Coordination Compounds Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which ligand pair in \( \ce{[Cr(H2O)4(CN)(SCN)]^+} \) could exhibit linkage isomerism if one were ambidentate?

\( \ce{SCN^-} \) is ambidentate (S or N binding), while \( \ce{CN^-} \) binds only via C. Thus, \( \ce{SCN^-} \) can form linkage isomers.

\( \ce{H2O} \) and \( \ce{CN^-} \)
\( \ce{SCN^-} \) and \( \ce{H2O} \)
\( \ce{SCN^-} \) and \( \ce{CN^-} \)
\( \ce{H2O} \) and \( \ce{SCN^-} \)
2

What is the oxidation state of platinum in \( \ce{[PtCl6]^2-} \)?

Let the oxidation state of Pt be \( x \). Each \( \ce{Cl^-} \) has a charge of -1, and there are 6 \( \ce{Cl^-} \) ligands, contributing -6. The overall charge is -2. Thus, \( x + (-6) = -2 \), so \( x = +4 \).

+4
+2
+3
+6
1

The hybridization of cobalt in \( \ce{[CoF6]^3-} \) is:

\( \ce{[CoF6]^3-} \) is octahedral (Co\(^{3+}\), \( d^6 \)). \( \ce{F^-} \) is a weak field ligand, forming a high-spin complex with outer orbitals, leading to \( sp^3d^2 \) hybridization.

\( d^2sp^3 \)
\( sp^3d^2 \)
\( sp^3 \)
\( dsp^2 \)
2

How many unpaired electrons are present in \( \ce{[Co(H2O)6]^3+} \)? (Co atomic number = 27)

Co\(^{3+}\) (\( d^6 \)) in \( \ce{[Co(H2O)6]^3+} \) (octahedral) with weak field \( \ce{H2O} \) is high spin: \( t_{2g}^4 e_g^2 \), giving 4 unpaired electrons.

2
4
0
6
2

Which of the following complexes has a fac-isomer?

\( \ce{[Cr(H2O)3Cl3]} \) (octahedral, \( \ce{[Ma3b3]} \)) can have a facial (fac) isomer where 3 \( \ce{Cl^-} \) ligands occupy one face.

\( \ce{[Cr(H2O)4Cl2]^+} \)
\( \ce{[Ni(NH3)4]^2+} \)
\( \ce{[Cr(H2O)3Cl3]} \)
\( \ce{[Fe(CN)6]^4-} \)
3

Which ligand produces the strongest crystal field splitting in an octahedral complex?

From the spectrochemical series (\( \ce{Cl^-} < \ce{H2O} < \ce{NH3} < \ce{CN^-} \)), \( \ce{CN^-} \) is the strongest field ligand, causing the largest \( \Delta_o \).

\( \ce{Cl^-} \)
\( \ce{H2O} \)
\( \ce{CN^-} \)
\( \ce{NH3} \)
3

Which complex exhibits a color due to a d-d transition involving \( e_g \) orbitals?

\( \ce{[Ni(H2O)6]^2+} \) (Ni\(^{2+}\), \( d^8 \)) in an octahedral field has \( t_{2g}^6 e_g^2 \), and the d-d transition from \( t_{2g} \) to \( e_g \) absorbs visible light (green color).

\( \ce{[Ni(CN)4]^2-} \)
\( \ce{[Co(CN)6]^3-} \)
\( \ce{[Ni(H2O)6]^2+} \)
\( \ce{[Fe(CN)6]^4-} \)
3

What is the primary valence of Ni in \( \ce{[Ni(NH3)6]Cl2} \)?

Primary valence is the oxidation state. In \( \ce{[Ni(NH3)6]Cl2} \), \( \ce{NH3} \) is neutral, and 2 \( \ce{Cl^-} \) balance \( \ce{[Ni(NH3)6]^2+} \), so Ni is +2.

1
3
2
4
3

The hybridization of Fe in \( \ce{[Fe(CN)6]^4-} \) is:

\( \ce{[Fe(CN)6]^4-} \) (Fe\(^{2+}\), \( d^6 \)) is octahedral with strong field \( \ce{CN^-} \), forming a low-spin complex. It uses inner d orbitals, giving \( d^2sp^3 \) hybridization.

\( sp^3d^2 \)
\( sp^3 \)
\( dsp^2 \)
\( d^2sp^3 \)
4

What is the shape of \( \ce{[PtCl4]^2-} \)?

\( \ce{[PtCl4]^2-} \) has Pt in \( +2 \) state (\( d^8 \)), with 4 ligands. It adopts a square planar geometry due to \( dsp^2 \) hybridization, typical for \( d^8 \) metals like Pt(II).

Tetrahedral
Square planar
Octahedral
Trigonal bipyramidal
2

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