated ligand types are now present.
The coordination entity is therefore heteroleptic.
Both ammonia and water are unidentate, so the coordination number remains six.
In Change III, cobalt changes oxidation state, but all six ammonia ligands remain attached.
The coordination-entity charge changes because ammonia is neutral.
However, the entity still contains only one ligand type.
It consequently remains homoleptic.
The question demonstrates that counter-ion identity, metal oxidation state and ligand diversity are distinct classification variables.
Only a change in the coordinated ligand types necessarily changes the homoleptic or heteroleptic designation in these stated cases.
317. The neutral molecules \(\mathrm{NH_3}\), \(\mathrm{H_2O}\) and CO can all behave as unidentate ligands, even though their electronic structures differ. A chemist prepares three complexes containing only one of these ligand species each. Which statement correctly identifies the Lewis acid–base roles and a consequence for coordination-number counting?
ⓐ. All three ligands donate a lone pair to a Lewis-acid metal; any additional metal–CO π bonding does not increase coordination number.
ⓑ. Each ligand accepts an electron pair from the metal, so the metal acts exclusively as a Lewis base.
ⓒ. Ammonia and water are Lewis bases, but carbon monoxide cannot be a ligand because it is electrically neutral.
ⓓ. Each ligand contributes two coordination sites because an electron pair contains two electrons.
Correct Answer: All three ligands donate a lone pair to a Lewis-acid metal; any additional metal–CO π bonding does not increase coordination number.
Explanation: A coordination bond is described as donation of an electron pair from a ligand to an appropriate acceptor orbital associated with the central metal.
In this Lewis acid–base description, the ligand acts as the electron-pair donor or Lewis base.
The central metal acts as the electron-pair acceptor or Lewis acid.
Ammonia ordinarily coordinates through nitrogen.
Water ordinarily coordinates through oxygen.
Carbon monoxide ordinarily coordinates through carbon in the standard metal carbonyl examples.
All three ligands are neutral, but neutrality does not prevent them from donating electron pairs.
A unidentate ligand makes one donor attachment to a particular metal centre.
The fact that a donated pair contains two electrons does not make it two separate donor atoms.
Carbon monoxide can also participate in metal-to-ligand pi back-donation.
This additional interaction affects bonding but is not counted as another independent coordination site.
The conventional coordination number is determined from the directly bonded ligand donor atoms.
Consequently, the same formal coordination-number rule applies to ordinary ammine, aqua and terminal carbonyl ligands.
The description does not require the three ligands to produce identical bond strengths, geometries or ligand-field splittings.
Their common unidentate character refers specifically to their actual donor-atom attachments.
318. A solution initially contains \(0.060\,\mathrm{mol}\) of \(\mathrm{Fe^{3+}}\) and \(0.100\,\mathrm{mol}\) of \(\mathrm{Ca^{2+}}\). A total of \(0.120\,\mathrm{mol}\) of fully deprotonated \(\mathrm{EDTA^{4-}}\) is supplied. Assume one EDTA ligand binds exactly one metal ion, complexation is quantitative, all iron(III) ions are complexed preferentially before any calcium(II) ions, and no other reactions occur. Which statement correctly gives the amounts of the two complexes and free calcium ions?
ⓐ. \(\mathrm{[Fe(EDTA)]^-}\): \(0.060\,\mathrm{mol}\); \(\mathrm{[Ca(EDTA)]^{2-}}\): \(0.060\,\mathrm{mol}\); free calcium: \(0.040\,\mathrm{mol}\)
ⓑ. \(\mathrm{[Fe(EDTA)]^-}\): \(0.060\,\mathrm{mol}\); \(\mathrm{[Ca(EDTA)]^{2-}}\): \(0.100\,\mathrm{mol}\); free calcium: zero
ⓒ. \(\mathrm{[Fe(EDTA)]^{2-}}\): \(0.060\,\mathrm{mol}\); \(\mathrm{[Ca(EDTA)]^-}\): \(0.060\,\mathrm{mol}\); free calcium: \(0.040\,\mathrm{mol}\)
ⓓ. \(\mathrm{[Fe(EDTA)]^-}\): \(0.040\,\mathrm{mol}\); \(\mathrm{[Ca(EDTA)]^{2-}}\): \(0.080\,\mathrm{mol}\); free calcium: \(0.020\,\mathrm{mol}\)
Correct Answer: \(\mathrm{[Fe(EDTA)]^-}\): \(0.060\,\mathrm{mol}\); \(\mathrm{[Ca(EDTA)]^{2-}}\): \(0.060\,\mathrm{mol}\); free calcium: \(0.040\,\mathrm{mol}\)
Explanation: \(\textbf{Binding stoichiometry:}\)
The stated complexation ratio is one metal ion per EDTA ligand.
The two reactions are:
\[
\mathrm{Fe^{3+}+EDTA^{4-}\rightarrow[Fe(EDTA)]^-}
\]
\[
\mathrm{Ca^{2+}+EDTA^{4-}\rightarrow[Ca(EDTA)]^{2-}}
\]
The coordination-entity charges follow from the metal-ion charges and the \(-4\) charge of EDTA.
\(\textbf{Preferential iron-binding stage:}\)
All \(0.060\,\mathrm{mol}\) of iron(III) ions are complexed first.
The EDTA consumed is:
\[
n_{\mathrm{EDTA,Fe}}=0.060\,\mathrm{mol}
\]
The amount remaining is:
\[
n_{\mathrm{EDTA,remaining}}=0.120-0.060
\]
\[
n_{\mathrm{EDTA,remaining}}=0.060\,\mathrm{mol}
\]
\(\textbf{Subsequent calcium-binding stage:}\)
The remaining EDTA binds calcium(II) in a \(1:1\) ratio.
