101. A cobalt(III) coordination compound has the formula \(\mathrm{[Co(NH_3)_4(H_2O)(CO_3)]Cl}\). The carbonato ligand coordinates through one oxygen donor atom, and all other ligands are unidentate. Which name and coordination number are correct?
ⓐ. Tetraammineaquacarbonatocobaltate(III) chloride; coordination number six
ⓑ. Tetraammineaquacarbonatocobalt(II) chloride; coordination number six
ⓒ. Tetraamminecarbonatodiaquacobalt(III) chloride; coordination number seven
ⓓ. Tetraammineaquacarbonatocobalt(III) chloride; coordination number six
Correct Answer: Tetraammineaquacarbonatocobalt(III) chloride; coordination number six
Explanation: \(\textbf{Interpretation of the ligands:}\)
The coordination sphere contains four ammonia molecules, one water molecule and one carbonato ligand.
Ammonia is named ammine, water is named aqua and coordinated carbonate is named carbonato.
The prefixes are determined by ligand multiplicities, giving tetraammine for the four ammonia molecules.
The other ligand types occur once each and require no numerical prefix.
\(\textbf{Alphabetical naming order:}\)
The ligand names are arranged alphabetically as ammine, aqua and carbonato.
Therefore, the ligand portion is tetraammineaquacarbonato.
\(\textbf{Charge verification:}\)
Cobalt has oxidation state \(+3\).
Ammonia and water are neutral ligands.
The carbonato ligand carries charge \(-2\).
Therefore:
\[
q=(+3)+4(0)+1(0)+(-2)
\]
\[
q=+1
\]
One chloride counter ion balances the positively charged coordination entity.
This is consistent with the complete compound formula.
\(\textbf{Coordination-number determination:}\)
The four ammonia molecules contribute four nitrogen donor atoms.
The water molecule contributes one oxygen donor atom.
The carbonato ligand is explicitly stated to use one oxygen donor atom.
Therefore:
\[
CN=4(1)+1(1)+1(1)
\]
\[
CN=6
\]
\(\textbf{Metal-name selection:}\)
The complex is a cation, so the metal is named cobalt rather than cobaltate.
Its oxidation state is represented by III.
Thus, the correct systematic name is tetraammineaquacarbonatocobalt(III) chloride.
The coordination number is six, while the carbonato ligand retains its formal charge of \(-2\).
Ligand charge and denticity must not be confused when determining the name and coordination number.
\(\textbf{Final answer:}\)
\[
CN=6
\]
102. A cobalt(III) compound contains two chloride ligands, two ammonia ligands and one en ligand inside its coordination sphere. The en ligand is neutral and didentate. One chloride ion remains outside the sphere. Which statement correctly describes its coordination formula and naming order?
ⓐ. The ligands are named chlorido, ammine and ethane-1,2-diamine, because chloride carries a negative charge.
ⓑ. The compound has formula \(\mathrm{[CoCl_2(en)(NH_3)_2]Cl_2}\), and the coordinated ligands are named in order of decreasing denticity.
ⓒ. The compound has formula \(\mathrm{[CoCl_2(en)(NH_3)_2]Cl}\), and the ligand names appear in the order ammine, chlorido and ethane-1,2-diamine.
ⓓ. The compound has formula \(\mathrm{[CoCl_2(en)(NH_3)_2]Cl}\), but its name must begin with ethane-1,2-diamine because en is didentate.
Correct Answer: The compound has formula \(\mathrm{[CoCl_2(en)(NH_3)_2]Cl}\), and the ligand names appear in the order ammine, chlorido and ethane-1,2-diamine.
Explanation: The coordination entity contains cobalt(III), two coordinated chloride ions, two ammonia molecules and one en molecule.
The coordinated chloride ions contribute a total charge of \(-2\).
Both ammonia and en are electrically neutral.
Consequently, the coordination entity has charge \(+1\).
One external chloride ion is therefore sufficient for electrical neutrality.
The compound is represented as \(\mathrm{[CoCl_2(en)(NH_3)_2]Cl}\).
In writing the coordination formula, the central metal is placed first.
The ligand formulae are then arranged according to the prescribed formula-ordering convention, including the use of the abbreviation en.
However, the systematic name is constructed using alphabetical order of ligand names rather than their charges or denticities.
The ligand names are ammine, chlorido and ethane-1,2-diamine.
Therefore, the naming order begins with ammine, followed by chlorido and ethane-1,2-diamine.
The complete name is diamminedichlorido(ethane-1,2-diamine)cobalt(III) chloride.
The coordination number is six because two chloride ligands and two ammonia molecules contribute four donor atoms, while en contributes two.
The fact that en is didentate does not give it priority over the other ligands in alphabetical naming.
Similarly, the negative charge of chlorido does not require it to be named before neutral ligands.
This example distinguishes formula ordering, nomenclature ordering, charge balance and coordination-number counting.
103. A mononuclear palladium(II) coordination compound contains one ammonia molecule, one bromido ligand, one chlorido ligand and one nitrogen-bonded nitrite ligand. All four ligands are unidentate, and potassium ions provide electrical neutrality. Which complete formula and systematic name are correct?
ⓐ. \(\mathrm{K[Pd(NH_3)BrCl(NO_2)]}\); amminebromidochloridonitrito-N-palladium(II)
ⓑ. \(\mathrm{K_2[Pd(NH_3)BrCl(NO_2)]}\); amminebromidochloridonitrito-N-palladate(II)
ⓒ. \(\mathrm{[Pd(NH_3)BrCl(NO_2)]K}\); amminebromidochloridonitrito-O-palladate(II)
ⓓ. \(\mathrm{K[Pd(NH_3)BrCl(NO_2)]}\); potassium amminebromidochloridonitrito-N-palladate(II)
Correct Answer: \(\mathrm{K[Pd(NH_3)BrCl(NO_2)]}\); potassium amminebromidochloridonitrito-N-palladate(II)
Explanation: \(\textbf{Coordination-sphere composition:}\)
Palladium has oxidation state \(+2\).
