Correct answer Carries: 4.
Wrong Answer Carries: -1.
What is the ratio of unidentate to bidentate ligands in \( \ce{[Co(NH3)3(en)Cl]^2+} \)?
In \( \ce{[Co(NH3)3(en)Cl]^2+} \), unidentate ligands = 3 \( \ce{NH3} \) + 1 \( \ce{Cl^-} \) = 4; bidentate = 1 en. Ratio = 4 : 1.
What is the geometry of \( \ce{[CoF4]^2-} \)?
\( \ce{[CoF4]^2-} \) (Co\(^{2+}\), \( d^7 \)) with 4 ligands is tetrahedral, using \( sp^3 \) hybridization due to weak field \( \ce{F^-} \).
Which of the following is a homoleptic complex?
\( \ce{[Fe(H2O)6]^2+} \) has only one type of ligand (\( \ce{H2O} \)), making it homoleptic, unlike \( \ce{[Fe(H2O)5Cl]^+} \) with two types.
Which complex has a metal ion that becomes diamagnetic when coordinated with 6 \( \ce{CN^-} \) ligands?
\( \ce{[Fe(H2O)6]^2+} \) (Fe\(^{2+}\), \( d^6 \)) is high spin with \( \ce{H2O} \) (4 unpaired electrons). With 6 \( \ce{CN^-} \) (strong field), it becomes \( \ce{[Fe(CN)6]^4-} \), low spin (0 unpaired electrons).
What is the geometry of the coordination sphere in \( \ce{[Ni(en)2]^2+} \)?
\( \ce{[Ni(en)2]^2+} \) (Ni\(^{2+}\), \( d^8 \)) with 2 bidentate en (4 donor atoms) is square planar, using \( dsp^2 \) hybridization.
The hybridization of cobalt in \( \ce{[CoF6]^3-} \) is:
\( \ce{[CoF6]^3-} \) (Co\(^{3+}\), \( d^6 \)) with weak field \( \ce{F^-} \) in an octahedral field is high spin, using outer orbitals (\( sp^3d^2 \)).
Which complex has a metal ion with a \( d^7 \) configuration in an octahedral field that is high spin?
\( \ce{[Co(H2O)6]^2+} \) (Co\(^{2+}\), \( d^7 \)) with weak field \( \ce{H2O} \) in an octahedral field is high spin (\( t_{2g}^5 e_g^2 \)).
Which complex has a secondary valence of 4 according to Werner’s theory?
Secondary valence is the coordination number. \( \ce{[Cu(NH3)4]SO4} \) has 4 \( \ce{NH3} \) ligands, giving a coordination number of 4.
The EAN of Co in \( \ce{[Co(NH3)6]^3+} \) is: (Co atomic number = 27)
Co\(^{3+}\) with 6 \( \ce{NH3} \) ligands. EAN = 27 - 3 + 2 × 6 = 36.
The magnetic moment of \( \ce{[MnCl4]^2-} \) is approximately: (Mn atomic number = 25)
Mn\(^{2+}\) (\( d^5 \)) in tetrahedral \( \ce{[MnCl4]^2-} \) (\( sp^3 \)) with weak field \( \ce{Cl^-} \) ligands is high spin (5 unpaired electrons). Magnetic moment = \( \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \) BM.
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