Co-ordination compounds practice · Mixed difficulty

Option A ($\mathrm{C}{\mathrm{N}^ - }$) is correct because cyanide ($CN^−$) is the strongest field ligand among the given options based on the spectrochemical series.

Option A is correct because $\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {SCN} \right)}_2}} \right]$ and $\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {NCS} \right)}_2}} \right]$ differ in the mode of attachment of the $SCN^-$ ligand, exhibiting linkage isomerism.

Option D is correct because the two complexes will yield different ions upon electrolysis, leading to different conductivity changes.

Option A ($\mathrm{NaO}{\mathrm{C}_2}{\mathrm{H}_5}$) is correct because it is an ionic compound (sodium ethoxide) with no direct metal-carbon bond, thus not an organometallic compound.

The correct answer is C (2). The complex $[Co{(N{H_3})_3}C{l_3}]$ can exist as two isomers: fac (facial) and mer (meridional).

Option C ($Cis-$ $\left[{Co}({en})_2 {Cl}_2\right]^{+}$) is correct because the cis isomer of this complex is chiral and can therefore exhibit optical isomerism.

The correct answer is C (${\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }$). Boron in ${\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }$ has four bond pairs and no lone pairs, resulting in a regular tetrahedral geometry.

Option C is correct because the coordination number of M is 6 and the oxidation number is +3, summing to 9.

Option A is correct because both the assertion and the reason are true, and the reason correctly explains why tetrahedral complexes do not exhibit geometrical isomerism.

Option A (Diamagnetic) is correct because in AgO, Ag is in the +2 oxidation state ($d^9$ configuration), but it forms a complex where the unpaired electron is involved in bonding or pairing, resulting in diamagnetism.

Option D ($\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{C}{\mathrm{l}_3}} \right]$) is correct because in this complex, all the chloride ions are coordinated inside the coordination sphere and hence do not ionize to give free chloride ions that can react with silver nitrate to form a white precipitate.

Option C ($[{(P{h_3}P)_3}RhCl]$) is correct because it accurately represents Wilkinson's catalyst, which is a rhodium complex with three triphenylphosphine ligands and one chloride ligand.

Option D (${\left[ {\mathrm{Mn}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$) is correct because it has the maximum number of unpaired electrons (5), leading to the highest paramagnetic character.

Option B (Tetraammine dichloro cobalt(III) chloride) is correct because it follows the IUPAC naming conventions for coordination complexes, listing ligands alphabetically before the metal and indicating the oxidation state of the metal.

Option D (Aluminium(III)chloride-4-ethanol) is correct because it correctly names the compound, indicating the aluminum cation, the chloride anion, and the four ethanol ligands.

Option B ($\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$) is correct because $Cr^{3+}$ has a $d^3$ configuration, leading to a larger crystal field splitting compared to other options with weaker ligand field stabilization.

Option A is correct because only three species, namely [NiCl4]2-, Na3[CoF6], and CsO2, are paramagnetic among the given compounds.

Option C (Lewis base) is correct because ligands donate electron pairs to the central metal atom, which is the definition of a Lewis base.

The correct answer is 33 because the EAN of Cr in $\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}$ is calculated as 24 (atomic number) - 3 (oxidation state) + 2(6) (electrons from ligands) = 33.

Option B (Potassium trioxalatoaluminate(III)) is correct because it follows the IUPAC naming rules for coordination compounds.

Option D (6) is correct because two chloride ions are outside the coordination sphere, and they will precipitate with silver nitrate, requiring 6 mL of 0.1 M AgNO3.

The correct answer is B ($\left[{Cr}({CO})_6\right]$). Carbonyl (CO) is a neutral ligand, so if the overall complex has no charge, the metal's oxidation state must be zero.

Option D (All are correct) is correct because all the given complexes, ${\left[ {\mathrm{Zn}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$, $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$, and ${\left[ {\mathrm{Cd}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$, possess tetrahedral shape with $s{p^3}$ hybridization.

The correct answer is D ($\mathrm{N}{\mathrm{H}_3}$). Ammonia is a borderline ligand and can form both low-spin and high-spin complexes depending on the metal ion and the geometry.

Option A (diaquasilver(I) dicyanidoargentate(I)) is correct because it follows the IUPAC naming rules for complex ions, with correct oxidation states.

Option A (D-penicillamine) is correct because it is a chelating agent used to remove excess copper and iron in chelate therapy, particularly in Wilson's disease.

Option B is correct because it follows IUPAC nomenclature rules, correctly naming the complex salt with proper oxidation state and hydration.

Option A (+6 and +3) is correct because in ${\mathrm{K}_3}\left[ {\mathrm{Cr}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]$, the coordination number of Cr is 6 (since each oxalate ligand is bidentate), and the oxidation state of Cr is +3.

Option B (Assertion and Reason both are true but Reason is not the correct explanation of Assertion) is correct because both the assertion and the reason are true statements, but the reason (lability of ligands) does not explain why chelated complexes are more stable.

Option C (6) is correct because the coordination number of Cobalt in $[Co(NH_3)_6]^{2+}$ is 6, as there are six ammonia ligands directly attached to the central cobalt ion.