The ammonia ligand is neutral.
The bromido, chlorido and nitrite ligands each carry charge \(-1\).
All four ligands are unidentate, giving coordination number four.
\(\textbf{Entity-charge calculation:}\)
\[
q=(+2)+0+(-1)+(-1)+(-1)
\]
\[
q=-1
\]
The coordination entity is therefore negatively charged.
One potassium ion balances its charge.
The complete formula is:
\[
\mathrm{K[Pd(NH_3)BrCl(NO_2)]}
\]
The nitrite ligand is specified to coordinate through nitrogen.
\(\textbf{Ligand-name construction:}\)
The ligand names are ammine, bromido, chlorido and nitrito-N.
They are placed alphabetically in that order.
Each ligand occurs once, so multiplicative prefixes are unnecessary.
The coordinated nitrite is distinguished from an oxygen-bonded nitrite ligand by its donor-atom designation.
\(\textbf{Metal name and complete compound name:}\)
Because the coordination entity is anionic, palladium takes the ending -ate, giving palladate.
Its oxidation state is shown as II.
The potassium counter ion is named first.
Thus, the systematic name is potassium amminebromidochloridonitrito-N-palladate(II).
Naming the metal palladium without the -ate ending would ignore the negative charge of the complex.
Using nitrito-O would incorrectly specify oxygen coordination rather than the given nitrogen coordination.
The correct name therefore combines ligand identity, donor-specific notation, alphabetical ordering and charge-dependent metal nomenclature.
104. A square-planar platinum(II) coordination entity contains four different unidentate ligands: one ammonia molecule, one water molecule, one bromido ligand and one chlorido ligand. Its formula is \(\mathrm{[Pt(NH_3)(H_2O)BrCl]}\). No information about the relative ligand positions is initially supplied. Which statement is correct?
ⓐ. Its systematic base name specifies one unique geometrical isomer because all four ligands are different.
ⓑ. Its systematic base name identifies the ligand composition, but the four different ligands permit three geometrical arrangements.
ⓒ. The presence of four different ligands makes the entity chiral, so only two optical isomers are possible.
ⓓ. Its coordination formula uniquely identifies the trans arrangement because platinum(II) is square planar.
Correct Answer: Its systematic base name identifies the ligand composition, but the four different ligands permit three geometrical arrangements.
Explanation: The coordination entity contains platinum(II), two neutral ligands and two negatively charged ligands.
Ammonia and water contribute zero charge, while bromido and chlorido contribute \(-1\) each.
Therefore, the complete coordination entity is electrically neutral.
Its coordination number is four because all four ligands are unidentate.
The systematic base name is ammineaquabromidochloridoplatinum(II).
This name identifies the metal oxidation state and the chemical identities of all coordinated ligands.
However, it does not by itself specify their relative positions around the square-planar metal centre.
In a square-planar entity containing four different unidentate ligands, three distinct geometrical arrangements are possible.
For example, ammonia may be opposite water, bromido or chlorido, giving three distinct choices of trans partner.
Once the ligand opposite ammonia is specified, the remaining two positions determine the other trans pair.
Interchanging the two remaining positions does not create an additional geometrical isomer because the resulting arrangement can be related by a suitable spatial symmetry operation.
Consequently, there are three distinct geometrical arrangements rather than six permutations or only one arrangement.
These square-planar arrangements do not exhibit ordinary optical isomerism merely because the four ligand identities are different.
Thus, the base coordination name establishes composition but does not uniquely identify one of the three geometrical arrangements.
An appropriate stereochemical descriptor or unambiguous spatial description is needed to specify a particular form.
105. Two coordination salts contain the same tris(ethane-1,2-diamine)iron coordination framework. In compound P, the metal is iron(III) and chloride is the counter ion. In compound Q, the metal is iron(II) and sulphate is the counter ion. Assume all en ligands are neutral and didentate, and both compounds dissociate ideally without ligand loss. Which statement correctly compares their formulae and ionic-species counts?
ⓐ. P is \(\mathrm{[Fe(en)_3]Cl_3}\), Q is \(\mathrm{[Fe(en)_3]SO_4}\), and P gives twice as many ionic species per formula unit as Q.
ⓑ. P is \(\mathrm{[Fe(en)_3]Cl_2}\), Q is \(\mathrm{[Fe(en)_3]SO_4}\), and both give three ionic species per formula unit.
ⓒ. P is \(\mathrm{[Fe(en)_3]Cl_3}\), Q is \(\mathrm{[Fe(en)_3](SO_4)_2}\), and P gives four ionic species per formula unit.
ⓓ. P is \(\mathrm{[Fe(en)_3]Cl}\), Q is \(\mathrm{[Fe(en)_3]SO_4}\), and Q gives twice as many ionic species per formula unit as P.
Correct Answer: P is \(\mathrm{[Fe(en)_3]Cl_3}\), Q is \(\mathrm{[Fe(en)_3]SO_4}\), and P gives twice as many ionic species per formula unit as Q.
Explanation: \(\textbf{Coordination framework:}\)
Both compounds contain three neutral en ligands.
Each en ligand coordinates through two nitrogen donor atoms.
Therefore:
\[
CN=3(2)=6
\]
The coordination number is the same in both compounds.
\(\textbf{Compound P:}\)
Iron has oxidation state \(+3\).
All three en ligands are neutral.
Hence, the coordination entity carries charge \(+3\).
Three chloride ions are required to balance this charge.
The complete formula is:
\[
\mathrm{[Fe(en)_3]Cl_3}
\]
Its ideal dissociation is:
\[
\mathrm{[Fe(en)_3]Cl_3\longrightarrow[Fe(en)_3]^{3+}+3Cl^-}
\]
Thus:
\[
N_P=1+3=4
\]
\(\textbf{Compound Q:}\)
Iron has oxidation state \(+2\).
The coordination entity therefore carries charge \(+2\).
One sulphate ion, carrying charge \(-2\), balances the complex cation.