Therefore:
\[
n_{\mathrm{CaEDTA}}=0.060\,\mathrm{mol}
\]
The initial calcium amount is \(0.100\,\mathrm{mol}\).
The amount remaining uncomplexed is:
\[
n_{\mathrm{Ca^{2+},free}}=0.100-0.060
\]
\[
n_{\mathrm{Ca^{2+},free}}=0.040\,\mathrm{mol}
\]
\(\textbf{Conservation checks:}\)
The total amount of EDTA incorporated is:
\[
0.060+0.060=0.120\,\mathrm{mol}
\]
The total amount of calcium remains:
\[
0.060+0.040=0.100\,\mathrm{mol}
\]
Both balances agree with the initial amounts.
The different complex-ion charges arise from the different metal oxidation states, not from changes in EDTA's specified formal charge.
The preferential binding order is an explicit experimental assumption rather than a universal conclusion based solely on the metals' charges.
\(\textbf{Final answer:}\)
\[
\boxed{n_{\mathrm{FeEDTA}}=0.060,\quad n_{\mathrm{CaEDTA}}=0.060,\quad n_{\mathrm{free\ Ca^{2+}}}=0.040\ \mathrm{mol}}
\]
319. A divalent metal ion forms an octahedral coordination cation containing six ammonia ligands. Under suitable conditions, excess en is added and the tris(en) complex forms. The reaction is represented by:
\[
\mathrm{[M(NH_3)_6]^{2+}+3en\rightleftharpoons[M(en)_3]^{2+}+6NH_3}
\]
Which statement correctly interprets this substitution and the chelate effect?
ⓐ. Three en molecules supply only three donor atoms, so the product cannot retain octahedral coordination.
ⓑ. The metal changes from oxidation state \(+2\) to \(+3\) because more ammonia molecules are released than en molecules consumed.
ⓒ. Three didentate en ligands replace six ammonia ligands without changing six metal–nitrogen bonds; chelation may favour the product.
ⓓ. The reactant and product are coordination isomers because the ligand molecules exchange places without changing the complete molecular composition.
Correct Answer: Three didentate en ligands replace six ammonia ligands without changing six metal–nitrogen bonds; chelation may favour the product.
Explanation: The initial complex contains six unidentate ammonia ligands.
They contribute six nitrogen donor attachments to M.
Each en molecule contains two nitrogen donor atoms capable of binding simultaneously to the same metal centre.
Three en molecules consequently supply six donor attachments.
The coordination number therefore remains six during the stated complete substitution.
Both ammonia and en are neutral ligands.
The oxidation state of M remains \(+2\), and the coordination-entity charge stays \(+2\).
The reaction also conserves ligand atoms: three en molecules enter, while six ammonia molecules leave.
In the product, the two donor atoms of each en belong to the same connected ligand.
Each en can therefore form a metal-containing chelate ring.
The product is a tris-chelate complex rather than a complex containing six independent nitrogen-donor molecules.
Chelate complexes often have enhanced stability relative to suitable analogous unidentate complexes.
The stated conversion is consistent with that general tendency.
However, the actual position of a chemical equilibrium depends on the particular metal, ligands and conditions.
The chelate effect does not imply that every such substitution proceeds quantitatively under every possible condition.
The reactant and product are not coordination isomers because the complete ligand compositions differ.
The comparison therefore requires donor-atom conservation, charge conservation, chelate connectivity and a qualified stability interpretation.
320. Consider three forms of a hypothetical ligand L containing two nitrogen donor atoms:
P: Both nitrogen atoms coordinate to the same metal M, forming a metal-containing ring.
Q: One nitrogen coordinates to M₁ and the other to a different metal M₂.
R: Only one nitrogen coordinates to M; the second nitrogen remains uncoordinated.
Which statement correctly describes the actual denticity contributions of L to individual metal centres?
ⓐ. L contributes two donor atoms to every metal in P, Q and R because it has two potential nitrogen donors.
ⓑ. L contributes two to M in P, one to each metal in Q, and one to M in R.
ⓒ. L contributes one to M in P because it is one ligand molecule, two to each metal in Q, and zero in R.
ⓓ. L contributes two to M in P and Q, but cannot be called a ligand in R because its second nitrogen remains free.
Correct Answer: L contributes two to M in P, one to each metal in Q, and one to M in R.
Explanation: A ligand's potential donor-atom count does not necessarily equal the number of attachments it makes to every individual metal.
In P, both nitrogen atoms coordinate simultaneously to M.
They therefore contribute two sites to M's coordination number.
The connected ligand also forms a metal-containing chelate ring.
In Q, the ligand connects two different metal centres.
Only one nitrogen donor atom attaches to M₁.
Only the other nitrogen donor atom attaches to M₂.
Thus, L contributes one coordination site to each individual metal.
It acts as a bridging ligand in the specified arrangement.
The complete ligand participates in two metal–donor attachments, but those attachments belong to different metals.
In R, one nitrogen atom coordinates to M and the other does not.
The actual coordination contribution to M is consequently one.
The ligand remains a ligand despite its second nitrogen not being coordinated.
Its attachment is simply unidentate in the stated structure.
The three cases therefore demonstrate different actual binding modes for a molecule possessing the same two potential donor atoms.
The distinction is important whenever coordination numbers are reconstructed from structural descriptions.
Counting all potential donor atoms without checking their actual attachments would overestimate the coordination number in Q and R.
