Co-ordination compounds practice · Easy

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.

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

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.

Option D ([CO(NH3)3Cl3]) is correct because it does not have any chloride ions outside the coordination sphere to react with AgNO3 to form a white precipitate of AgCl.

Option A is correct because when gold dissolves in aqua regia, it forms the complex ion $[AuCl_4]^−$, which is named tetrachloridoaurate(III).

The correct answer is C. The complex dissociates into three ions: one complex ion and two chloride ions.

Option C (6) is correct because the coordination number is the number of ligands directly attached to the central metal atom, and in this case, there are two ethylenediamine (en) ligands (each donating 2) and two bromide ions (each donating 1).

Option B (${\left[ {\mathrm{Fe}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}$) is correct because it obeys the 18-electron rule, having 18 valence electrons.

Option A (2) is correct because the complex dissociates into two ions: K+ and [Pt(NH3)5Cl5]-.

Option D (${\mathrm{C}{\mathrm{u}^{2 + }}}$) is correct because ${\mathrm{C}{\mathrm{u}^{2 + }}}$ exhibits Jahn-Teller distortion, which enhances the stability of its complexes.

Option C (4) is correct because tetrahedral complexes have a coordination number of 4, meaning the central metal atom is directly bonded to four ligands.

Option C ($Cis - { {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}{{\left( {\mathrm{en}} \right)}_2}} }^{3 + }$) is correct because only cis-octahedral complexes can exhibit optical activity due to their non-superimposable mirror images.

Option C (4) is correct because tetrahedral complexes have a coordination number of 4.

Option A (Ionisation) is correct because the isomers differ in the counter ion that is coordinated to the metal center, leading to different ions in solution.

Option D (Hexadentate ligand) is correct because ${\left[ {\mathrm{EDTA}} \right]^{4 - }}$ can donate six pairs of electrons to the central metal atom, making it a hexadentate ligand.

Option C is correct because ionization isomerism involves the exchange of ligands between the coordination sphere and the ionization sphere.

Option D (2 and 4) is correct because the oxidation state of Ni is +2 and the coordination number is 4.

Option C ($\mathrm{Mn}$) is correct because manganese forms polynuclear carbonyls such as $\mathrm{M}{\mathrm{n}_2}{\left( {\mathrm{CO}} \right)_{10}}$.

The correct answer is D (5) because the complex [MCl2Br2]SO4 can have five isomers: three geometrical isomers (cis, trans, and another trans) and two ionization isomers.

Option D (All of the above) is correct because metals in coordination complexes possess primary valencies satisfied by anions, secondary valencies satisfied by donor molecules, and a coordination number.

Option C (Both $\mathrm{N}{\mathrm{H}_3}\mathrm{\;and\;C}{\mathrm{l}^ - }$) is correct because both ammonia and chloride ions are directly bonded to the central metal ion.

The correct answer is ${\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}$ because Ni in this complex has $s{p^3}{d^2}$ hybridization, which is an outer orbital complex.

Option A (Both (I) and (II) are paramagnetic) is correct because both complexes have unpaired electrons due to F- being a weak field ligand, resulting in high-spin configurations.

Option C (Ferric ion) is correct because ${\mathrm{K}_4}[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} ]$ reacts with ferric ions ($Fe^{3+}$) to form Prussian blue, a characteristic colored precipitate.

Option C (Sodium hexanitrito cobaltate (III)) is correct because it follows the IUPAC naming conventions for coordination compounds, indicating the ligands and oxidation state of the central metal atom.

Option D 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 ${\left[ {\mathrm{Cd}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$, have tetrahedral shape with $s{p^3}$ hybridization.

Option A (Colourless and diamagnetic) is correct because Sc3+ has a d0 electronic configuration, meaning it has no unpaired electrons and therefore does not absorb visible light and is diamagnetic.