Co-ordination compounds practice · Mixed difficulty

Option D (All are correct) is correct because CO, NO2-, and CN- are all strong field ligands that produce high crystal field splitting.

Option B ($4.7 \times 10^{-14}$) is correct because the overall dissociation constant is the inverse of the overall stability constant ($\beta_4$).

Option A ([Pt(NH3)6]Cl4) has the highest molar conductivity because it dissociates into the largest number of ions in solution.

Option B is correct because cobalt hexaammine cation, [Co(NH3)6]+3, has six ligands coordinated to the central cobalt atom, resulting in an octahedral geometry.

Option A is correct because for a square planar complex $[M(F)(Cl)(SCN)(N{O_2})]$, the number of isomers is 4!/1 = 24/2 = 12, since there is no symmetry.

The correct answer is C (Excitation of electrons from $d - d$) because the color of the complex ${\left[ {\mathrm{Ti}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ is due to $d-d$ transitions.

The correct answer is C ($Co(NH_3)_4 Cl_3 < K_2Pt.Cl_6 < Cr(NH_3)_6 Cl_3 < Pt(NH_3)_6 Cl_4$) because the conductivity depends on the number of ions produced in the solution.

Option A (Assertion and Reason both are true, Reason is correct explanation of Assertion) is correct because chelating ligands form more stable complexes (chelates) with metal ions, effectively removing toxic metal ions from the system.

Option D (All are correct) is correct because all the given complexes have weak field ligands or are in a high oxidation state, leading to high spin configurations.

Option D (Pentaamminechloridocobalt(III) chloride) is correct because it dissociates into 3 ions: [Co(NH3)5Cl]2+ and 2 Cl- ions.

Option D ( Potassium carbonylpentacyanoferrate (II)) is correct because it correctly names the ligands and the oxidation state of iron according to IUPAC nomenclature.

Option D (In acidic solutions, protons coordinate with ammonia molecules forming $\mathrm{NH}_4^ +$ ions and $\mathrm{N}{\mathrm{H}_3}$ molecules are not available) is correct because the presence of protons in acidic solution consumes ammonia, preventing it from acting as a ligand.

Option A (6) is correct because the sum of geometrical isomers for the given complexes is 6.

Option A (${SCN}^{-} < {F}^{-} < {CN}^{-} < {CO}$) is correct because it represents the ligands in increasing order of their field strength as per the spectrochemical series.

Option B (33) is correct because the Effective Atomic Number (EAN) is calculated by adding the metal's atomic number to the electrons donated by the ligands.

The correct answer is C (${Cl}^{-} < {F}^{-} < {H}_2 {O} < {NH}_3 < {CN}^{-} < {CO}$) because it correctly represents the increasing order of crystal field splitting strength based on the spectrochemical series.

Option A is correct. The complex $[CoCl_3(NH_3)_3]$ does not ionize in solution because all the ions are within the coordination sphere.

Option B (${I^ - } < O{H^ - } < NC{S^ - }$) is correct because it represents the correct increasing order of field strength for the given ligands based on the spectrochemical series.

Option D is correct. The existence of two different colored complexes is due to Ionisation isomerism.

Option C (6) is correct because triethylenediamine (en) is a bidentate ligand, meaning each en molecule coordinates through two nitrogen atoms. Thus, three en molecules coordinate to the cobalt(III) ion, resulting in a coordination number of 6.

Option A (${I^ - } < C{l^ - } < {H_2}O < C{N^ - }$) is the correct order because it follows the spectrochemical series, which ranks ligands by their field strength.

Option D is correct. The assertion is false because the complex with ethylenediamine (en) is more stable due to the chelate effect, and the reason is true because the complex with en exhibits the chelate effect.

Option A is correct because ${\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]^ + }$ can exist as cis and trans isomers, showing geometrical isomerism.

Option B ($\mathrm{C}{\mathrm{u}^2} + 4\mathrm{CN}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}\mathrm{\;};\left( {K = 2.0 \times {{10}^{27}}} \right)$) is correct because the stability constant (K) is highest for this reaction, indicating the strongest ligand.

Option B ($Mg$) is correct because the central atom in chlorophyll is magnesium (Mg).

Option B (Hexaamine cobalt (III) tris (oxalato ) chromate(III)) is correct because it correctly names both the cationic and anionic complexes, including the correct oxidation states of cobalt and chromium.

Option B (Tetraaquodichlorochromium(III) chloride) is correct because it follows the IUPAC naming conventions for coordination complexes, correctly naming the ligands, central metal, oxidation state, and counter ion.

Option A (In the formation of '${X}$', ${CuSO}_4$ acts as reducing agent) is correct because in the reaction, $CuSO_4$ is reduced to $CuCN$, hence it acts as an oxidizing agent, not a reducing agent.

Option D is correct. The complex exhibits linkage isomerism because the $NO_2$ ligand can coordinate through either the nitrogen or oxygen atom.

Option A (287 × 10^-3) is correct because only one chloride ion is outside the coordination sphere, leading to the precipitation of only that one chloride ion as AgCl.