201. The fundamental distinction between structural isomerism and stereoisomerism is that:
ⓐ. structural isomers always have different molecular formulas
ⓑ. structural: connectivity; stereo: spatial arrangement
ⓒ. stereoisomers contain different central metals
ⓓ. structural isomers can occur only in neutral complexes
Correct Answer: structural: connectivity; stereo: spatial arrangement
Explanation: Structural isomerism involves a change in which atoms or ions are directly connected. This may involve exchange between the coordination sphere and counter-ion region or use of different donor atoms. Stereoisomers preserve the same metal–ligand connections. Their distinction arises from different three-dimensional arrangements around the metal. Charge and neutrality do not define the boundary between these two broad classes.
202. Two octahedral entities contain the same metal and the same six ligands. Every ligand is bonded through the same donor atom in both entities, but two identical ligands are adjacent in one and opposite in the other. These entities are:
ⓐ. stereoisomers
ⓑ. coordination isomers
ⓒ. ionisation isomers
ⓓ. hydrate isomers
Correct Answer: stereoisomers
Explanation: The same ligands remain bonded to the same metal through the same donor atoms. Metal–ligand connectivity is therefore unchanged. The difference is only whether the two identical ligands occupy adjacent or opposite positions. Such a difference corresponds to cis–trans geometrical isomerism. Geometrical isomerism belongs to the broader class of stereoisomerism.
203. A coordinated sulfate ion and an external bromide ion exchange positions to produce a second compound with the same overall elemental composition. The change represents:
ⓐ. only a rotation of the original structure
ⓑ. a change in oxidation state of the metal
ⓒ. changed connectivity, giving structural isomerism
ⓓ. a change in coordination number without isomerism
Correct Answer: changed connectivity, giving structural isomerism
Explanation: In one compound, sulfate is directly coordinated while bromide acts as a counter ion. In the second, bromide becomes coordinated and sulfate moves outside the coordination sphere. The identity of the species directly bonded to the metal has changed. This is a connectivity change rather than a simple spatial rearrangement. The pair therefore belongs to structural isomerism, specifically the ionisation type.
204. Study the classification table.
| Pair description | Proposed broad classification |
|---|
| P. Coordinated ion and counter ion exchange places | Structural isomerism |
| Q. Same bonds, but identical ligands are adjacent or opposite | Stereoisomerism |
| R. Same structure viewed after rotation | Distinct stereoisomers |
| S. Same ambidentate ligand binds through different donor atoms | Structural isomerism |
The inconsistent row is:
ⓐ. P
ⓑ. R
ⓒ. Q
ⓓ. S
Correct Answer: R
Explanation: Exchange between a coordinated ion and a counter ion changes connectivity and is structural isomerism. Adjacent-versus-opposite ligand placement preserves connectivity and is stereoisomerism. Different donor atoms of an ambidentate ligand also change the metal–ligand connection, placing linkage isomerism under structural isomerism. A rotated view of the same superimposable structure is not a new isomer. Row R incorrectly treats a viewing change as a chemical distinction.
205. Assertion: Connectivity should be checked before deciding whether a pair of coordination compounds shows geometrical isomerism.
Reason: Geometrical isomers must have the same metal–ligand connectivity and differ only in spatial arrangement.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Geometrical isomerism is a stereochemical difference. It requires the same ligands to remain attached to the same metal through the same donor atoms. If connectivity differs, the compounds belong to a structural-isomerism category instead. Checking connectivity first prevents an ionisation or linkage pair from being mislabelled as cis–trans isomers. The Reason therefore gives the basis for the procedure stated in the Assertion.
206. A student compares two compounds with the same molecular formula. The observations are:
Case 1: The same atoms are directly bonded, but the ligand positions cannot be made identical by rotation.
Case 2: A ligand bonded to the metal in one compound is outside the coordination sphere in the other.
The correct classifications for Case 1 and Case 2 are respectively:
ⓐ. stereoisomerism and structural isomerism
ⓑ. structural isomerism and stereoisomerism
ⓒ. identical compounds and optical isomerism
ⓓ. ionisation isomerism and identical compounds
Correct Answer: stereoisomerism and structural isomerism
Explanation: Case 1 preserves all direct bonding relationships, so connectivity is unchanged. The non-superimposable spatial difference therefore gives stereoisomerism. Case 2 changes whether a species is directly coordinated or lies outside the sphere. That change alters connectivity and gives structural isomerism. The classification is determined by bonding first and spatial arrangement second.
207. Consider the following statements about \(\mathrm{[Co(NH_3)_5Br]SO_4}\) and \(\mathrm{[Co(NH_3)_5SO_4]Br}\).
Statement I: They have the same overall elemental composition.
Statement II: They release different counter ions in water.