The complete formula is:
\[
\mathrm{[Fe(en)_3]SO_4}
\]
Its ideal dissociation is:
\[
\mathrm{[Fe(en)_3]SO_4\longrightarrow[Fe(en)_3]^{2+}+SO_4^{2-}}
\]
Thus:
\[
N_Q=1+1=2
\]
\(\textbf{Comparison:}\)
The number of ionic species produced per formula unit is:
\[
\frac{N_P}{N_Q}=\frac{4}{2}=2
\]
Both compounds retain the same coordination number and ligand multiplicity.
However, their metal oxidation states and counter-ion charges differ.
The names tris(ethane-1,2-diamine)iron(III) chloride and tris(ethane-1,2-diamine)iron(II) sulphate therefore correspond to different ionic formulations.
The multiplicative prefix tris refers to three en ligand molecules in each compound, not to the ionic-species count.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{N_P}{N_Q}=2}
\]
106. Consider the following proposed formula–name–coordination-number combinations.
| Option | Coordination formula | Proposed systematic name | Proposed coordination number |
| A | \(\mathrm{[Co(NH_3)_4(C_2O_4)]Cl}\) | Tetraammineoxalatocobalt(III) chloride | 6 |
| B | \(\mathrm{K_2[Ni(CN)_4]}\) | Potassium tetracyanidonickel(II) | 4 |
| C | \(\mathrm{[Cr(en)_3]Cl_3}\) | Hexa(ethane-1,2-diamine)chromium(III) chloride | 6 |
| D | \(\mathrm{[Co(NH_3)_5Cl]Cl_2}\) | Pentaamminechloridocobalt(II) chloride | 6 |
Which option is fully consistent with the formula, nomenclature and coordination-number rules?
ⓐ. Option A
ⓑ. Option B
ⓒ. Option C
ⓓ. Option D
Correct Answer: Option A
Explanation: In option A, cobalt is in oxidation state \(+3\).
The four ammonia molecules are neutral, while the oxalato ligand carries charge \(-2\).
The coordination entity therefore has charge \(+1\), balanced by one external chloride ion.
The oxalato ligand is didentate, so four ammonia donor atoms and two oxalato donor atoms give coordination number six.
The ligand names are tetraammine and oxalato, arranged alphabetically.
Therefore, tetraammineoxalatocobalt(III) chloride is consistent with the formula.
Option B has the correct coordination number because four cyanido ligands coordinate to nickel.
However, the bracketed entity is negatively charged, so the metal must be named nickelate rather than nickel.
The correct name is potassium tetracyanidonickelate(II).
Option C also has coordination number six, but it contains only three en ligand molecules.
The prefix hexa incorrectly counts the six nitrogen donor atoms as six complete en ligands.
The appropriate name is tris(ethane-1,2-diamine)chromium(III) chloride.
Option D contains one coordinated chloride ligand and two chloride counter ions.
The resulting coordination-entity charge is \(+2\), so cobalt must have oxidation state \(+3\), not \(+2\).
Its correct name is pentaamminechloridocobalt(III) chloride.
Thus, only option A satisfies all three requirements simultaneously.
The exercise illustrates why a correct coordination number alone does not establish that the corresponding systematic name is correct.
107. A metal ion \(\mathrm{M^{3+}}\) forms the coordination entity \(\mathrm{[M(NH_3)_2(C_2O_4)_2]^-}\). A second entity is obtained by replacing one oxalato ligand with two neutral ammonia molecules, while the metal oxidation state remains unchanged and all donor atoms specified remain coordinated. Which statement correctly describes the resulting ligand-name multiplicities, coordination-entity charge and coordination number?
ⓐ. The product contains tetraammine and monooxalato, has charge \(+1\), and coordination number six.
ⓑ. The product contains diammine and oxalato, has charge \(+1\), and coordination number four.
ⓒ. The product contains tetraammine and dioxalato, has charge \(-1\), and coordination number eight.
ⓓ. The product contains tetraammine and oxalato, has charge \(+1\), and coordination number six.
Correct Answer: The product contains tetraammine and oxalato, has charge \(+1\), and coordination number six.
Explanation: \(\textbf{Initial coordination composition:}\)
The original entity contains two neutral ammonia molecules and two didentate oxalato ligands.
The oxalato ligands each carry charge \(-2\).
Therefore:
\[
q_{\mathrm{initial}}=(+3)+2(0)+2(-2)
\]
\[
q_{\mathrm{initial}}=-1
\]
The coordination number is:
\[
CN_{\mathrm{initial}}=2(1)+2(2)=6
\]
\(\textbf{Ligand replacement:}\)
One oxalato ligand is removed and two ammonia molecules enter the coordination sphere.
The removed oxalato ligand contributes two oxygen donor atoms.
The two entering ammonia molecules collectively contribute two nitrogen donor atoms.
Thus, the total number of donor attachments remains unchanged.
\(\textbf{Resulting ligand composition:}\)
The original entity contains two ammonia molecules.
After two additional ammonia molecules enter, the total becomes four.
One oxalato ligand remains coordinated.
The resulting entity is:
\[
\mathrm{[M(NH_3)_4(C_2O_4)]^+}
\]
\(\textbf{Charge calculation:}\)
The metal retains oxidation state \(+3\).
Four ammonia molecules contribute zero formal charge.
One oxalato ligand contributes charge \(-2\).
Therefore:
\[
q_{\mathrm{final}}=(+3)+4(0)+(-2)
\]
\[
q_{\mathrm{final}}=+1
\]
\(\textbf{Coordination-number verification:}\)
Four ammonia molecules contribute four donor attachments.
One didentate oxalato ligand contributes two.
Thus:
\[
CN_{\mathrm{final}}=4(1)+1(2)=6
\]
\(\textbf{Nomenclature consequence:}\)
Four coordinated ammonia molecules require the prefix tetra, giving tetraammine.
One oxalato ligand is named oxalato without an additional multiplicative prefix.
The ligand-name portion is therefore tetraammineoxalato.
The metal name retains its ordinary form because the resulting entity is cationic.
The replacement simultaneously changes ligand multiplicities and reverses the sign of the coordination-entity charge while conserving coordination number.
