Co-ordination compounds practice · Mixed difficulty

Option D (5 ions) is correct because the complex dissociates into one complex ion, $[\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}]^{4+}$, and four chloride ions, $4\mathrm{Cl}^-$.

Option D is correct because potassium ferrocyanide, K4[Fe(CN)6], dissociates into 5 ions: 4 K+ ions and 1 [Fe(CN)6]4- ion.

Option A (1) is correct because the magnetic moment of 1.7 BM indicates the presence of one unpaired electron in the complex.

Option A (${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^ + }$) is correct because the positive charge on the complex reduces back-bonding, leading to a stronger C-O bond.

Option A is correct. EDTA is a hexadentate ligand, meaning it can occupy six coordination sites. Therefore, all six coordinate sites of Mg2+ are occupied by EDTA.

The correct answer is D (Linear) because the complex ${\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]^ + }$ has a coordination number of 2, resulting in a linear geometry.

Option B is correct. The complexes are linkage isomers because the $NO_2$ ligand can coordinate through either the nitrogen (nitro) or the oxygen (nitrito) atom.

Option B ([Co(en){{(N{H_3})}_2}C{l_2}]Cl) is correct because it can exhibit cis-trans isomerism, and the cis isomer can further exhibit optical isomerism, leading to four isomers.

Option A (2) is correct because the iron complex formed, likely containing oxalate ligands, can exhibit optical isomerism due to its non-superimposable mirror images.

Option A (Tetramminedichloroplatinum(IV) tetrachloro platinate(II)) is correct because it accurately names both the cationic and anionic complexes, including the correct oxidation states of platinum in each.

Option B is correct. Both the assertion and reason are true, but the reason is not the correct explanation for the assertion because the reducing nature is due to the ability to be oxidized to a more stable state, not merely the presence of unpaired electrons.

Option B (${\left[ {\mathrm{Ru}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$) is correct because Ru has a higher charge and CN- is a strong field ligand, leading to a larger crystal field splitting.

Option B is correct. $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_4}$ will give a white precipitate with $\mathrm{AgN}{\mathrm{O}_3}$ because all four chloride ions are outside the coordination sphere and will precipitate as AgCl.

Option B is correct. ${\left[ {\mathrm{Ni}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ has the minimum paramagnetic behavior because it has the least number of unpaired electrons among the given options.

Option A (III > I > II) is correct because the crystal field splitting energy depends on the charge of the metal ion and the strength of the ligand. Higher charge and stronger ligands lead to greater splitting.

Option D ($V{(CO)_6}$) is correct because it is the only complex that is paramagnetic (odd number of electrons) and thus readily accepts an electron from sodium.

Option C (${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$) is correct because CN- is a strong field ligand, leading to a high crystal field splitting energy and a more stable complex.

Option D (Less availability of $\mathrm{A}{\mathrm{g}^ + }$ ion as $\mathrm{Cu}$ cannot displace $\mathrm{Ag}$ from $\mathrm{Ag}{\left( {\mathrm{CN}} \right)_2}$) is correct because the complex ion ${\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }$ provides a lower concentration of free $Ag^+$ ions, leading to a more uniform and adherent silver plating.

Option D ($\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$) is correct because the oxidation state of Cr in this compound is +6.

Option A (1) is correct because the change in the number of CO and NO ligands must balance to maintain the same EAN.

Option C ($\mathrm{Mn}$) is correct because Manganese is known to form polynuclear carbonyl complexes due to its ability to form metal-metal bonds.

The correct answer is Tetrammine dichloride platinum [IV] chloride because it correctly names the complex according to IUPAC nomenclature rules, including the oxidation state of platinum.

Option B (${\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^ + }$) is correct because it can exhibit both geometrical (cis and trans) and optical isomerism (for the cis isomer).

Option A ($\mathrm{Fe}{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)_2}$) is correct because it accurately represents ferrocene, where iron is coordinated to two cyclopentadienyl rings in a pentahapto (η5) fashion.

Option C is correct because $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}$ can exhibit linkage isomerism due to the ambidentate ligand $NO_2^-$.

Option B (Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$) is correct because both the assertion and the reason are true statements, but the reason (presence of a center of symmetry in octahedral complexes) does not explain why tetrahedral complexes have more intense absorption bands.

Option C ($Cis - {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}{{\left( {\mathrm{en}} \right)}_2}} \right]^{3 + }}$) is correct because the cis isomer of this complex lacks a plane of symmetry and is therefore chiral.

Option B (0.001) is correct because only the chloride ions outside the coordination sphere will precipitate as AgCl.

Option D ($\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$) is correct because the ferric ions are complexed within the coordination compound and do not exist as free ions in solution, thus not giving normal tests for ferric ions.

Option A (Assertion and Reason both are true, Reason is correct explanation of Assertion) is correct because linkage isomerism arises when an ambidentate ligand coordinates to a metal ion through different donor atoms.