Statement III: Cobalt has oxidation state \(+3\) in both compounds.
Statement IV: They differ only in the spatial arrangement of identical ligands.
The valid statements are:
ⓐ. I and IV only
ⓑ. I, II and III
ⓒ. II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III
Explanation: Both formulas contain the same total quantities of cobalt, ammonia, bromine, and sulfate. Their difference is the placement of bromido and sulfate inside or outside the coordination sphere. Charge balance gives cobalt oxidation state \(+3\) in both compounds. Their metal–ligand connectivity changes, so they are structural isomers rather than stereoisomers. Statement IV is therefore incorrect.
208. A \(0.025\,\mathrm{mol}\) sample of \(\mathrm{[Co(NH_3)_5SO_4]Br}\) is treated with excess \(\mathrm{AgNO_3}\). If \(M(\mathrm{AgBr})=187.8\,\mathrm{g\,mol^{-1}}\), the mass of \(\mathrm{AgBr}\) formed is:
ⓐ. \(2.348\,\mathrm{g}\)
ⓑ. \(9.390\,\mathrm{g}\)
ⓒ. \(7.512\,\mathrm{g}\)
ⓓ. \(4.695\,\mathrm{g}\)
Correct Answer: \(4.695\,\mathrm{g}\)
Explanation: \( \textbf{Amount of coordination compound:} \)
\[
n=0.025\,\mathrm{mol}
\]
\( \textbf{External bromide ions per formula unit:} \)
\[
1
\]
\( \textbf{Precipitation reaction:} \)
\[
\mathrm{Ag^+ + Br^- \rightarrow AgBr(s)}
\]
\( \textbf{Mole ratio:} \)
\[
1\,\mathrm{mol}\,\mathrm{Br^-}:1\,\mathrm{mol}\,\mathrm{AgBr}
\]
\( \textbf{Moles of bromide released:} \)
\[
n(\mathrm{Br^-})=0.025\times1=0.025\,\mathrm{mol}
\]
\( \textbf{Moles of silver bromide:} \)
\[
n(\mathrm{AgBr})=0.025\,\mathrm{mol}
\]
\( \textbf{Mass relation:} \)
\[
m=nM
\]
\( \textbf{Substitution:} \)
\[
m=0.025\times187.8
\]
\( \textbf{Calculated mass:} \)
\[
m=4.695\,\mathrm{g}
\]
\( \textbf{Result:} \) The sample forms \(4.695\,\mathrm{g}\) of \(\mathrm{AgBr}\).
209. Equal molar solutions of the two ionisation isomers may contain the same total number of ions because each formula unit can dissociate into:
ⓐ. one neutral molecule only
ⓑ. three complex ions
ⓒ. one metal ion and six free ligands
ⓓ. one complex and one counter ion
Correct Answer: one complex and one counter ion
Explanation: \(\mathrm{[Co(NH_3)_5Br]SO_4}\) gives one complex cation and one sulfate ion. \(\mathrm{[Co(NH_3)_5SO_4]Br}\) gives one complex cation and one bromide ion. Each compound therefore produces two ions per formula unit under ideal dissociation. Their ion identities differ even though the ion count is the same. Selective precipitation tests are consequently more informative than ion count alone.
210. Study the expected qualitative-test results.
| Row | Compound | Reagent | Expected observation |
|---|
| P | \(\mathrm{[Co(NH_3)_5SO_4]Br}\) | \(\mathrm{AgNO_3}\) | \(\mathrm{AgBr}\) precipitate |
| Q | \(\mathrm{[Co(NH_3)_5Br]SO_4}\) | \(\mathrm{BaCl_2}\) | \(\mathrm{BaSO_4}\) precipitate |
| R | \(\mathrm{[Co(NH_3)_5Br]SO_4}\) | \(\mathrm{AgNO_3}\) | \(\mathrm{AgBr}\) from free bromide |
| S | \(\mathrm{[Co(NH_3)_5SO_4]Br}\) | \(\mathrm{BaCl_2}\) | No immediate sulfate precipitate |
The inconsistent row is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: R
Explanation: Row P is valid because bromide is the counter ion and reacts with \(\mathrm{Ag^+}\). Row Q is valid because external sulfate reacts with \(\mathrm{Ba^{2+}}\). Row S is also consistent because sulfate is coordinated rather than freely ionisable. In Row R, bromido lies inside the coordination sphere. It cannot be treated as free \(\mathrm{Br^-}\) in the immediate precipitation test.
211. Assertion: Ionisation isomers can give different precipitates with suitable reagents.
Reason: The identity of the ion released from outside the coordination sphere differs between the isomers.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: One ionisation isomer may release bromide while another releases sulfate. These ions respond differently to selective reagents such as \(\mathrm{AgNO_3}\) and \(\mathrm{BaCl_2}\). The coordinated species do not behave as freely dissolved counter ions in the immediate test. Different counter ions therefore produce different precipitates. The Reason directly accounts for the observation stated in the Assertion.
