Co-Ordination Compounds MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: The starting coordination entities have the same formula but different relative ligand positions. In the facial form, the three B ligands occupy three mutually adjacent octahedral positions. Every B–B relationship is therefore cis. Removing any one B ligand leaves two B ligands that remain adjacent. Because the question prohibits rearrangement of the other ligands, the resulting \(\mathrm{MA_4B_2}\) entity must be cis. The choice of which facial B ligand is replaced does not change this conclusion. In the meridional form, two of the three B ligands are trans. The third B ligand is cis to each member of that trans pair. If the third B ligand is replaced, the original trans pair remains. The product is then trans-\(\mathrm{MA_4B_2}\). Alternatively, one member of the trans B pair may be replaced. The two remaining B ligands are then adjacent. The product is cis-\(\mathrm{MA_4B_2}\). Thus, the meridional starting structure permits two geometrically distinct substitution products. The facial starting structure permits only one under the fixed-position condition. The products have the same ligand composition and coordination number, but their geometrical possibilities retain information about the positions occupied in the starting structures. The reasoning requires tracing the surviving ligand relationships rather than counting ligand molecules alone.

Explanation: \(\textbf{Identifying the donor attachments:}\) Each cobalt centre has four directly coordinated ammonia molecules. These contribute four nitrogen donor atoms. One coordinated chloride ligand contributes another donor atom. The bridging oxalato ligand contributes one oxygen donor atom to each cobalt centre. Thus, the coordination number at each metal is: \[ CN=4+1+1 \] \[ CN=6 \] The bridging oxalato ligand supplies two donor attachments across the complete dinuclear entity. However, those attachments are distributed between two different metal centres. They must not both be counted towards the coordination number of either one cobalt centre. \(\textbf{Calculating the complete entity charge:}\) The two cobalt(III) centres contribute: \[ q_{\mathrm{metals}}=2(+3)=+6 \] The two coordinated chlorido ligands contribute: \[ q_{\mathrm{chlorido}}=2(-1)=-2 \] The bridging oxalato ligand contributes: \[ q_{\mathrm{oxalato}}=-2 \] All ammonia ligands are neutral. Therefore: \[ q_{\mathrm{entity}}=(+6)+(-2)+(-2) \] \[ q_{\mathrm{entity}}=+2 \] \(\textbf{Chelation and magnetic interpretation:}\) A chelate ring requires multiple donor attachments from the same ligand to the same metal centre. Here, the two oxalato donor attachments lead to different cobalt centres. Under the specified connectivity, oxalato is therefore a bridging ligand rather than a two-donor chelate around either cobalt. Each cobalt(III) centre has configuration \(d^6\). The specified low-spin occupancy is \(t_{2g}^{6}e_g^{0}\). Both metal centres consequently have their d electrons paired. The coordination and charge conclusions depend on the actual donor connectivity, not merely on the fact that oxalato is ordinarily capable of didentate chelation. \(\textbf{Final answer:}\) \[ q_{\mathrm{entity}}=+2,\quad CN_{\mathrm{each\ Co}}=6 \]

Explanation: P contains the coordination cation \(\mathrm{[Co(NH_3)_6]^{3+}}\). Its six ammonia ligands are neutral. The complex-ion charge is therefore \(+3\). Under ideal dissociation, P produces one complex cation and three chloride ions. Its total ionic-species count is four. Q contains the coordination anion \(\mathrm{[Co(CN)_6]^{3-}}\). Each cyanido ligand contributes charge \(-1\). Cobalt(III) contributes \(+3\), giving the coordination anion charge \(-3\). Under ideal dissociation, Q produces three potassium ions and one complex anion. Its ionic-species count is also four. In an applied electric field, positive ions move towards the negatively charged cathode. Negative ions move towards the positively charged anode. Consequently, the cobalt-containing ion from P moves towards the cathode. The cobalt-containing ion from Q moves towards the anode. Both cobalt centres possess the same \(d^6\) electron count. Under the specified low-spin conditions, both are diamagnetic. However, diamagnetism does not determine the direction of electrophoretic migration. That direction is governed by the electrical charge of the moving coordination ion. The comparison demonstrates why identical metal oxidation states, coordination numbers, magnetic classifications and ideal ionic-species counts need not imply identical charge signs for coordination entities.

