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The d- and f- Block Elements Set-4 MCQ Online Practice Test | Class 12 | Quiz Tweek

The d- and f- Block Elements Set-4

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

The d- and f-block chapter is crucial for both CBSE and competitive exams because it links electronic configurations to real chemical behavior—such as oxidation states, magnetic properties, stability of coordination compounds, ligand-field splitting, and special stability of lanthanoids/actinoids. Most board and JEE/NEET questions test whether you can correctly predict (i) spin states (high/low spin), (ii) CFSE and pairing effects, (iii) magnetic moments using spin-only formula, and (iv) why certain ions (like Eu²⁺ and Yb²⁺) show unusual stability.

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Q1. Consider the octahedral complex $[Fe(CN)_6]^{4-}$ . Assuming CN $^-$ is a strong-field ligand, how many unpaired electrons does the complex contain?

(A)

(B)

0 (diamagnetic)

(C)

(D)

Q2. Determine the oxidation state of cobalt and the number of unpaired electrons in the octahedral complex $[CoF_6]^{3-}$ , given that $F^-$ is a weak-field ligand.

(A)

Co is $+3$ , $d^6$ , 4 unpaired electrons

(B)

Co is $+2$ , $d^7$ , 3 unpaired electrons

(C)

Co is $+3$ , $d^6$ , 0 unpaired electrons

(D)

Co is $+3$ , $d^6$ , 2 unpaired electrons

Q3. Which of the following best explains why $[Cu(en)_2]^{2+}$ is generally more stable than $[Cu(NH_3)_4]^{2+}$ in aqueous solution? ( $en$ = ethylenediamine)

(A)

$[Cu(en)_2]^{2+}$ — $en$ is bidentate (forms chelate rings) whereas $NH_3$ is monodentate

(B)

Both complexes have similar stability because both provide four nitrogen donors

(C)

$[Cu(en)_2]^{2+}$ — $en$ is a stronger $\sigma$ -donor than $NH_3$

(D)

$[Cu(en)_2]^{2+}$ — chelate effect (formation of stable five-membered rings and favourable entropy) increases stability

Q4. Europium and ytterbium are more frequently encountered in the +2 oxidation state than most other lanthanoids. Which of the following provides the best explanation?

(A)

Their +2 state is favoured because Eu and Yb have relatively large ionic radii compared to others in the series

(B)

The +2 state is stabilised mainly by lattice energy effects because Eu $^{2+}$ and Yb $^{2+}$ are unusually small

(C)

Eu $^{2+}$ has a $4f^7$ (half-filled) and Yb $^{2+}$ has a $4f^{14}$ (filled) configuration; these electronic configurations offer additional stability, making reduction from +3 to +2 easier

(D)

The +2 oxidation state is stabilised due to strong participation of 5d orbitals in bonding for Eu and Yb

Q5. For $Fe^{2+}$ ( $d^6$ ) in octahedral coordination, predict the magnetic behaviour at room temperature for $[Fe(CN)_6]^{4-}$ and $[FeF_6]^{4-}$ respectively, given that CN $^-$ is a strong-field ligand and F $^-$ a weak-field ligand.

(A)

$[Fe(CN)_6]^{4-}$ — diamagnetic (low-spin, 0 unpaired); $[FeF_6]^{4-}$ — paramagnetic (high-spin, 4 unpaired)

(B)

Both complexes are diamagnetic.

(C)

Both complexes are paramagnetic with 4 unpaired electrons.

(D)

$[Fe(CN)_6]^{4-}$ — paramagnetic; $[FeF_6]^{4-}$ — diamagnetic

Q6. Using the spin-only formula $\mu_{so}=\sqrt{n(n+2)}\,\mu_B$ , where $n$ is the number of unpaired electrons, calculate the magnetic moment (in $\mu_B$ ) of a high-spin octahedral $Fe^{3+}$ ion.

(A)

$4.90\ \mu_B$

(B)

$3.87\ \mu_B$

(C)

$5.92\ \mu_B$

(D)

$2.83\ \mu_B$

Q7. Which of the following octahedral complexes has the largest ligand field stabilization energy (CFSE)? (Assume low-spin where applicable and neglect pairing energy; express CFSE in terms of $\Delta_o$ .)

