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The d- and f- Block Elements Set-3

Practice Important MCQs for The d- and f- Block Elements — Chemistry, Class 12.

The d- and f-block chapter is crucial for CBSE Class 12 as well as competitive exams because it links electronic configurations with magnetic properties, oxidation states, crystal field stabilization, and trends across transition and inner-transition series. Mastery of CFSE, magnetic moment calculations, lanthanide contraction, and f-block stability factors directly helps solve numericals and concept-based MCQs efficiently.

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Q1. The spin-only magnetic moment is given by $\mu_{s-o}=\sqrt{n(n+2)}\ \mu_B$ , where $n$ is the number of unpaired electrons. For $Cr^{3+}$ (atomic number 24) in an octahedral complex, what is the expected spin-only magnetic moment (in $\mu_B$ )?

(A)

$4.90\ \mu_B$

(B)

$3.87\ \mu_B$

(C)

$1.73\ \mu_B$

(D)

$2.83\ \mu_B$

Q2. Consider an octahedral complex of $Fe^{2+}$ (d $^6$ ). Given $\Delta_o=200\ \mathrm{kJ\,mol^{-1}}$ and pairing energy $P=150\ \mathrm{kJ\,mol^{-1}}$ . Using octahedral CFSE contributions ( $-0.4\Delta_o$ per $t_{2g}$ electron and $+0.6\Delta_o$ per $e_g$ electron) and $P$ per electron pair, which spin state is energetically favoured and how many unpaired electrons will the complex have?

(A)

High-spin; 4 unpaired electrons

(B)

High-spin; 2 unpaired electrons

(C)

Low-spin; 2 unpaired electrons

(D)

Low-spin; 0 unpaired electrons

Q3. Among the following trivalent lanthanoid ions, which is most readily reduced from $+3$ to $+2$ in aqueous medium due to extra stability of the resulting $4f$ configuration?

(A)

$Eu^{3+}$

(B)

$Ce^{3+}$

(C)

$Gd^{3+}$

(D)

$Yb^{3+}$

Q4. A Ni(II) complex shows an experimental magnetic moment of $3.20\ \mu_B$ per Ni atom. Which coordination geometry for Ni $^{2+}$ (d $^8$ ) best explains this observed magnetic moment?

(A)

Square planar; diamagnetic (0 unpaired electrons)

(B)

Octahedral (high-spin); 2 unpaired electrons; spin-only $\mu_{s-o}\approx2.83\ \mu_B$

(C)

Tetrahedral; 2 unpaired electrons; spin-only $\mu_{s-o}\approx2.83\ \mu_B$

(D)

Octahedral (high-spin); 4 unpaired electrons; spin-only $\mu_{s-o}\approx4.90\ \mu_B$

Q5. Compare the crystal field splitting parameter $\Delta_o$ for the complexes $[Fe(CN)_6]^{4-}$ (Fe $^{2+}$ ) and $[Fe(CN)_6]^{3-}$ (Fe $^{3+}$ ). Which statement is correct and why?

(A)

$\Delta_o([Fe(CN)_6]^{4-})>\Delta_o([Fe(CN)_6]^{3-})$ , because the extra electron in Fe $^{2+}$ increases electron–electron repulsion and thus splitting

(B)

$\Delta_o([Fe(CN)_6]^{3-})>\Delta_o([Fe(CN)_6]^{4-})$ , because the higher oxidation state (Fe $^{3+}$ ) contracts M–L distance and increases M–L overlap (larger $\Delta_o$ )

(C)

$\Delta_o$ is the same for both since the ligand is identical

(D)

Magnitude of $\Delta_o$ depends only on geometry, not on oxidation state, so it cannot be compared here

Q6. The hexaaquairon(II) ion $[Fe(H_2O)_6]^{2+}$ is a high-spin octahedral complex of Fe(II) ( $d^6$ ). Using the spin-only formula $\mu_{so}=\sqrt{n(n+2)}\,\mu_B$ , where $n$ is the number of unpaired electrons, the magnetic moment of $[Fe(H_2O)_6]^{2+}$ (in $\mu_B$ ) is closest to:

(A)

$5.92\ \mu_B$

(B)

$4.90\ \mu_B$

(C)

$0\ \mu_B$

(D)

$1.73\ \mu_B$

Q7. For an octahedral $d^4$ complex the crystal field stabilization energies (CFSE) of the high-spin ( $t_{2g}^3e_g^1$ ) and low-spin ( $t_{2g}^4e_g^0$ ) arrangements can be estimated using $t_{2g}=-0.4\Delta_o$ and $e_g=+0.6\Delta_o$ per electron. If $\Delta_o=180\ \mathrm{kJ\,mol^{-1}}$ and pairing energy $P=160\ \mathrm{kJ\,mol^{-1}}$ , which electronic state is favoured and how many unpaired electrons will it have?

(A)

High-spin; 4 unpaired electrons

(B)

High-spin; 2 unpaired electrons

(C)

Low-spin; 2 unpaired electrons

(D)

Low-spin; 0 unpaired electrons

Q8. Consider two octahedral $d^3$ complexes: $[V(H_2O)_6]^{2+}$ (V $^{2+}$ with H $_2$ O ligands) and $[Cr(NH_3)_6]^{3+}$ (Cr $^{3+}$ with NH $_3$ ligands). Which complex is expected to have the larger octahedral crystal field splitting $\Delta_o$ , and why?

(A)

$[Cr(NH_3)_6]^{3+}$ — because NH $_3$ is a stronger field ligand than H $_2$ O and Cr $^{3+}$ has higher effective nuclear charge than V $^{2+}$ , both increasing $\Delta_o$

(B)

$[V(H_2O)_6]^{2+}$ — because V $^{2+}$ has larger ionic radius which increases metal–ligand overlap and thus $\Delta_o$

(C)

Both have approximately the same $\Delta_o$ since a $d^3$ configuration gives the same splitting regardless of metal or ligand

(D)

Cannot predict without experimental absorption ( $\lambda_{\max}$ ) data

Q9. Assertion (A): Fractional crystallisation cannot separate adjacent lanthanides because their ionic radii are identical. Reason (R): Lanthanide contraction arises because 4f electrons are poorly shielding, causing a gradual decrease in ionic radii across the series.

(A)

Both A and R are true and R explains A

(B)

Both A and R are true but R does not explain A

(C)

A is true but R is false

(D)

A is false but R is true

Q10. For trivalent lanthanide ions the spin-only magnetic moment is $\mu_{so}=\sqrt{n(n+2)}\,\mu_B$ , but experimental moments often deviate due to orbital contribution. Which of the following Ln $^{3+}$ ions is expected to have an experimental effective magnetic moment closest to the spin-only value? (Given electronic configurations: Ce $^{3+}$ : $4f^1$ ; Gd $^{3+}$ : $4f^7$ ; Dy $^{3+}$ : $4f^9$ ; Sm $^{3+}$ : $4f^5$ .)

(A)

Gd $^{3+}$ ( $4f^7$ )

(B)

Ce $^{3+}$ ( $4f^1$ )

(C)

Dy $^{3+}$ ( $4f^9$ )

(D)

Sm $^{3+}$ ( $4f^5$ )

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