Explanation: \(\textbf{Oxidation state and configuration of P:}\) Carbon monoxide is a neutral ligand in conventional oxidation-state calculations. Nickel tetracarbonyl is also electrically neutral. Therefore, nickel has oxidation state zero in P. Its ordinary metal d-electron count is \(d^{10}\). The standard valence-bond description of \(\mathrm{Ni(CO)_4}\) uses \(sp^3\) hybridisation. The coordination geometry is tetrahedral. All its d electrons are paired, so P is diamagnetic. \(\textbf{Oxidation state and configuration of Q:}\) Each cyanido ligand carries charge \(-1\). Four cyanido ligands contribute total charge \(-4\). The coordination entity has overall charge \(-2\). Therefore: \[ x+4(-1)=-2 \] \[ x=+2 \] Nickel(II) has configuration \(d^8\). The strong-field cyanido environment produces the ordinary square-planar nickel(II) arrangement. Its elementary valence-bond description uses \(dsp^2\) hybridisation. The relevant d electrons are paired, so Q is also diamagnetic. \(\textbf{Integrated distinction:}\) Both compounds have coordination number four. Both are diamagnetic. However, they differ in metal oxidation state, d-electron count, hybridisation and coordination geometry. The presence of four directly coordinated ligands therefore does not uniquely determine geometry. Similarly, diamagnetism does not imply that two metal centres possess identical d-electron configurations. The chemical identity and charge of the ligands must be combined with the electronic configuration of the metal. This comparison illustrates why nickel tetracarbonyl and tetracyanidonickelate(II) cannot be assigned the same bonding description merely because both contain four ligands.

Explanation: Haemoglobin is an important example of biological coordination chemistry. Its iron centre is associated with a surrounding ligand environment. The biological function involves reversible interaction with oxygen. In the question's simplified model, oxygen occupies an available coordination position. The previously attached haem donor atoms remain coordinated. Thus, oxygen binding does not require the complete breakdown of the metal-containing coordination structure. The incoming oxygen changes the immediate surroundings of the iron centre. However, the model explicitly keeps the formal iron oxidation state unchanged. Therefore, a change in coordination environment must not automatically be identified as a redox change. A coordination process and an oxidation–reduction process describe different chemical changes. The presence of an additional attached ligand also does not require replacement of iron by another metal. On oxygen release, the original coordination position becomes available again under the stipulated reversible model. This provides a coordination-based interpretation of transport without requiring the metal to leave the biological molecule. The simplified formal oxidation-state assignment should not be confused with a complete description of electron-density distribution in actual oxyhaemoglobin. The question concerns the logically distinct roles of metal–ligand attachment, reversible oxygen binding and formal redox bookkeeping.

Explanation: \(\textbf{Formal oxidation-state assignment:}\) Carbon monoxide is treated as a neutral ligand in ordinary oxidation-state bookkeeping for metal carbonyls. The complete entity \(\mathrm{[Cr(CO)_6]}\) is also neutral. If chromium has oxidation state \(x\), charge balance gives: \[ x+6(0)=0 \] \[ x=0 \] \(\textbf{Coordination-number assignment:}\) Each CO ligand coordinates through its carbon donor atom and contributes one metal–ligand attachment. Six CO ligands therefore give coordination number six. The presence of both \(\sigma\) donation and \(\pi\) back-donation does not double the coordination number. \(\textbf{Why back-donation does not alter the formal charge:}\) Back-donation is a bonding interaction between occupied metal orbitals and suitable CO acceptor orbitals. It does not require assigning CO a formal \(-1\) charge in the conventional oxidation-state calculation. Thus, CO remains a neutral ligand, chromium is in oxidation state zero, and the coordination number is six.