212. Samples \(P\) and \(Q\) have the same composition \(\mathrm{Co(NH_3)_5BrSO_4}\). Sample \(P\) gives \(\mathrm{AgBr}\) with \(\mathrm{AgNO_3}\), whereas sample \(Q\) gives \(\mathrm{BaSO_4}\) with \(\mathrm{BaCl_2}\). The formulas of \(P\) and \(Q\) are respectively:
ⓐ. \(\mathrm{[Co(NH_3)_5Br]SO_4}\) and \(\mathrm{[Co(NH_3)_5SO_4]Br}\)
ⓑ. \(\mathrm{[Co(NH_3)_5SO_4]Br}\) and \(\mathrm{[Co(NH_3)_5Br]SO_4}\)
ⓒ. \(\mathrm{[Co(NH_3)_4Br_2]SO_4}\) and \(\mathrm{[Co(NH_3)_4SO_4]Br_2}\)
ⓓ. \(\mathrm{[Co(NH_3)_5]BrSO_4}\) and \(\mathrm{[Co(NH_3)_5BrSO_4]}\)
Correct Answer: \(\mathrm{[Co(NH_3)_5SO_4]Br}\) and \(\mathrm{[Co(NH_3)_5Br]SO_4}\)
Explanation: Formation of \(\mathrm{AgBr}\) shows that sample \(P\) contains free bromide outside the square brackets. Sulfate must therefore be coordinated in \(P\). Formation of \(\mathrm{BaSO_4}\) shows that sample \(Q\) contains sulfate as the counter ion. Bromido must consequently be coordinated in \(Q\). The test results identify the two ionisation-isomer formulas directly.
213. Solvate isomerism arises when:
ⓐ. the oxidation state of the metal changes
ⓑ. different metals exchange ligand sets
ⓒ. identical ligands occupy cis and trans positions
ⓓ. solvent differs inside and outside the sphere
Correct Answer: solvent differs inside and outside the sphere
Explanation: Solvate isomers differ in whether solvent molecules are coordinated to the metal or present outside the coordination sphere. The overall composition may remain unchanged. When the solvent is water, the phenomenon is specifically called hydrate isomerism. Coordinated solvent contributes to the coordination number, whereas external solvent acts as solvent of crystallisation. No change in metal oxidation state is required.
214. In the hydrate-isomer formula
\[
\mathrm{[Cr(H_2O)_{6-x}Cl_x]Cl_{3-x}\cdot xH_2O},
\]
the number of immediately ionisable chloride ions is:
ⓐ. \(6-x\)
ⓑ. \(x\)
ⓒ. \(3-x\)
ⓓ. \(3+x\)
Correct Answer: \(3-x\)
Explanation: The \(x\) chlorido ligands inside the square brackets are directly coordinated to chromium. They are not counted as immediate chloride counter ions. The chloride ions outside the brackets carry the subscript \(3-x\). These external ions dissociate in water and react with \(\mathrm{Ag^+}\). Increasing \(x\) therefore decreases the number of immediately precipitable chloride ions.
215. If only the water of crystallisation is removed from \(\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}\), without changing the coordination sphere, the product is:
ⓐ. \(\mathrm{[Cr(H_2O)_6]Cl_3}\)
ⓑ. \(\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot H_2O}\)
ⓒ. \(\mathrm{CrCl_3+5H_2O}\)
ⓓ. \(\mathrm{[Cr(H_2O)_5Cl]Cl_2}\)
Correct Answer: \(\mathrm{[Cr(H_2O)_5Cl]Cl_2}\)
Explanation: The dot-separated water molecule lies outside the coordination sphere. Its removal does not automatically break any chromium–oxygen bond inside the brackets. Five aqua ligands and one chlorido ligand therefore remain coordinated to chromium. The two chloride counter ions also remain unchanged. Conversion into another hydrate isomer would require ligand rearrangement rather than simple loss of lattice water.
216. Study the hydrate-isomer data.
| Row | Formula | Ionisable \(\mathrm{Cl^-}\) per formula unit |
|---|
| P | \(\mathrm{[Cr(H_2O)_6]Cl_3}\) | \(3\) |
| Q | \(\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}\) | \(2\) |
| R | \(\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}\) | \(1\) |
| S | \(\mathrm{[Cr(H_2O)_3Cl_3]\cdot3H_2O}\) | \(3\) |
The inconsistent row is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: The first formula has three chloride counter ions outside the brackets. The second has two, and the third has one. In Row S, all three chlorido ligands are written inside the coordination sphere. No chloride counter ion remains outside the brackets. Its number of immediately ionisable chloride ions should therefore be zero rather than three.