\(\textbf{Final answer:}\)
\[
q_{\mathrm{final}}=+1,\quad CN_{\mathrm{final}}=6
\]
108. A coordination compound is described by the systematic name potassium diaquadicyanidoferrate(II). Assume each coordinated water and cyanido ligand is unidentate. A student proposes the formula \(\mathrm{K_2[Fe(H_2O)_2(CN)_2]}\). Which statement correctly evaluates the proposed formula and the stated name?
ⓐ. The formula is correct because two coordinated cyanido ligands require two potassium counter ions.
ⓑ. The formula is incorrect because the coordination entity is neutral; the name cannot correctly include potassium as a counter ion.
ⓒ. The formula is incorrect because iron must have oxidation state \(+4\) to balance the two cyanido ligands.
ⓓ. The formula is correct, but the coordination number is six because water contains two hydrogen atoms.
Correct Answer: The formula is incorrect because the coordination entity is neutral; the name cannot correctly include potassium as a counter ion.
Explanation: \(\textbf{Interpretation of the named coordination entity:}\)
The name specifies iron in oxidation state \(+2\).
Two aqua ligands and two cyanido ligands are directly coordinated.
Each aqua ligand is neutral.
Each cyanido ligand carries charge \(-1\).
\(\textbf{Entity-charge calculation:}\)
\[
q=(+2)+2(0)+2(-1)
\]
\[
q=2-2=0
\]
Therefore, the bracketed coordination entity is electrically neutral.
\(\textbf{Counter-ion consistency:}\)
A neutral coordination entity does not require potassium ions to balance its charge.
The proposed formula includes two potassium ions outside the coordination sphere.
Those ions contribute a total charge of \(+2\).
Because no corresponding anionic component is present, the proposed formula violates electrical neutrality.
Therefore, \(\mathrm{K_2[Fe(H_2O)_2(CN)_2]}\) is not a valid neutral salt formulation under the stated assumptions.
\(\textbf{Correct coordination formulation:}\)
The coordination entity itself may be represented as:
\[
\mathrm{[Fe(H_2O)_2(CN)_2]}
\]
Its systematic base name is diaquadicyanidoiron(II).
The neutral entity uses iron rather than ferrate in its name.
\(\textbf{Coordination-number verification:}\)
Two water molecules contribute two oxygen donor atoms.
Two cyanido ligands contribute two carbon donor atoms.
Hence:
\[
CN=2(1)+2(1)=4
\]
The coordination number is four, not six.
The crucial error is not the number of coordinated ligands but the assumption that any coordination entity containing negatively charged ligands must itself be an anion.
The positive metal oxidation state can exactly balance the ligand charges, producing a neutral coordination entity.
Therefore, both the proposed potassium-containing formula and the corresponding anionic-complex name are inconsistent with the specified composition.
109. Consider the following statements about isomerism in coordination compounds.
Statement I: Structural isomers may differ in the identities of atoms directly bonded to the central metal.
Statement II: Geometrical isomers must differ in the total numbers of coordinated ligand molecules.
Statement III: Optical isomers have the same coordination connectivity but are non-superimposable mirror images.
Statement IV: Two coordination compounds having different overall elemental compositions can be classified as linkage isomers if both contain ambidentate ligands.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I and III only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I and III only
Explanation: Statement I is correct because structural isomerism involves a difference in chemical bonding or coordination connectivity while the overall chemical composition remains unchanged.
For example, linkage isomers differ in the donor atom through which an ambidentate ligand attaches to the central metal.
Ionisation and coordination isomerism provide other examples involving changes in the distribution of coordinated groups.
Statement II is incorrect because geometrical isomers retain the same chemical formula and ligand composition.
Their difference concerns the relative spatial positions occupied by ligands around the metal.
The cis and trans forms of a suitable square-planar or octahedral complex illustrate this distinction.
Statement III is correct because optical isomers are non-superimposable mirror images.
They possess the same relevant bonding connectivity but differ in their three-dimensional arrangements.
Such mirror-image stereoisomers are called enantiomers.
Statement IV is incorrect because isomers must have the same overall chemical composition.
The presence of an ambidentate ligand in two unrelated compounds does not make them linkage isomers.
For linkage isomerism, an appropriate change in donor-atom attachment must occur without changing the overall composition.
Thus, Statements I and III correctly distinguish structural isomerism from stereoisomerism.
110. Two cobalt(III) compounds P and Q have the same overall composition, \(\mathrm{Co(NH_3)_5BrSO_4}\). Their coordination formulae are:
P: \(\mathrm{[Co(NH_3)_5(SO_4)]Br}\)
Q: \(\mathrm{[Co(NH_3)_5Br]SO_4}\)
Assume that under the stated rapid precipitation-test conditions, the coordination spheres remain intact. Which observation and ionic-species comparison are simultaneously correct?
ⓐ. P gives an immediate \(\mathrm{BaSO_4}\) precipitate, Q gives \(\mathrm{AgBr}\), and both produce three ions per formula unit.
ⓑ. P gives an immediate \(\mathrm{AgBr}\) precipitate, Q gives \(\mathrm{BaSO_4}\), and P produces more ions than Q.
ⓒ. P gives an immediate \(\mathrm{AgBr}\) precipitate, Q gives \(\mathrm{BaSO_4}\), and both produce two ions per formula unit.
ⓓ. Neither produces an immediate precipitate with these reagents because both bromide and sulphate are coordinated.
Correct Answer: P gives an immediate \(\mathrm{AgBr}\) precipitate, Q gives \(\mathrm{BaSO_4}\), and both produce two ions per formula unit.
Explanation: \(\textbf{Coordination sphere of P:}\)
P contains five neutral ammonia ligands and one coordinated sulphato ligand.
Cobalt has oxidation state \(+3\), while coordinated sulphate contributes charge \(-2\).
Therefore:
\[
q_P=(+3)+5(0)+(-2)=+1
\]
One bromide ion occurs outside the coordination sphere.