Explanation: \(\textbf{Initial metal–EDTA composition:}\) The solution initially contains: \[ n_{\mathrm{NiEDTA},i}=0.060\,\mathrm{mol} \] Nickel(II) is initially present almost entirely in its EDTA coordination entity. The amount of free nickel(II) is specified to be negligible. EDTA acts as a hexadentate ligand in its usual coordination mode and can form stable chelate complexes with suitable metal ions. \(\textbf{Metal-ion exchange reaction:}\) The supplied reaction is: \[ \mathrm{[Ni(EDTA)]^{2-}+Fe^{3+}\rightarrow[Fe(EDTA)]^-+Ni^{2+}} \] One mole of iron(III) replaces one mole of coordinated nickel(II). The amount of iron(III) added is: \[ n_{\mathrm{Fe^{3+}}}=0.040\,\mathrm{mol} \] Since the initial nickel–EDTA amount is \(0.060\,\mathrm{mol}\), iron(III) is the limiting reactant. The amount of nickel–EDTA consumed is: \[ n_{\mathrm{NiEDTA,consumed}}=0.040\,\mathrm{mol} \] \(\textbf{Amount of released nickel(II):}\) Each mole of nickel–EDTA consumed releases one mole of nickel(II). Therefore: \[ n_{\mathrm{Ni^{2+},free}}=0.040\,\mathrm{mol} \] The amount of intact nickel–EDTA remaining after the exchange is: \[ n_{\mathrm{NiEDTA},f}=0.060-0.040 \] \[ n_{\mathrm{NiEDTA},f}=0.020\,\mathrm{mol} \] \(\textbf{Concentration of free nickel(II):}\) The final solution volume is: \[ V=2.00\,\mathrm{L} \] Therefore: \[ [\mathrm{Ni^{2+}}]_{\mathrm{free}}=\frac{n_{\mathrm{Ni^{2+},free}}}{V} \] Substituting: \[ [\mathrm{Ni^{2+}}]_{\mathrm{free}}=\frac{0.040}{2.00} \] \[ [\mathrm{Ni^{2+}}]_{\mathrm{free}}=0.020\,\mathrm{M} \] \(\textbf{Qualitative identification using DMG:}\) Dimethylglyoxime is an important analytical reagent for identifying nickel(II). In a suitable ammoniacal medium, nickel(II) reacts with dimethylglyoxime to form an insoluble red nickel dimethylglyoximate coordination compound. The precipitated nickel(II) complex is commonly represented as: \[ \mathrm{[Ni(dmgH)_2]} \] Here, each coordinated dimethylglyoximate ligand is singly deprotonated and acts as a didentate ligand through two nitrogen donor atoms. The resulting nickel complex has a square-planar coordination environment. Since the exchange reaction releases nickel(II), the specified DMG test gives the characteristic red precipitate. \(\textbf{Charge-balance verification:}\) The EDTA exchange reaction is electrically balanced. The total charge on the reactant side is: \[ q_{\mathrm{reactants}}=-2+3=+1 \] The total charge on the product side is: \[ q_{\mathrm{products}}=-1+2=+1 \] Thus, the equation conserves charge as well as the number of metal ions and EDTA ligands. \(\textbf{Integrated interpretation:}\) The total amount of nickel does not change during the metal-exchange reaction. Instead, \(0.040\,\mathrm{mol}\) of nickel changes from the coordinated form to the released ionic form. The remaining \(0.020\,\mathrm{mol}\) of nickel–EDTA is still coordinated immediately before the DMG test. The qualitative observation is therefore associated with the availability of released nickel(II), rather than an increase in the total nickel content. \(\textbf{Final answer:}\) The free nickel(II) concentration is \(0.020\,\mathrm{M}\), the remaining nickel–EDTA amount is \(0.020\,\mathrm{mol}\), and the DMG test produces the characteristic red precipitate.

Explanation: Both complexes contain four chlorido ligands. Each chlorido ligand has charge \(-1\). Their total ligand charge is therefore \(-4\). Since each coordination ion has charge \(-2\), the central metal oxidation state is \(+2\). Nickel(II) has a \(3d^8\) configuration. Platinum(II) has a \(5d^8\) configuration. Thus, the formal d-electron counts are identical, although the metals belong to different transition series. The ordinary tetrachloridonickelate(II) ion has tetrahedral geometry. Its \(d^8\) electron arrangement contains two unpaired electrons. It is therefore paramagnetic. The ordinary tetrachloridoplatinate(II) ion has square-planar geometry. Its electronic arrangement favours pairing of the d electrons. It is diamagnetic. Both entities nevertheless have coordination number four. The difference demonstrates that coordination geometry cannot be determined solely by counting ligands. Metal identity, the extent of d-orbital interactions and the associated electronic energy pattern also matter. A fixed formal d-electron count does not guarantee the same electron pairing in geometrically different coordination environments. The student's conclusion therefore overgeneralises from ligand composition and oxidation state while ignoring differences in the metal centres.

