Every Book Back multiple-choice question from Coordination Chemistry (12th Standard Chemistry, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
The sum of primary valence and secondary valence of the metal M in the complex \([M(en)_2(Ox)]Cl\) is
- A. 3
- B. 6
- C. -3
- D. 9Correct
Explanation. The primary valence is the oxidation state of the metal, which is +3 here. The secondary valence is the coordination number, which is 6 due to two bidentate ethylenediamine and one bidentate oxalate ligands. Their sum is nine.
Q2
An excess of silver nitrate is added to 100ml of a 0.01M solution of pentaaquachloridochromium(III)chloride. The number of moles of AgCl precipitated would be
- A. 0.02
- B. 0.002Correct
- C. 0.01
- D. 0.2
Explanation. The complex formula is \([Cr(H_{2}O)_{5}Cl]Cl_{2}\), which contains two ionizable chloride ions. Since 0.1 liters of 0.01 molar solution contains 0.001 moles of complex, adding excess silver nitrate precipitates twice that amount, resulting in 0.002 moles of silver chloride.
Q3
A complex has a molecular formula \(MSO_4Cl \cdot 6H_2O\). The aqueous solution of it gives white precipitate with Barium chloride solution and no precipitate is obtained when it is treated with silver nitrate solution. If the secondary valence of the metal is six, which one of the following correctly represents the complex?
- A. \([M(H_2O)_4Cl]SO_4 \cdot 2H_2O\)
- B. \([M(H_2O)_6]SO_4\)
- C. \([M(H_2O)_5Cl]SO_4 \cdot H_2O\)Correct
- D. \([M(H_2O)_3Cl]SO_4 \cdot 3H_2O\)
Explanation. A precipitate with barium chloride means sulfate is outside the coordination sphere. No reaction with silver nitrate means chloride is inside. To maintain a coordination number of six, one chloride and five water molecules must act as ligands to the metal.
Q4
Oxidation state of Iron and the charge on the ligand NO in \([Fe(H_2O)_5NO]SO_4\) are
- A. +2 and 0 respectively
- B. +3 and 0 respectively
- C. +3 and -1 respectively
- D. +1 and +1 respectivelyCorrect
Explanation. In the brown ring complex, iron exists in an unusual +1 oxidation state. The nitric oxide ligand transfers one electron to the iron and exists as a nitrosylium cation with a +1 charge to balance the overall charge of the coordination entity.
Q5
As per IUPAC guidelines, the name of the complex \([Co(en)_2(ONO)Cl]Cl\) is
- A. Chlorobisethylenediaminenitritocobalt(III) chloride
- B. Chloridobis(ethane-1,2-diamine)nitro -Ocobaltate(III) chloride
- C. Chloridobis(ethane-1,2-diammine)nitrito -Ocobalt(II) chloride
- D. Chloridobis(ethane-1,2-diammine)nitrito \(\kappa\) -Ocobalt(III)chlorideCorrect
Explanation. Ligands are named alphabetically: chlorido then ethane-1,2-diamine. The bidentate ligand uses 'bis'. The nitrito group linked through oxygen is indicated by kappa-O. Cobalt is in the +3 oxidation state, and the metal is named normally in a cationic complex.
Q6
IUPAC name of the complex \(K_3[Al(C_2O_4)_3]\) is
- A. Potassiumtrioxalatoaluminium(III)
- B. Potassiumtrioxalatoaluminate(II)
- C. Potassiumtrisoxalatoaluminate(III)
- D. Potassiumtrioxalatoaluminate(III)Correct
Explanation. The complex is anionic, so the aluminum name must end in 'ate'. Three oxalate ligands are named trioxalato. The oxidation state of aluminum is +3, calculated by balancing the charges of three potassium ions and three dianionic oxalate ligands.