Its ideal dissociation is:
\[
\mathrm{[Co(NH_3)_5(SO_4)]Br\longrightarrow[Co(NH_3)_5(SO_4)]^++Br^-}
\]
The free bromide counter ion can produce \(\mathrm{AgBr}\) with silver ions.
The coordinated sulphato ligand is not treated as a freely dissociated sulphate ion under the stated conditions.
\(\textbf{Coordination sphere of Q:}\)
Q contains five ammonia ligands and one coordinated bromido ligand.
Its coordination-entity charge is:
\[
q_Q=(+3)+5(0)+(-1)=+2
\]
One sulphate ion occurs outside the coordination sphere.
The dissociation is:
\[
\mathrm{[Co(NH_3)_5Br]SO_4\longrightarrow[Co(NH_3)_5Br]^{2+}+SO_4^{2-}}
\]
The free sulphate counter ion can produce \(\mathrm{BaSO_4}\) with barium ions.
\(\textbf{Isomerism and ionic comparison:}\)
The two compounds have the same overall composition but differ in which anion occurs inside the coordination sphere.
They are therefore ionisation isomers.
P releases bromide as its counter ion, whereas Q releases sulphate.
Each compound nevertheless dissociates into one complex cation and one simple anion.
Thus, both produce two ionic species per formula unit.
Their different precipitation behaviour is more informative than their identical ideal ionic-species counts.
The key distinction is the identity of the external counter ion, not the total number of ions produced.
111. Three hydrated chromium(III) chloride compounds have the same empirical composition, \(\mathrm{CrCl_3\cdot6H_2O}\). An unknown sample X produces one mole of \(\mathrm{AgCl}\) per mole of compound under the usual Werner precipitation conditions. Independent structural information establishes that chromium has coordination number six and that exactly two water molecules occur outside its coordination sphere. Which formula and ideal ionic-species count are consistent with these observations?
ⓐ. \(\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}\); two ionic species
ⓑ. \(\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}\); three ionic species
ⓒ. \(\mathrm{[Cr(H_2O)_6]Cl_3}\); four ionic species
ⓓ. \(\mathrm{[Cr(H_2O)_3Cl_3]\cdot3H_2O}\); one molecular species
Correct Answer: \(\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}\); two ionic species
Explanation: \(\textbf{Interpretation of chloride precipitation:}\)
The empirical composition contains three chloride ions per chromium atom.
One mole of \(\mathrm{AgCl}\) per mole of compound indicates one chloride counter ion under the specified conditions.
Therefore, two chloride ions must remain coordinated to chromium.
\(\textbf{Distribution of water molecules:}\)
The empirical formula contains six water molecules.
Two are stated to occur outside the coordination sphere.
The remaining number of coordinated water molecules is:
\[
N_{\mathrm{coordinated\ water}}=6-2
\]
\[
N_{\mathrm{coordinated\ water}}=4
\]
\(\textbf{Construction of the coordination entity:}\)
Four water molecules and two chloride ions are directly coordinated.
All are unidentate in the stated arrangement.
Hence:
\[
CN=4(1)+2(1)=6
\]
This agrees with the specified coordination number.
\(\textbf{Charge verification:}\)
Chromium has oxidation state \(+3\).
The four coordinated water molecules are neutral.
The two coordinated chloride ions contribute charge \(-2\).
Therefore:
\[
q=(+3)+4(0)+2(-1)
\]
\[
q=+1
\]
One external chloride ion is sufficient for electrical neutrality.
The complete formula is:
\[
\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}
\]
\(\textbf{Ideal dissociation:}\)
\[
\mathrm{[Cr(H_2O)_4Cl_2]Cl\longrightarrow[Cr(H_2O)_4Cl_2]^++Cl^-}
\]
One complex cation and one chloride anion are produced.
The two waters of crystallisation are not counted as ionic species.
Therefore, the ideal ionic-species count is two.
The observations collectively identify the third conventional hydrate arrangement of this empirical composition.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{coordinated\ H_2O}}=4,\quad N_{\mathrm{external\ Cl^-}}=1,\quad N_{\mathrm{ions}}=2}
\]
112. The octahedral coordination entity \(\mathrm{[Co(en)_2Cl_2]^+}\) contains two didentate en ligands and two chloride ligands. Both en ligands form their normal chelate rings. If optical enantiomers are counted separately, what are the numbers of geometrical classes and total distinct stereoisomers, respectively?
ⓐ. Two geometrical classes and two stereoisomers
ⓑ. Three geometrical classes and three stereoisomers
ⓒ. Two geometrical classes and four stereoisomers
ⓓ. Two geometrical classes and three stereoisomers
Correct Answer: Two geometrical classes and three stereoisomers
Explanation: \(\textbf{Possible geometrical arrangements:}\)
The two chloride ligands can occupy adjacent or opposite octahedral positions.
Adjacent chloride ligands give the cis arrangement.
Opposite chloride ligands give the trans arrangement.
Therefore, there are two geometrical classes.
\(\textbf{Optical behaviour of the cis form:}\)
The cis arrangement contains two en chelate rings and two adjacent chloride ligands.
Its three-dimensional chelate arrangement can exist in two non-superimposable mirror-image forms.
These forms constitute one enantiomeric pair.
Thus, the cis geometrical class contributes two distinct stereoisomers.
\(\textbf{Optical behaviour of the trans form:}\)
In the trans arrangement, the chloride ligands occupy opposite octahedral positions.
For this standard symmetrical complex, the trans arrangement is achiral.
Its mirror image is superimposable on the original structure.
Therefore, the trans class contributes one stereoisomer.
\(\textbf{Total stereoisomer count:}\)
\[
N_{\mathrm{geometrical\ classes}}=2
\]
\[
N_{\mathrm{stereoisomers}}=2_{\mathrm{cis}}+1_{\mathrm{trans}}
\]
\[
N_{\mathrm{stereoisomers}}=3
\]
The number of geometrical classes is therefore not the same as the total number of stereoisomers.
Counting only cis and trans would omit the optical distinction within the cis class.
Conversely, assigning an enantiomeric pair to the trans form would incorrectly treat an achiral structure as chiral.