Explanation: The two compounds contain identical overall ligand and counter-ion inventories. However, the locations of bromide and sulphate differ. In P, five ammonia molecules and one sulphato ligand are coordinated to cobalt. The sulphato ligand is specified to be unidentate, so the coordination number is six. The coordinated sulphato ligand carries charge \(-2\), while cobalt(III) contributes \(+3\). The coordination entity therefore has charge \(+1\), balanced by one external bromide ion. Consequently, P supplies freely available bromide under the stated conditions. It gives AgBr with silver nitrate but does not supply free sulphate for the specified barium test. In Q, five ammonia ligands and one bromido ligand occupy the coordination sphere. The complex cation has charge \(+2\). One external sulphate ion balances this charge. Therefore, Q gives BaSO4 but does not supply free bromide for the specified silver nitrate test. The relationship between P and Q is ionisation isomerism. The different precipitation products arise from exchange of coordinated and external ions, not from different cobalt oxidation states. Both cobalt centres retain coordination number six. The two independent analytical tests distinguish the isomers even though their overall elemental compositions are identical.

Explanation: Both compounds contain the same divalent metal ion. The ligand charges are zero, so both coordination entities carry charge \(+2\). Each en ligand supplies two nitrogen donor atoms. Three en molecules therefore provide six metal–nitrogen donor attachments. Six ammonia molecules also supply six nitrogen donor attachments. Thus, the coordination numbers are identical. However, the bonding arrangements differ because en forms chelate rings while ammonia binds as independent unidentate ligands. The greater formation stability of P is consistent with the general chelate effect. Thermodynamic formation stability describes the equilibrium favourability of complex formation. The rate of ligand substitution instead depends on the kinetic pathway and relevant activation-energy barrier. An equilibrium stability comparison does not automatically determine that barrier. Crystal-field splitting is another distinct quantity. It describes the energetic separation of appropriate metal d-orbital sets in a coordination environment. Although ligand identity can influence both stability and splitting, their numerical relationship cannot be inferred from the stated stability observation alone. Consequently, the student has combined three different properties as though they were interchangeable. Only the chelate-related thermodynamic comparison is directly supported.

Explanation: Statement I is correct. The three ammonia molecules and the en ligand are neutral. The coordinated chloride contributes charge \(-1\). The coordination cation carries charge \(+2\), as indicated by its two external chloride counter ions. Chromium therefore has oxidation state \(+3\). Statement II is correct because coordination number depends on actual donor attachments. The three ammonia ligands contribute three donor atoms. The en ligand contributes two nitrogen donor atoms. The coordinated chloride contributes one donor atom. Together these provide six coordination positions. Statement III is incorrect because only two chloride ions occur outside the square brackets. The third chloride ion is coordinated directly to chromium. Under conditions preserving the coordination sphere, it is not counted among the freely dissociable chloride ions. Statement IV is correct because the coordination entity contains ammonia, en and chlorido ligands. It is therefore heteroleptic. Its charge is \(+2\), which is balanced by two external chloride ions. The example demonstrates that oxidation state, coordination number, ligand diversity and the number of ionisable counter ions must be determined independently. The correct combination is I, II and IV.

Explanation: The starting coordination entity contains four ammonia ligands and two chlorido ligands. All six ligands are unidentate, so the coordination number is six. The chlorido ligands are initially trans. They occupy opposite octahedral positions. The remaining four positions form the plane perpendicular to their common axis. Among those four positions, adjacent pairs are available. An en ligand is didentate and must bind through two nitrogen donor atoms. Its donor atoms occupy adjacent rather than opposite octahedral positions. Therefore, substitution is possible if two adjacent ammonia ligands leave. The en ligand occupies the two vacancies and forms a chelate ring. The two remaining ammonia molecules retain their original positions. Since neither chloride ligand moves, their trans relationship is preserved. The product is consequently trans-\(\mathrm{[Co(en)(NH_3)_2Cl_2]^+}\). Its coordination number remains six because two ammonia donor attachments are replaced by two en donor attachments. Cobalt remains in oxidation state \(+3\), and en is neutral. The product therefore retains charge \(+1\). The result illustrates why chelate formation constrains which donor positions can be replaced, but need not change the relative positions of other coordinated ligands.