Q7
A magnetic moment of 1.73BM will be shown by one among the following (NEET)
- A. \(TiCl_4\)
- B. \([CoCl_6]^{4-}\)
- C. \([Cu(NH_3)_4]^{2+}\)Correct
- D. \([Ni(CN)_4]^{2-}\)
Explanation. A magnetic moment of 1.73 Bohr Magnetons indicates the presence of one unpaired electron. The copper complex contains \(Cu^{2+}\) with a \(d^9\) electronic configuration, which naturally has one unpaired electron regardless of the coordination geometry or field strength.
Q8
Crystal field stabilization energy for high spin \(d^5\) octahedral complex is
- A. \(-0.6 \Delta_o\)
- B. 0Correct
- C. \(2(P - \Delta_o)\)
- D. \(2(P + \Delta_o)\)
Explanation. In a high-spin \(d^5\) octahedral configuration, three electrons occupy the \(t_{2g}\) orbitals and two occupy the \(e_{g}\) orbitals. The stabilization from \(t_{2g}\) \((-1.2 \Delta_o)\) and destabilization from \(e_{g}\) \((+1.2 \Delta_o)\) cancel each other out, resulting in zero energy.
Q9
In which of the following coordination entities the magnitude of \(\Delta_o\) will be maximum?
- A. \([Co(CN)_6]^{3-}\)Correct
- B. \([Co(C_2O_4)_3]^{3-}\)
- C. \([Co(H_2O)_6]^{3+}\)
- D. \([Co(NH_3)_6]^{3+}\)
Explanation. The magnitude of crystal field splitting depends on the ligand strength in the spectrochemical series. The cyanide ion is a very strong field ligand, producing the largest splitting energy compared to oxalate, water, or ammine ligands for the same metal ion.
Q10
Which one of the following will give a pair of enantiomorphs?
- A. \([Cr(NH_3)_6][Co(CN)_6]\)
- B. \([Co(en)_2Cl_2]\)Correct
- C. \([Pt(NH_3)_4][PtCl_4]\)
- D. \([Co(NH_3)_4Cl_2]NO_2\)
Explanation. Enantiomorphs are non-superimposable mirror images formed by chiral complexes. The cis-isomer of \([Co(en)_2Cl_2]\) lacks symmetry elements and is optically active, existing as a pair of enantiomers, whereas the other options are either symmetric or constitutional isomers.
Q11
Which type of isomerism is exhibited by \([Pt(NH_{3})_{2}Cl_{2}]\)?
- A. Coordination isomerism
- B. Linkage isomerism
- C. Optical isomerism
- D. Geometrical isomerismCorrect
Explanation. Square planar complexes of the form \(MA_{2}B_{2}\), such as this diamminedichloridoplatinum(II) complex, can exist as cis and trans geometrical isomers depending on the spatial arrangement of the ligands.
Q12
How many geometrical isomers are possible for \([Pt(py)(NH_{3})(Br)(Cl)]\)?
- A. 3Correct
- B. 4
- C. 0
- D. 15
Explanation. For a square planar complex with four different monodentate ligands (type MABCD), three geometrical isomers are possible by fixing one ligand and rotating the other three positions.
Q13
Which one of the following pairs represents linkage isomers?
- A. \([Cu(NH_{3})_{4}][PtCl_{4}]\) and \([Pt(NH_{3})_{4}][CuCl_{4}]\)
- B. \([Co(NH_{3})_{5}(NO_{2})]SO_{4}\) and \([Co(NH_{3})_{5}(ONO)]SO_{4}\)
- C. \([Co(NH_{3})_{4}(NCS)_{2}]Cl\) and \([Co(NH_{3})_{4}(SCN)_{2}]Cl\)Correct
- D. both (b) and (c)
Explanation. Linkage isomerism occurs when an ambidentate ligand like \(SCN^{-}\) bonds through different donor atoms. Pair (c) shows the ligand bonding via nitrogen in one and sulfur in the other.
Q14
Which kind of isomerism is possible for a complex \([Co(NH_{3})_{4}Br_{2}]Cl\)?