(A)

$[Fe(CN)_6]^{4-}$ (Fe $^{2+}$ , $d^6$ , low-spin)

(B)

$[Cr(NH_3)_6]^{3+}$ (Cr $^{3+}$ , $d^3$ )

(C)

$[Mn(H_2O)_6]^{2+}$ (Mn $^{2+}$ , $d^5$ , high-spin)

(D)

$[CoF_6]^{3-}$ (Co $^{3+}$ , $d^6$ , high-spin)

Q8. Gd $^{3+}$ exhibits an effective magnetic moment close to the spin-only value ( $\approx 7.94\ \mu_B$ ) while Sm $^{3+}$ shows an anomalously small observed moment. Which statement best explains this difference?

(A)

Gd $^{3+}$ has its orbital angular momentum quenched by the crystal field, whereas Sm $^{3+}$ retains full orbital contribution.

(B)

Gd $^{3+}$ forms more covalent bonds increasing its observed moment, while Sm $^{3+}$ is purely ionic.

(C)

Sm $^{3+}$ has all $4f$ electrons paired in its ground state, giving a very low moment.

(D)

Gd $^{3+}$ has a half‑filled $4f^7$ configuration with $L\!=\!0$ , so the moment is essentially spin-only; Sm $^{3+}$ ( $4f^5$ ) has significant spin–orbit coupling with partial cancellation between spin and orbital contributions, producing a small observed $\mu_{\rm eff}$ .

Q9. Which of the following gaseous/solvated metal ions is expected to form the most stable complex with CO (i.e., strongest M→CO π-backbonding) under comparable conditions?

(A)

$Ti^{3+}$ ( $d^1$ )

(B)

$Fe^{2+}$ ( $d^6$ )

(C)

$Zn^{2+}$ ( $d^{10}$ )

(D)

$Sc^{3+}$ ( $d^0$ )

Q10. For $Fe^{2+}$ ( $d^6$ ) in octahedral coordination, predict the magnetic behaviour at room temperature for $[Fe(CN)_6]^{4-}$ and $[FeF_6]^{4-}$ respectively, given that CN $^-$ is a strong-field ligand and F $^-$ a weak-field ligand.

(A)

$[Fe(CN)_6]^{4-}$ — diamagnetic (low-spin, 0 unpaired); $[FeF_6]^{4-}$ — paramagnetic (high-spin, 4 unpaired)

(B)

Both complexes are diamagnetic.

(C)

Both complexes are paramagnetic with 4 unpaired electrons.

(D)

$[Fe(CN)_6]^{4-}$ — paramagnetic; $[FeF_6]^{4-}$ — diamagnetic

Q11. For Mn^{2+} in an aqueous high-spin complex the electronic configuration is $d^5$ . Using the spin-only formula $\mu_{\text{spin}}=\sqrt{n(n+2)}$ (where $n$ is the number of unpaired electrons), determine the number of unpaired electrons and the spin-only magnetic moment (in Bohr magneton).

(A)

4 unpaired electrons; $\mu_{\text{spin}}=\sqrt{4(4+2)}=\sqrt{24}\approx 4.90\ \mathrm{BM}$

(B)

5 unpaired electrons; $\mu_{\text{spin}}=\sqrt{5(5+2)}=\sqrt{35}\approx 5.92\ \mathrm{BM}$

(C)

3 unpaired electrons; $\mu_{\text{spin}}=\sqrt{3(3+2)}=\sqrt{15}\approx 3.87\ \mathrm{BM}$

(D)

1 unpaired electron; $\mu_{\text{spin}}=\sqrt{1(1+2)}=\sqrt{3}\approx 1.73\ \mathrm{BM}$

Q12. Consider octahedral complexes $[Co(L_1)_6]^{3+}$ and $[Co(L_2)_6]^{3+}$ of $Co^{3+}$ ( $d^6$ ). For $L_1$ , $\Delta_o=210\ \mathrm{kJ\,mol^{-1}}$ ; for $L_2$ , $\Delta_o=150\ \mathrm{kJ\,mol^{-1}}$ . The pairing energy for $Co^{3+}$ is $P=180\ \mathrm{kJ\,mol^{-1}}$ . Using the criterion $\Delta_o>P$ for low-spin and the spin-only formula $\mu=\sqrt{n(n+2)}$ , identify the spin state (low- or high-spin) and the spin-only magnetic moment (in BM) for each complex.