Explanation: \(\textbf{Initial orbital occupancy:}\) A \(d^3\) ion in its ordinary octahedral ground state has configuration: \[ t_{2g}^{3}e_g^{0} \] Each \(t_{2g}\) electron contributes \(-0.4\Delta_o\). Therefore: \[ E_i=3(-0.4\Delta_o) \] \[ E_i=-1.2\Delta_o \] Substituting: \[ E_i=-1.2(18\,000) \] \[ E_i=-21\,600\,\mathrm{cm^{-1}} \] \(\textbf{Occupancy after excitation:}\) One electron moves into the higher-energy \(e_g\) set. The excited occupancy becomes: \[ t_{2g}^{2}e_g^{1} \] The remaining two \(t_{2g}\) electrons can occupy separate lower orbitals. The promoted electron occupies an \(e_g\) orbital. Thus, all three electrons may remain unpaired. \(\textbf{Excited-state crystal-field energy:}\) \[ E_f=2(-0.4\Delta_o)+1(+0.6\Delta_o) \] \[ E_f=-0.2\Delta_o \] Substituting: \[ E_f=-0.2(18\,000) \] \[ E_f=-3\,600\,\mathrm{cm^{-1}} \] \(\textbf{Required excitation energy:}\) \[ \Delta E=E_f-E_i \] \[ \Delta E=-3\,600-(-21\,600) \] \[ \Delta E=18\,000\,\mathrm{cm^{-1}} \] \(\textbf{Electronic interpretation:}\) The increase equals the separation between the two orbital sets. The number of d electrons remains three because excitation does not remove an electron. The number of unpaired electrons can also remain three in the specified arrangement. Therefore, a measurable d–d excitation need not produce a change in the ordinary unpaired-electron count. \(\textbf{Final answer:}\) \[ \boxed{E_i=-21\,600,\quad E_f=-3\,600,\quad \Delta E=18\,000\ \mathrm{cm^{-1}}} \]

Explanation: P has three electrons in the three lower-energy \(t_{2g}\) orbitals. They occupy separate orbitals in the ordinary ground-state arrangement. Thus, P has three unpaired electrons, and its claim is correct. Q has six electrons in the three \(t_{2g}\) orbitals. Each lower orbital contains two electrons of opposite spin. The entity is therefore diamagnetic, so Q is correct. R has configuration \(t_{2g}^{4}e_g^{2}\). Four electrons occupy the three \(t_{2g}\) orbitals, creating one pair and leaving two electrons unpaired in that set. The two \(e_g\) electrons occupy separate orbitals in the high-spin arrangement. They contribute two additional unpaired electrons. Consequently, R has four unpaired electrons in total. The claimed count of two ignores the \(e_g\) contribution. S has nine d electrons arranged as \(t_{2g}^{6}e_g^{3}\). The lower set is fully paired. Three electrons occupy the two upper orbitals, so one orbital contains a pair and the other contains a single electron. Therefore, S has one unpaired electron, and its claim is correct. The incorrect assignment is R alone. The comparison requires examining how electrons occupy individual orbitals within each set rather than treating the superscript as a direct unpaired-electron count.

Explanation: \(\textbf{Charge-neutrality requirement:}\) The chromium-containing coordination cation has charge \(+3\). The iron-containing coordination anion has charge \(-4\). The least common multiple of the charge magnitudes is twelve. Therefore, four chromium cations are required: \[ 4(+3)=+12 \] Three iron-containing anions are required: \[ 3(-4)=-12 \] The charges cancel. The smallest neutral formula is: \[ \mathrm{[Cr(NH_3)_6]_4[Fe(CN)_6]_3} \] \(\textbf{Chromium electronic contribution:}\) Chromium(III) has configuration \(d^3\). Its ordinary octahedral occupancy is: \[ t_{2g}^{3}e_g^{0} \] Each chromium centre has three unpaired electrons. There are four chromium centres per formula unit. Therefore: \[ N_{\mathrm{unpaired,Cr}}=4(3) \] \[ N_{\mathrm{unpaired,Cr}}=12 \] \(\textbf{Iron electronic contribution:}\) Iron(II) has configuration \(d^6\). The specified low-spin arrangement is: \[ t_{2g}^{6}e_g^{0} \] All six d electrons are paired. Each iron centre contributes zero unpaired electrons. The three iron centres therefore contribute: \[ N_{\mathrm{unpaired,Fe}}=3(0)=0 \] \(\textbf{Combined total:}\) \[ N_{\mathrm{unpaired,total}}=12+0 \] \[ N_{\mathrm{unpaired,total}}=12 \] \(\textbf{Interpretation:}\) The smallest formula-unit ratio is determined by charge neutrality, not by the numbers of unpaired electrons. The magnetic count must be evaluated after the metal-centre multiplicities have been established. The complete salt is electrically neutral even though it contains paramagnetic chromium-containing coordination ions. The calculated twelve is an unpaired-electron count across the specified formula unit, not its measured bulk magnetic moment. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{[Cr(NH_3)_6]_4[Fe(CN)_6]_3},\quad N_{\mathrm{unpaired}}=12} \]