- A. geometrical and ionizationCorrect
- B. geometrical and optical
- C. optical and ionization
- D. geometrical only
Explanation. This octahedral complex exhibits geometrical isomerism (cis/trans). It also shows ionization isomerism by exchanging the \(Br^{-}\) ligands in the coordination sphere with the \(Cl^{-}\) counter ion.
Q15
Which one of the following complexes is not expected to exhibit isomerism?
- A. \([Ni(NH_{3})_{4}(H_{2}O)_{2}]^{2+}\)
- B. \([Pt(NH_{3})_{2}Cl_{2}]\)
- C. \([Co(NH_{3})_{5}SO_{4}]Cl\)
- D. \([FeCl_{6}]^{3-}\)Correct
Explanation. A homoleptic octahedral complex like \([FeCl_{6}]^{3-}\), which contains only one type of monodentate ligand, cannot have different spatial arrangements or structural variations to form isomers.
Q16
A complex in which the oxidation number of the metal is zero is
- A. \(K_{4}[Fe(CN)_{6}]\)
- B. \([Fe(CN)_{3}(NH_{3})_{3}]\)
- C. \([Fe(CO)_{5}]\)Correct
- D. both (b) and (c)
Explanation. In metal carbonyls such as \([Fe(CO)_{5}]\), the carbon monoxide ligands are neutral molecules, and the metal atom maintains an oxidation state of zero.
Q17
Formula of tris(ethane-1,2-diamine)iron(II) phosphate is
- A. \([Fe(en)_{3}]PO_{4}\)
- B. \([Fe(en)_{3}](PO_{4})_{3}\)
- C. \([Fe(en)_{3}](PO_{4})_{2}\)
- D. \([Fe(en)_{3}]_{3}(PO_{4})_{2}\)Correct
Explanation. The iron(II) complex cation with three neutral ethylenediamine ligands has a \(+2\) charge. Balancing this with phosphate ions, which have a \(-3\) charge, requires three complex cations for every two phosphate ions.
Q18
Which of the following is paramagnetic in nature?
- A. \([Zn(NH_{3})_{4}]^{2+}\)
- B. \([Co(NH_{3})_{6}]^{3+}\)
- C. \([Ni(H_{2}O)_{6}]^{2+}\)Correct
- D. \([Ni(CN)_{4}]^{2-}\)
Explanation. In \([Ni(H_{2}O)_{6}]^{2+}\), the \(Ni^{2+}\) ion has a \(d^{8}\) configuration. With water as a weak field ligand, the electrons remain unpaired in the \(e_{g}\) level, resulting in paramagnetism.
Q19
Fac-mer isomerism is shown by
- A. \([Co(en)_{3}]^{3+}\)
- B. \([Co(NH_{3})_{4}Cl_{2}]^{+}\)
- C. \([Co(NH_{3})_{3}Cl_{3}]\)Correct
- D. \([Co(NH_{3})_{5}Cl]SO_{4}\)
Explanation. Facial-meridional (fac-mer) isomerism is a specific type of geometrical isomerism occurring in octahedral complexes with a general formula of \(MA_{3}B_{3}\).
Q20
Choose the correct statement.
- A. Square planar complexes are more stable than octahedral complexes
- B. The spin only magnetic moment of \([CuCl_{4}]^{2-}\) is \(1.732\text{ BM}\) and it has square planar structure.
- C. Crystal field splitting energy \(\Delta_{o}\) of \([FeF_{6}]^{4-}\) is higher than the \(\Delta_{o}\) of \([Fe(CN)_{6}]^{4-}\)
- D. The crystal field stabilization energy of \([V(H_{2}O)_{6}]^{2+}\) is higher than that of \([Ti(H_{2}O)_{6}]^{2+}\)Correct
Explanation. Vanadium(II) is a \(d^{3}\) system with a magnitude of CFSE \(= 1.2 \Delta_{o}\), while Titanium(II) is \(d^{2}\) with CFSE \(= 0.8 \Delta_{o}\). Thus, the vanadium complex has higher stabilization energy.