\(\textbf{Final answer:}\)
There are two geometrical classes and three distinct stereoisomers.
113. A pure sample of one optical isomer of \(\mathrm{[Co(en)_3]^{3+}}\) rotates plane-polarised light clockwise, while its mirror-image isomer rotates it anticlockwise by the same magnitude under identical measurement conditions. A solution is prepared containing equal amounts of these two enantiomers. Which conclusion is correct?
ⓐ. The solution shows zero net optical rotation even though both constituent coordination entities are chiral.
ⓑ. The solution rotates plane-polarised light clockwise because the cobalt centre has oxidation state \(+3\).
ⓒ. The solution becomes a new achiral coordination entity because the two enantiomers combine chemically.
ⓓ. The solution shows twice the rotation of either pure enantiomer because both contain three chelate rings.
Correct Answer: The solution shows zero net optical rotation even though both constituent coordination entities are chiral.
Explanation: Optical isomers are non-superimposable mirror images called enantiomers.
The two forms of \(\mathrm{[Co(en)_3]^{3+}}\) arise from the three-dimensional arrangement of its three didentate en ligands.
Each en ligand forms a chelate ring, and the complete arrangement can possess opposite handedness.
The two pure enantiomers rotate plane-polarised light in opposite directions under identical conditions.
For equal concentrations and identical optical paths, their specific contributions to the observed rotation have equal magnitudes and opposite signs.
When equal amounts are mixed, these rotations cancel.
The resulting mixture is called a racemic mixture.
A racemic mixture shows zero net optical rotation even though its individual molecular or ionic components remain chiral.
The enantiomers do not need to react chemically with each other for the cancellation to occur.
The oxidation state of cobalt also does not determine the direction of optical rotation.
A zero optical rotation therefore does not necessarily imply that every chemical species in the solution is achiral.
This distinction is essential when interpreting the optical behaviour of coordination compounds and their enantiomeric mixtures.
114. Consider the coordination salts P and Q:
P: \(\mathrm{[Co(en)_3][Cr(C_2O_4)_3]}\)
Q: \(\mathrm{[Cr(en)_3][Co(C_2O_4)_3]}\)
Both metals are in oxidation state \(+3\). Each en ligand is neutral and didentate, while each oxalato ligand has charge \(-2\) and is didentate. Which statement correctly describes the relationship between P and Q?
ⓐ. They are ionisation isomers because their external counter ions are different.
ⓑ. They are linkage isomers because en coordinates through nitrogen and oxalate through oxygen.
ⓒ. They are coordination isomers because the ligand groups are redistributed between the two metal centres.
ⓓ. They are geometrical isomers because the same metal–ligand bonds have different spatial orientations.
Correct Answer: They are coordination isomers because the ligand groups are redistributed between the two metal centres.
Explanation: \(\textbf{Coordination entities in P:}\)
The first entity contains cobalt(III) coordinated to three neutral en ligands.
Its charge is \(+3\).
The second entity contains chromium(III) coordinated to three oxalato ligands.
Its charge is:
\[
q_{\mathrm{Cr(ox)_3}}=(+3)+3(-2)=-3
\]
Therefore, the two entities combine in a \(1:1\) ratio to give an electrically neutral salt.
\(\textbf{Coordination entities in Q:}\)
In Q, the en ligands are attached to chromium(III) instead of cobalt(III).
The oxalato ligands are attached to cobalt(III) instead of chromium(III).
The charges of the new complex cation and anion remain \(+3\) and \(-3\), respectively.
\(\textbf{Overall composition:}\)
Both compounds contain one cobalt atom, one chromium atom, three en ligands and three oxalato ligands.
Their overall chemical compositions are therefore identical.
\(\textbf{Change in coordination connectivity:}\)
The essential difference is which metal is coordinated by which ligand type.
In P, cobalt is coordinated by en and chromium by oxalate.
In Q, chromium is coordinated by en and cobalt by oxalate.
This redistribution of ligands between complex cationic and anionic entities containing different metal centres is coordination isomerism.
It differs from ionisation isomerism, which involves an interchange between a coordinated ligand and an external counter ion.
It also differs from linkage isomerism, where an ambidentate ligand changes its donor atom.
Both salts retain octahedral coordination numbers of six around each metal.
The isomerism consequently arises from a change in metal–ligand connectivity rather than a change in coordination number or overall composition.
115. Two octahedral entities P and Q each contain three ammonia ligands and three chloride ligands coordinated to a trivalent metal M. All six ligands are unidentate. The following geometrical information is supplied.
| Entity | Number of ammonia–M–ammonia angles equal to \(180^\circ\) | Number of ammonia–M–ammonia angles equal to \(90^\circ\) |
| P | 0 | 3 |
| Q | 1 | 2 |
Which statement correctly identifies the geometrical arrangements?
ⓐ. P is meridional and Q is facial; both entities have charge zero.
ⓑ. Both are meridional because each contains one set of three identical ligands.
ⓒ. Both are facial because each contains three ammonia ligands.
ⓓ. P is facial and Q is meridional; both entities have charge zero.
Correct Answer: P is facial and Q is meridional; both entities have charge zero.
Explanation: \(\textbf{Interpretation of the octahedral angles:}\)
In an octahedral complex, adjacent coordination positions subtend angles of \(90^\circ\).
Opposite positions subtend angles of \(180^\circ\).
Three ammonia ligands generate three distinct ammonia–metal–ammonia pairs.
Their relative angles can therefore be used to distinguish facial from meridional arrangements.
\(\textbf{Identification of P:}\)
In P, all three ammonia–metal–ammonia angles are \(90^\circ\).
Therefore, every ammonia pair occupies adjacent positions.
The three ammonia donor atoms consequently lie at the corners of one triangular face of the octahedron.
P is the facial isomer.
\(\textbf{Identification of Q:}\)
In Q, one ammonia pair occupies opposite positions, giving an angle of \(180^\circ\).
The remaining two ammonia pairs occupy adjacent positions.
This is characteristic of the meridional arrangement.