217. Equal \(0.010\,\mathrm{mol}\) samples of \(\mathrm{[Cr(H_2O)_6]Cl_3}\), \(\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}\), and \(\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}\) are mixed and treated with excess \(\mathrm{AgNO_3}\). If \(M(\mathrm{AgCl})=143.5\,\mathrm{g\,mol^{-1}}\), the total mass of \(\mathrm{AgCl}\) is:
ⓐ. \(4.305\,\mathrm{g}\)
ⓑ. \(5.740\,\mathrm{g}\)
ⓒ. \(8.610\,\mathrm{g}\)
ⓓ. \(12.915\,\mathrm{g}\)
Correct Answer: \(8.610\,\mathrm{g}\)
Explanation: \( \textbf{First hydrate external chlorides:} \)
\[
3
\]
\( \textbf{AgCl from the first sample:} \)
\[
0.010\times3=0.030\,\mathrm{mol}
\]
\( \textbf{Second hydrate external chlorides:} \)
\[
2
\]
\( \textbf{AgCl from the second sample:} \)
\[
0.010\times2=0.020\,\mathrm{mol}
\]
\( \textbf{Third hydrate external chlorides:} \)
\[
1
\]
\( \textbf{AgCl from the third sample:} \)
\[
0.010\times1=0.010\,\mathrm{mol}
\]
\( \textbf{Total moles of precipitate:} \)
\[
n_{\text{total}}=0.030+0.020+0.010
\]
\[
n_{\text{total}}=0.060\,\mathrm{mol}
\]
\( \textbf{Mass calculation:} \)
\[
m=nM
\]
\[
m=0.060\times143.5
\]
\[
m=8.610\,\mathrm{g}
\]
\( \textbf{Result:} \) The mixed samples produce \(8.610\,\mathrm{g}\) of \(\mathrm{AgCl}\).
218. A compound of composition \(\mathrm{CrCl_3\cdot6H_2O}\) gives two moles of \(\mathrm{AgCl}\) per mole of compound and produces three ions per formula unit on ideal dissociation. Its formula is:
ⓐ. \(\mathrm{[Cr(H_2O)_6]Cl_3}\)
ⓑ. \(\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}\)
ⓒ. \(\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O}\)
ⓓ. \(\mathrm{[Cr(H_2O)_3Cl_3]\cdot3H_2O}\)
Correct Answer: \(\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}\)
Explanation: Two moles of \(\mathrm{AgCl}\) indicate two chloride counter ions outside the coordination sphere. One complex cation together with two chloride ions gives three ions on ideal dissociation. The third chloride required by the total composition must therefore be coordinated. Five water molecules remain inside the coordination sphere, and one lies outside as water of crystallisation. These observations uniquely support the second formula.
219. Which pair represents linkage isomerism?
ⓐ. \(\mathrm{[Co(NH_3)_5(NO_2)]Cl_2}\) and \(\mathrm{[Co(NH_3)_5(ONO)]Cl_2}\)
ⓑ. \(\mathrm{[Co(NH_3)_5Br]SO_4}\) and \(\mathrm{[Co(NH_3)_5SO_4]Br}\)
ⓒ. Cis-\(\mathrm{[Pt(NH_3)_2Cl_2]}\) and trans-\(\mathrm{[Pt(NH_3)_2Cl_2]}\)
ⓓ. \(\mathrm{[Co(NH_3)_6][Cr(CN)_6]}\) and \(\mathrm{[Cr(NH_3)_6][Co(CN)_6]}\)
Correct Answer: \(\mathrm{[Co(NH_3)_5(NO_2)]Cl_2}\) and \(\mathrm{[Co(NH_3)_5(ONO)]Cl_2}\)
Explanation: Both compounds contain the same cobalt centre, five ammine ligands, one \(\mathrm{NO_2^-}\) ligand, and two chloride counter ions. In the first compound, the ambidentate ligand is bonded through nitrogen. In the second, it is bonded through oxygen. The atom directly attached to cobalt therefore changes while the overall composition remains constant. This donor-atom change identifies the pair as linkage isomers.
220. Consider the following statements about linkage isomerism.
Statement I: The same ambidentate ligand is present in both isomers.
Statement II: The donor atom bonded to the metal changes.
Statement III: The metal oxidation state must change.
Statement IV: The overall elemental composition remains the same.
The valid statements are:
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Linkage isomers contain the same ambidentate ligand in both structures. The distinction arises because alternative donor atoms form the metal–ligand bond. The formal charge of the ligand remains unchanged, so the metal oxidation state need not change. The total elemental composition also remains identical. Statement III therefore introduces a condition that is not required for linkage isomerism.
