Co-ordination compounds practice · Mixed difficulty

Option D (Assertion is false, Reason is true) is correct because the complex does not give a white precipitate with AgNO3, indicating the chloride ions are not ionizable.

Option B (Assertion and Reason both are true but Reason is not the correct explanation of Assertion) is correct because both complexes are reducing agents due to their ability to form more stable complex ions after gaining an electron, but the presence of unpaired electrons is not the direct reason.

Option A (+6 and +3) is correct because the coordination number of Cr is 6 (three bidentate ligands) and its oxidation state is +3.

Option A is correct because a positive charge on the metal center increases the back-bonding from the metal to the CO ligand, strengthening the C–O bond.

Option D is correct. [Cr(SCN)₂(NH₃)₄]⁺ exhibits both geometrical (cis-trans) and linkage isomerism, leading to a maximum number of isomers.

Option A (Of the ${d^4}$ species, manganese (III) is strongly reducing while $C{r^{2 + }}$ is strongly oxidising) is false because Mn(III) is oxidizing and Cr(II) is reducing.

Option A ($\mathrm{Al(OC_2H_5)_3}$) is correct because in aluminum ethoxide, the aluminum atom is bonded to oxygen atoms, not directly to carbon.

Option A is correct because, according to Werner's theory, primary valencies (oxidation state) are ionizable, corresponding to the counter ions in the coordination complex.

The correct answer is D (If both the Assertion and Reason are incorrect) because the complex with ethylenediamine is more stable due to chelation, and its geometry is octahedral, not trigonal bipyramidal.

Statement I and statement II are correct and statement II is the correct explanation of statement I because cis isomers allow for bidentate ligands to occupy adjacent positions, facilitating the reaction.

Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because while both the assertion and reason are true, the reason doesn't explain why hexachloroplatinate is a complex anion.

Option C ($[Co{(N{H_3})_5}Br]B{r_2}$) is correct because it is the only complex that produces 3 ions in solution and 2 moles of AgBr with 2 moles of AgNO₃.

Option B (If both (A) and (R) are correct but (R) is not correct explanation for (A)) is correct because both the assertion and reason are true, but the hybridization doesn't directly explain the diamagnetic nature.

Option B (${\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{3 - }}$) is correct because CuS${\mathrm{O}_4}$ reacts with KCN to form the complex ion ${\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{3 - }}$.

Option C ($e_{\mathrm{g}}=t_{2 \mathrm{~g}}$) is correct because in a spherically symmetrical field, all d-orbitals remain degenerate, meaning they have the same energy.

Option C ($\left[ {CrC{l_2}{{\left( {{H_2}O} \right)}_4}} \right]Cl.2{H_2}O$) is correct because 1/3 of the chlorine is precipitated, indicating one Cl- ion is outside the coordination sphere.

Option D (Assertion is false, but Reason is true) is correct because the chelate effect makes [Pt(en)2Cl2]2+ more stable than [Pt(NH3)4Cl2]2+.

Option C ([Cr(NH3)3Cl3]) is correct because it is a neutral complex that does not dissociate into ions in aqueous solution, resulting in the lowest electrical conductance.

Option D is correct because the assertion is false (Cu2+ and Cd2+ do form stable complexes with CN-), but the reason is true ([Cu(CN)4]3- is not decomposed by H2S).

The correct answer is D ([Cu(NH3)4]2+ : sp3) because [Cu(NH3)4]2+ has a dsp2 hybridization, not sp3.

Option D (IV $>$ I $>$ II $>$ III) is correct because the wavelength of light absorbed is inversely proportional to the ligand field strength, and the spectrochemical series dictates the order of ligand field strength.

The correct answer is B (${sp}^3, {dsp}^2$) because ${Ni}({CO})_4$ is tetrahedral and ${sp}^3$ hybridized, while $\left[{Ni}({CN})_4\right]^{2-}$ is square planar and ${dsp}^2$ hybridized.

The correct answer is A (6) because the complex compound [Cu(NH3)4][PtCl4] can form six different isomers through ligand exchange between the two coordination complexes.

The correct answer is A because both statements are correct, and Statement II correctly explains why SnCl3- acts as a good ligand while SnCl2 does not.

Option C is correct because the assertion is true, but the reason is false. $\left[{PtCl}_2\left(e{n}_2\right)\right]^{+2}$ shows geometrical isomerism (cis and trans), but the trans isomer does not show optical activity because it has a plane of symmetry.

Option C (Zero) is correct because ${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$ is a low spin complex with no unpaired electrons.

Option B ($4,5,6$) is correct because it provides the correct number of carbonyl ligands for Ni, Fe, and Cr to satisfy the Effective Atomic Number (EAN) rule.

Option D (All of these) is correct because each reagent (BaCl2, AgNO3, PbCl2) will react differently with the two complexes to form different precipitates, allowing them to be distinguished.

Option C ($\left[{Ni}({Cl})_4\right]^{2-}$) is correct because it forms a tetrahedral complex with weak field ligands, resulting in unpaired electrons and thus paramagnetism.

The correct answer is C ([Co(C2O4)3]3-) because in this complex, cobalt has a d6 configuration with strong field ligands, leading to pairing of electrons and a diamagnetic nature.