(A)

$[Co(L_1)_6]^{3+}$ : high-spin, $\mu\approx 4.90\ \mathrm{BM}$ ; $[Co(L_2)_6]^{3+}$ : low-spin, $\mu=0\ \mathrm{BM}$

(B)

Both $[Co(L_1)_6]^{3+}$ and $[Co(L_2)_6]^{3+}$ are low-spin; both $\mu=0\ \mathrm{BM}$

(C)

Both $[Co(L_1)_6]^{3+}$ and $[Co(L_2)_6]^{3+}$ are high-spin; both $\mu\approx 4.90\ \mathrm{BM}$

(D)

$[Co(L_1)_6]^{3+}$ : low-spin, $\mu=0\ \mathrm{BM}$ ; $[Co(L_2)_6]^{3+}$ : high-spin, $\mu\approx 4.90\ \mathrm{BM}$

Q13. An octahedral $Ti^{3+}$ complex ( $d^1$ ) shows an absorption band at $650\ \mathrm{nm}$ assigned to the $t_{2g}\rightarrow e_g$ transition (i.e. $\Delta_o$ ). (a) Calculate $\Delta_o$ in $\mathrm{kJ\,mol^{-1}}$ using $\Delta_o=\dfrac{N_A h c}{\lambda}$ . (b) Calculate the CFSE for $d^1$ (CFSE $=-0.4\Delta_o$ ). (c) State the expected spin-only magnetic moment (in BM). (Use $h c N_A\approx 1.196\times10^{-1}\ \mathrm{J\,m\,mol^{-1}}$ .)

(A)

$\Delta_o\approx 184\ \mathrm{kJ\,mol^{-1}};\ \mathrm{CFSE}\approx -73.6\ \mathrm{kJ\,mol^{-1}};\ \mu_{\text{spin}}\approx 1.73\ \mathrm{BM}$

(B)

$\Delta_o\approx 153\ \mathrm{kJ\,mol^{-1}};\ \mathrm{CFSE}\approx -61.2\ \mathrm{kJ\,mol^{-1}};\ \mu_{\text{spin}}\approx 2.83\ \mathrm{BM}$

(C)

$\Delta_o\approx 276\ \mathrm{kJ\,mol^{-1}};\ \mathrm{CFSE}\approx -110.4\ \mathrm{kJ\,mol^{-1}};\ \mu_{\text{spin}}=0\ \mathrm{BM}$

(D)

$\Delta_o\approx 92\ \mathrm{kJ\,mol^{-1}};\ \mathrm{CFSE}\approx -36.8\ \mathrm{kJ\,mol^{-1}};\ \mu_{\text{spin}}\approx 1.73\ \mathrm{BM}$

Q14. For the half-reaction $Ag^+ + e^- \rightarrow Ag(s)$ the standard reduction potential is $E^\circ=+0.80\ \mathrm{V}$ . In a solution where $[Ag(NH_3)_2^+]=1.0\ \mathrm{M}$ and $[NH_3]=1.0\ \mathrm{M}$ , the complex formation constant is $K_f=\dfrac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.7\times10^7$ . Using $[Ag^+]=\dfrac{[Ag(NH_3)_2^+]}{K_f[NH_3]^2}$ and the Nernst equation $E=E^\circ+0.05916\log[Ag^+]$ at 298 K, calculate the reduction potential $E$ (in V) of the $Ag^+/Ag$ couple under these conditions.

(A)

$E=+0.80\ \mathrm{V}$

(B)

$E\approx +0.60\ \mathrm{V}$

(C)

$E\approx +0.373\ \mathrm{V}$

(D)

$E\approx +0.12\ \mathrm{V}$

Q15. A $Cu^{2+}$ ion in an octahedral complex undergoes a tetragonal elongation (Jahn–Teller distortion) along the $z$ -axis. Which of the following correctly gives the qualitative energy ordering of the five $d$ -orbitals after elongation (highest → lowest) and identifies the orbital that contains the single unpaired electron in the $d^9$ configuration? (Assume axial ligands move farther away.)

(A)

$d_{z^2} > d_{x^2-y^2} > d_{xy} > d_{xz}\approx d_{yz}$ ; unpaired electron in $d_{z^2}$

(B)

$d_{x^2-y^2} > d_{xy} > d_{z^2} > d_{xz}\approx d_{yz}$ ; unpaired electron in $d_{x^2-y^2}$

(C)

$d_{x^2-y^2} > d_{z^2} > d_{xz}\approx d_{yz} > d_{xy}$ ; unpaired electron in $d_{xz}/d_{yz}$

(D)

$d_{xy} > d_{x^2-y^2} > d_{z^2} > d_{xz}\approx d_{yz}$ ; unpaired electron in $d_{xy}$

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