Explanation: \(\textbf{Initial salt dissociation:}\) Every formula unit of P contains one cobalt coordination cation and two external chloride ions. The initial amount of coordination cations is: \[ n_{\mathrm{Co,total}}=0.080\,\mathrm{mol} \] The initial free chloride amount is: \[ n_{\mathrm{Cl^-,initial}}=2(0.080) \] \[ n_{\mathrm{Cl^-,initial}}=0.160\,\mathrm{mol} \] \(\textbf{Extent of aqua substitution:}\) Seventy-five percent of the cobalt coordination cations undergo substitution. Therefore: \[ n_{\mathrm{converted}}=0.75(0.080) \] \[ n_{\mathrm{converted}}=0.060\,\mathrm{mol} \] The amount of P-type cations remaining is: \[ n_P=0.080-0.060 \] \[ n_P=0.020\,\mathrm{mol} \] The amount of Q-type cations formed is: \[ n_Q=0.060\,\mathrm{mol} \] \(\textbf{Additional free chloride:}\) Each substitution releases one previously coordinated chloride ion. Therefore: \[ n_{\mathrm{Cl^-,released}}=0.060\,\mathrm{mol} \] \(\textbf{Final chloride inventory:}\) \[ n_{\mathrm{Cl^-,final}}=0.160+0.060 \] \[ n_{\mathrm{Cl^-,final}}=0.220\,\mathrm{mol} \] \(\textbf{Charge-balance verification:}\) The remaining P-type cations carry charge \(+2\). Their total positive-charge amount is: \[ 2(0.020)=0.040\,\mathrm{mol\ of\ charge} \] The Q-type cations carry charge \(+3\). They contribute: \[ 3(0.060)=0.180\,\mathrm{mol\ of\ charge} \] The total is \(0.220\,\mathrm{mol}\) of positive charge equivalents. This is balanced by the \(0.220\,\mathrm{mol}\) of free chloride ions. Both coordination cations remain six-coordinate, and cobalt retains oxidation state \(+3\). The change in free chloride therefore arises from ligand substitution rather than redox chemistry. \(\textbf{Final answer:}\) \[ \boxed{n_P=0.020,\quad n_Q=0.060,\quad n_{\mathrm{Cl^-,free}}=0.220\ \mathrm{mol}} \]

Explanation: Both coordination entities contain iron(III). The metal-ion d-electron configuration is consequently \(d^5\) in both cases. In P, all six coordinated water molecules are neutral. The coordination entity therefore carries charge \(+3\). Its high-spin octahedral configuration is \(t_{2g}^{3}e_g^{2}\). All five d electrons occupy separate orbitals. Thus, P contains five unpaired electrons. In Q, all six cyanido ligands are anionic. They contribute total charge \(-6\), giving the iron(III) coordination entity charge \(-3\). The strong-field cyanido environment favours low-spin occupation. The corresponding configuration is \(t_{2g}^{5}e_g^{0}\). Two lower orbitals contain electron pairs, and one contains a single electron. Thus, Q contains one unpaired electron. Both entities remain paramagnetic, although their spin-only magnetic moments differ. The opposite signs of their coordination-entity charges arise from their ligand charges. They do not require iron to be oxidised or reduced. The example illustrates why oxidation state, entity charge and electron-pairing behaviour must be determined independently.

