Co-ordination compounds practice · Mixed difficulty

Option C is correct because Lithium tetrahydridoaluminate, commonly known as Lithium aluminum hydride, is correctly represented as Li[AlH₄].

Option A (2) is correct because the iron in hemoglobin is in the +2 oxidation state, allowing it to bind reversibly with oxygen.

Option A (5.92) is correct because the magnetic moment is calculated using the number of unpaired electrons in the complex ion, which is 5 for [FeF6]3-.

Option A (${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$) is correct because cyanide (CN$^-$) is a strong field ligand, resulting in a larger crystal field splitting energy (${\Delta _0}$).

Option C ($\mathrm{ds}{\mathrm{p}^2}$) is correct because the complex is square planar, requiring dsp² hybridization due to the strong field ligand.

Option C (Tetradentate) is correct because the ligand has one nitrogen atom and three amine groups, each capable of donating a lone pair, resulting in four donor atoms.

Option C (As anticancer agent) is correct because cis-[PtCl2(NH3)2] (Cisplatin) is a well-known anticancer drug.

Option B ($\left[{Ni}({CN})_4\right]^{2-}$) is correct because it is a square planar complex with strong field ligands, resulting in pairing of electrons and thus diamagnetism.

Option A is correct because none of the ligands in the set ($\mathrm{C}_2 \mathrm{O}_4^{2-}$, ethylene diammine, $\mathrm{H}_2 \mathrm{O}$) are ambidentate ligands.

The correct answer is 35 because the Effective Atomic Number (EAN) is calculated by subtracting the oxidation state of the central metal atom from its atomic number and adding twice the number of ligands.

Option D (Dianionic complex with coordination number four) is correct because lead chromate ($PbCrO_4$) reacts with hot NaOH to form the tetrahydroxoplumbate(II) ion, which is a dianionic complex, $[Pb(OH)_4]^{2-}$, with a coordination number of four.

The correct answer is B (Diamagnetic, diamagnetic and paramagnetic) because Li2O and Na2O2 are diamagnetic while KO2 is paramagnetic due to the presence of unpaired electrons in O2-.

Option D (4) is correct because in $CuSO_4.5H_2O$, four water molecules are directly coordinated to the copper ion, forming the complex $[Cu(H_2O)_4]^{2+}$.

Option C ($\left[{Co}({NH}_3)_4 {Cl}_2\right] {Cl}$) is correct because it has one ionizable chloride ion, which precipitates with silver nitrate, and it exhibits cis-trans isomerism.

Option A (Fe(CO)5) is correct because the oxidation state of Fe in this compound is 0, which is the lowest among the given options.

Option A (If both assertion and reason are true and the reason is the correct explanation of the assertion) is correct because isomorphous substances have similar constitution and chemical formulae, leading to similar crystal shapes and the ability to grow in each other's saturated solutions.

Option B (A and C only) is correct because Chlorophyll contains Mg, not Co, and Photography involves AgBr, not [Ag(CN)₂]⁻.

Option A (${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }$) is correct because Mn in this complex has a +7 oxidation state, meaning it has no d-electrons.

The correct answer is B (Difluorido bis-(ethylene diamine) chromium (III) chloride) because it follows the IUPAC naming conventions for coordination compounds.

Option D ($\left[{CO}\left({NH}_3\right)_3 {Cl}_3\right]$) is correct because a complex with zero molar conductance does not ionize in solution.

Option C (Give the test for ${SO}_4^{2-}$ ion) is correct because the sulfate ion is a simple ion outside the coordination complex and will give its characteristic test.

Option A, $\left[ {CO{{(N{H_3})}_3}C{l_3}} \right]$, is the correct answer because $M{a_3}{b_3}$ complexes in octahedral geometry can exist as fac- and mer- isomers, but neither is chiral and thus does not exhibit optical isomerism.

Option C (number of ligand donor atoms to which the metal is directly bonded) is correct because the coordination number is defined as the number of ligands directly attached to the central metal ion through coordinate bonds.

Option A is correct because it represents the correct order of ligand field strength in the spectrochemical series.

Option C is correct because complexes of $\mathrm{MX}_6$ and $\mathrm{MX}_3 \mathrm{L}$ type do not show geometrical isomerism, but the reason that geometrical isomerism is not shown by complexes of coordinate number -6 is false, as some complexes with coordination number 6 do exhibit geometrical isomerism.

Option D (All) is correct because Diphos, Tetrameen, and tn are all bidentate ligands, meaning they can coordinate to a metal ion through two donor atoms.

Option D is correct because the complex $\left[\mathrm{Co}(\mathrm{SCN})_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}$ does not exhibit optical isomerism due to the presence of a plane of symmetry.

Option A ($36$) is correct because the Effective Atomic Number (EAN) is calculated as the atomic number of the metal minus its oxidation state, plus twice the coordination number.

Option C (+3) is correct because the oxidation number of Cr in the complex is +3, calculated by balancing the charges of the ligands and the overall complex.

Option A (High spin complexes) is correct because halide ligands are weak field ligands, resulting in small crystal field splitting and favoring high spin complexes.