Hence, Q is the meridional isomer.
\(\textbf{Charge verification:}\)
The central metal has oxidation state \(+3\).
All ammonia ligands are neutral, while the three coordinated chlorides contribute charge \(-3\).
Therefore:
\[
q=(+3)+3(0)+3(-1)=0
\]
Both coordination entities are neutral.
The facial–meridional distinction therefore does not require any change in metal oxidation state or coordination-entity charge.
The decisive evidence is the distribution of the pairwise angular relationships among the three identical ligands.
\(\textbf{Final answer:}\)
P is facial, Q is meridional, and both coordination entities are neutral.
116. An octahedral cobalt(III) coordination entity has the composition \(\mathrm{[Co(NH_3)_4(NO_2)_2]^+}\). Each nitrite ligand can independently coordinate through nitrogen or oxygen, and both nitrite ligands are unidentate in every arrangement. Assume that all such donor attachments are permitted and count distinct geometrical and linkage arrangements, but not optical isomers. How many distinct structures are possible?
ⓐ. Four
ⓑ. Five
ⓒ. Eight
ⓓ. Six
Correct Answer: Six
Explanation: \(\textbf{Possible donor-atom combinations:}\)
Nitrite is an ambidentate ligand capable of coordinating through nitrogen or oxygen.
The coordination entity contains two nitrite ligands.
Both may coordinate through nitrogen, both may coordinate through oxygen, or one may coordinate through nitrogen while the other coordinates through oxygen.
These give three distinct linkage-composition patterns:
\[
\mathrm{(N,N),\quad(N,O),\quad(O,O)}
\]
The arrangements \(\mathrm{(N,O)}\) and \(\mathrm{(O,N)}\) are not separate linkage-composition patterns because the two nitrite ligands are chemically identical before donor attachment.
\(\textbf{Geometrical possibilities:}\)
Four ammonia ligands occupy four of the six octahedral positions.
The two nitrite ligands occupy the remaining two positions.
For each linkage-composition pattern, the two nitrite donor groups may occupy adjacent or opposite positions.
These correspond to cis and trans geometrical arrangements.
\(\textbf{Counting distinct structures:}\)
For the N,N linkage pattern, there are cis and trans arrangements.
For the O,O linkage pattern, there are also cis and trans arrangements.
For the mixed N,O linkage pattern, the differently bonded nitrite groups can likewise occupy adjacent or opposite positions.
Thus:
\[
N_{\mathrm{structures}}=2+2+2
\]
\[
N_{\mathrm{structures}}=6
\]
\(\textbf{Chemical consistency:}\)
The cobalt oxidation state is \(+3\).
Each nitrite ligand contributes charge \(-1\), regardless of the donor atom used.
All six structures therefore have coordination-entity charge \(+1\).
The coordination number also remains six because every ammonia and nitrite ligand contributes one donor attachment.
The structural differences arise from the independent possibilities of donor-atom selection and geometrical arrangement.
The count excludes additional distinctions that were not requested.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{structures}}=6}
\]
117. Four square-planar arrangements P, Q, R and S contain a central metal M and four different unidentate ligands A, B, C and D. Their positions are specified below, with the coordination positions lying along the positive and negative \(x\) and \(y\) axes.
| Arrangement | \(+x\) | \(+y\) | \(-x\) | \(-y\) |
| P | A | B | C | D |
| Q | D | A | B | C |
| R | A | C | B | D |
| S | A | B | D | C |
Which conclusion is correct after allowing rigid rotations of the complete square-planar entity?
ⓐ. P and Q are equivalent, while R and S represent two other geometrical arrangements.
ⓑ. P and R are equivalent, while Q and S represent two other geometrical arrangements.
ⓒ. Q and S are equivalent, while P and R represent two other geometrical arrangements.
ⓓ. All four are distinct because none has identical ligand labels at corresponding axes.
Correct Answer: P and Q are equivalent, while R and S represent two other geometrical arrangements.
Explanation: \(\textbf{Criterion for geometrical equivalence:}\)
Two spatial descriptions do not necessarily represent different coordination isomers merely because the ligands are assigned different Cartesian directions.
A rigid rotation of the complete coordination entity does not alter its chemical identity.
For square-planar complexes with four different unidentate ligands, the identities of opposite ligand pairs provide a convenient equivalence test.
\(\textbf{Arrangement P:}\)
Ligand A is opposite C.
Ligand B is opposite D.
Therefore, the two trans pairs are A–C and B–D.
\(\textbf{Arrangement Q:}\)
Ligand A occupies \(+y\), while C occupies \(-y\).
Thus, A and C remain opposite.
Similarly, B and D occupy opposite \(x\)-axis positions.
Q has the same two trans pairs as P.
A quarter-turn rotation of P produces Q.
Therefore, P and Q are equivalent.
\(\textbf{Arrangement R:}\)
A and B occupy opposite \(x\)-axis positions.
C and D occupy opposite \(y\)-axis positions.
Its trans pairs are A–B and C–D.
This differs from P and Q.
\(\textbf{Arrangement S:}\)
A and D occupy opposite \(x\)-axis positions.
B and C occupy opposite \(y\)-axis positions.
Its trans pairs are A–D and B–C.
This is different from both previous trans-pair patterns.
\(\textbf{Final comparison:}\)
P and Q describe the same geometrical arrangement.
R and S describe the remaining two geometrical arrangements.
Thus, the four coordinate descriptions represent only three distinct square-planar geometrical structures.
The result illustrates why molecular rotations must be considered before counting isomers.
118. Three isolated chromium(III) compounds P, Q and R all have the empirical composition \(\mathrm{CrCl_3\cdot6H_2O}\). Their structures are specified as follows:
P: cis-\(\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}\)
Q: trans-\(\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}\)
R: \(\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}\)
Which statement correctly identifies the isomeric relationships?
ⓐ. P and Q are hydrate isomers, whereas P and R are geometrical isomers.
ⓑ. All three are geometrical isomers because chromium has coordination number six in each.
ⓒ. P and Q are geometrical isomers, whereas P and R are hydrate isomers.