Explanation: \(\textbf{Distinguishing the stereochemical contributions:}\) The cis arrangement of the stated octahedral coordination entity can exhibit optical isomerism. Its two enantiomers contribute equal and opposite rotations under identical conditions. The trans form is achiral and contributes no optical rotation. \(\textbf{Total cis fraction:}\) The trans fraction is 0.30. Therefore, the combined fraction of the two cis enantiomers is: \[ x+y=1-0.30 \] \[ x+y=0.70 \] Here, \(x\) is the cis-(+) fraction and \(y\) is the cis-(−) fraction. \(\textbf{Optical-rotation equation:}\) The pure cis-(+) enantiomer gives \(+10.0^\circ\) at the reference concentration. The cis-(−) enantiomer gives the corresponding negative rotation. Thus: \[ \alpha=10.0x-10.0y \] Substituting the observed rotation: \[ -3.0=10.0(x-y) \] \[ x-y=-0.30 \] \(\textbf{Solving the simultaneous constraints:}\) \[ x+y=0.70 \] \[ x-y=-0.30 \] Adding: \[ 2x=0.40 \] \[ x=0.20 \] Therefore: \[ y=0.70-0.20 \] \[ y=0.50 \] \(\textbf{Verification:}\) The three fractions sum to unity: \[ 0.20+0.50+0.30=1.00 \] The predicted rotation is: \[ \alpha=10.0(0.20-0.50) \] \[ \alpha=-3.0^\circ \] The negative rotation establishes that the cis-(−) contribution exceeds the cis-(+) contribution. However, the separately supplied trans fraction is essential for finding their individual percentages. The chloride counter-ion composition remains the same for all three stereoisomers. \(\textbf{Final answer:}\) \[ \boxed{\text{cis-(+)}=20\%,\quad\text{cis-(−)}=50\%,\quad\text{trans}=30\%} \]

Explanation: Both coordination entities have overall charge \(-2\). Each contains four anionic ligands, contributing combined charge \(-4\). Therefore, nickel has oxidation state \(+2\) in both cases. Nickel(II) has configuration \(3d^8\). The d-electron count is consequently identical. However, the coordinated ligands are chemically different. Cyanido is a strong-field ligand in the standard spectrochemical series. The ordinary tetracyanidonickelate(II) ion adopts a square-planar arrangement. Its electrons are paired in the elementary bonding description, making it diamagnetic. Chlorido is a weaker-field ligand. The ordinary tetrachloridonickelate(II) ion adopts tetrahedral geometry. Its \(d^8\) configuration contains two unpaired electrons. It is consequently paramagnetic. The difference in magnetic behaviour therefore does not require oxidation or reduction of nickel. Instead, it arises from different ligand environments and the associated geometrical and orbital arrangements. Coordination number four alone does not determine which geometry will occur. The example shows that identical metal oxidation states can coexist with different hybridisation and magnetic classifications.

Explanation: \(\textbf{Initial coordination inventory:}\) The solution contains: \[ n_{\mathrm{complex}}=0.050\,\mathrm{mol} \] Each coordination entity has four ammonia ligands. Therefore, the total coordinated ammonia amount is: \[ n_{\mathrm{NH_3}}=4(0.050) \] \[ n_{\mathrm{NH_3}}=0.200\,\mathrm{mol} \] \(\textbf{Protonation of released ammonia:}\) Every ammonia molecule accepts one hydrogen ion: \[ \mathrm{NH_3+H^+\rightarrow NH_4^+} \] One mole of ammonia consequently consumes one mole of hydrogen ions. Thus: \[ n_{\mathrm{H^+,consumed}}=0.200\,\mathrm{mol} \] \(\textbf{Ammonium formation:}\) Each protonated ammonia molecule produces one ammonium ion. Therefore: \[ n_{\mathrm{NH_4^+,formed}}=0.200\,\mathrm{mol} \] \(\textbf{Copper electronic structure:}\) Copper remains in oxidation state \(+2\) throughout the process. Its metal-ion d-electron configuration consequently remains \(3d^9\). Protonation and removal of coordinated ammonia do not automatically reduce or oxidise the metal. The ordinary \(d^9\) configuration contains one unpaired electron. \(\textbf{Coordination-environment interpretation:}\) The copper coordination environment changes because ammonia donor attachments are replaced by water-based coordination. This can alter the electronic absorption spectrum and observed colour. However, the stipulated unchanged oxidation state preserves the formal copper d-electron count. The reaction therefore couples acid–base chemistry with ligand substitution without requiring a redox process. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{H^+,consumed}}=0.200\,\mathrm{mol},\quad n_{\mathrm{NH_4^+,formed}}=0.200\,\mathrm{mol},\quad d^9} \]