ⓓ. All three are linkage isomers because chloride and water have different donor atoms.
Correct Answer: P and Q are geometrical isomers, whereas P and R are hydrate isomers.
Explanation: All three compounds have the same overall empirical composition.
This satisfies the essential compositional requirement for isomerism.
In P and Q, four water molecules and two chloride ions are directly coordinated to chromium.
Both compounds also contain one external chloride counter ion and two waters of crystallisation.
Their coordination connectivity is therefore identical.
However, P has its two coordinated chlorides in adjacent positions, while Q has them in opposite positions.
They consequently represent cis and trans geometrical isomers.
The relationship between P and R is different.
P contains four coordinated water molecules, two coordinated chloride ions and two waters outside the coordination sphere.
R contains five coordinated water molecules, one coordinated chloride ion and one water molecule outside the coordination sphere.
The numbers of directly coordinated water and chloride ligands therefore differ.
This change in the distribution of water between the coordination sphere and the crystal lattice is hydrate isomerism.
Hydrate isomerism is a water-containing form of solvate isomerism and belongs to structural isomerism.
The external chloride counts also differ, giving different ideal ionisation behaviour under ordinary dissolution.
Thus, geometrical and hydrate isomerism can describe different relationships among compounds sharing the same empirical composition.
The appropriate classification depends on whether the difference concerns spatial ligand positions or coordination connectivity.
119. Three cobalt(III) compounds contain the same total numbers of cobalt, ammonia, nitrogen, oxygen and sulphur atoms:
P: \(\mathrm{[Co(NH_3)_5(NO_2)]SO_4}\), with nitrite coordinated through nitrogen
Q: \(\mathrm{[Co(NH_3)_5(ONO)]SO_4}\), with nitrite coordinated through oxygen
R: \(\mathrm{[Co(NH_3)_5(SO_4)]NO_2}\), with sulphate coordinated and nitrite acting as the counter ion
Which statement correctly describes their isomeric relationships?
ⓐ. P and Q are ionisation isomers, while P and R are linkage isomers.
ⓑ. P and Q are linkage isomers, while P and R are ionisation isomers.
ⓒ. All three are linkage isomers because each contains a nitrite ion.
ⓓ. All three are geometrical isomers because each contains six coordinated ligand groups.
Correct Answer: P and Q are linkage isomers, while P and R are ionisation isomers.
Explanation: \(\textbf{Comparison of P and Q:}\)
Both compounds contain five coordinated ammonia molecules and one coordinated nitrite ligand.
The sulphate ion occurs outside the coordination sphere in both cases.
In P, nitrite is attached to cobalt through nitrogen.
In Q, nitrite is attached through oxygen.
Therefore, the two structures differ in the donor atom of the same ambidentate ligand.
P and Q are linkage isomers.
\(\textbf{Comparison of P and R:}\)
In P, nitrite is coordinated and sulphate acts as the external counter ion.
In R, sulphate is coordinated and nitrite acts as the external counter ion.
This involves an interchange between a coordinated ligand and a potential ligand functioning as a counter ion.
The corresponding relationship is ionisation isomerism.
\(\textbf{Charge verification:}\)
For P and Q:
\[
q=(+3)+5(0)+(-1)=+2
\]
The resulting complex cation is balanced by one sulphate ion.
For R:
\[
q=(+3)+5(0)+(-2)=+1
\]
The resulting complex cation is balanced by one nitrite ion.
All three compounds are electrically neutral overall.
\(\textbf{Coordination-number consistency:}\)
The five ammonia molecules contribute five donor atoms.
One nitrite or one coordinated sulphato group contributes the sixth donor attachment in the specified monodentate arrangements.
Thus, each cobalt centre has coordination number six.
The same overall composition therefore supports two different structural-isomer relationships.
The classification depends on whether donor-atom selection changes within a coordinated ligand or an exchange occurs between coordinated and external ions.
120. An octahedral coordination entity \(\mathrm{[Fe(NH_3)_2(CN)_4]^-}\) contains two neutral ammonia ligands and four carbon-bonded cyanido ligands. All ligands are monodentate. Which statement correctly combines its oxidation state, metal-name form and geometrical-isomer possibilities?
ⓐ. Iron(II), ferrate naming and two geometrical isomers
ⓑ. Iron(III), iron naming and one geometrical arrangement
ⓒ. Iron(II), iron naming and one geometrical arrangement
ⓓ. Iron(III), ferrate naming and two geometrical isomers
Correct Answer: Iron(III), ferrate naming and two geometrical isomers
Explanation: \(\textbf{Oxidation-state determination:}\)
The coordination entity carries charge \(-1\).
Each ammonia ligand is neutral.
Each cyanido ligand carries charge \(-1\).
Let the oxidation state of iron be \(x\).
\[
x+2(0)+4(-1)=-1
\]
\[
x-4=-1
\]
\[
x=+3
\]
Thus, iron has oxidation state \(+3\).
\(\textbf{Metal-name selection:}\)
The coordination entity is negatively charged.
Therefore, iron takes the established anionic-complex name ferrate.
The oxidation state is represented by the Roman numeral III.
The corresponding coordination-entity name is diamminetetracyanidoferrate(III).
\(\textbf{Geometrical possibilities:}\)
The entity has coordination number six and octahedral geometry.
Two ammonia ligands occupy two of the six coordination positions.
They may occupy adjacent positions, producing the cis arrangement.
Alternatively, they may occupy opposite positions, producing the trans arrangement.
The four cyanido ligands fill the remaining positions.
Both arrangements retain the same ligand composition and bonding connectivity.
They therefore constitute two geometrical isomers.
\(\textbf{Combined interpretation:}\)
The negative coordination-entity charge does not mean iron has a negative oxidation state.
The -ate ending follows the entity charge, while the Roman numeral follows the metal oxidation state.
The existence of cis–trans isomerism follows from the octahedral arrangement of two ammonia and four cyanido ligands.
Thus, iron(III), ferrate nomenclature and two geometrical isomers are simultaneously consistent.