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The d- and f- Block Elements Set-9 MCQ Online Practice Test | Class 12 | Quiz Tweek

The d- and f- Block Elements Set-9

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

The d- and f- block elements chapter is crucial because it forms the backbone of many core topics in CBSE and competitive exams—like magnetic properties, oxidation states, crystal field effects, stability of oxidation states, and characteristic electronic configurations. Mastery of these ideas helps you solve both numerical and conceptual questions quickly and accurately, especially those involving spin-only moments, high-spin/low-spin transitions, CFSE, and lanthanoid/actinoid behavior.

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Q1. Using the spin-only formula $\mu_s=\sqrt{n(n+2)}\ \mu_B$ , the magnetic moment of the $V^{3+}$ ion is closest to:

(A)

$1.73\ \mu_B$

(B)

$3.87\ \mu_B$

(C)

$2.83\ \mu_B$

(D)

$4.90\ \mu_B$

Q2. Given standard reduction potentials at 298 K: $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+})=+0.77\ \mathrm{V}$ and $E^\circ(\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+})=-0.41\ \mathrm{V}$ , calculate the standard cell potential $E^\circ_{\text{cell}}$ for the spontaneous reaction between $\mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{3+}$ and the equilibrium constant $K$ at 298 K. (Use $E^\circ=\dfrac{0.05916}{n}\log K$ .)

(A)

$E^\circ_{\text{cell}}=1.18\ \mathrm{V},\ K\approx 8.7\times10^{19}$

(B)

$E^\circ_{\text{cell}}=1.18\ \mathrm{V},\ K\approx 1.3\times10^{10}$

(C)

$E^\circ_{\text{cell}}=-1.18\ \mathrm{V},\ K\approx 8.7\times10^{19}$

(D)

$E^\circ_{\text{cell}}=0.36\ \mathrm{V},\ K\approx 2.3\times10^{6}$

Q3. For two octahedral complexes of $Co^{3+}$ (d $^6$ ): $[\mathrm{Co(L1)_6}]^{3+}$ with $\Delta_0=180\ \mathrm{kJ\,mol^{-1}}$ and $[\mathrm{Co(L2)_6}]^{3+}$ with $\Delta_0=360\ \mathrm{kJ\,mol^{-1}}$ . The pairing energy for $Co^{3+}$ is $P=240\ \mathrm{kJ\,mol^{-1}}$ . Which statement about their magnetic behavior and approximate spin-only magnetic moments is correct?

(A)

L1 is diamagnetic ( $0\ \mu_B$ ); L2 is paramagnetic with $\approx 4.90\ \mu_B$

(B)

Both L1 and L2 are paramagnetic with $\approx 2.83\ \mu_B$

(C)

L1 is paramagnetic with $\approx 2.83\ \mu_B$ ; L2 is paramagnetic with $\approx 4.90\ \mu_B$

(D)

L1 is paramagnetic (4 unpaired, $\approx 4.90\ \mu_B$ ); L2 is low-spin diamagnetic ( $0\ \mu_B$ )

Q4. Nickel(II) ( $\mathrm{Ni}^{2+}$ , d $^8$ ) forms $[\mathrm{Ni(CN)_4}]^{2-}$ (square planar) and $[\mathrm{NiCl}_4]^{2-}$ (tetrahedral). Which statement correctly compares their magnetic behaviour (number of unpaired electrons) and spin-only magnetic moments?

(A)

Both complexes are diamagnetic ( $0\ \mu_B$ )

(B)

$[\mathrm{Ni(CN)_4}]^{2-}$ is diamagnetic ( $0\ \mu_B$ ) and $[\mathrm{NiCl}_4]^{2-}$ is paramagnetic with $\approx 2.83\ \mu_B$ (2 unpaired)

(C)

$[\mathrm{Ni(CN)_4}]^{2-}$ is paramagnetic with $\approx 2.83\ \mu_B$ and $[\mathrm{NiCl}_4]^{2-}$ is diamagnetic

(D)

$[\mathrm{Ni(CN)_4}]^{2-}$ has 1 unpaired ( $\approx 1.73\ \mu_B$ ) while $[\mathrm{NiCl}_4]^{2-}$ has 2 unpaired ( $\approx 2.83\ \mu_B$ )

Q5. A saturated solution of $\mathrm{AgCl(s)}$ is treated with $0.10\ \mathrm{M}$ $\mathrm{NH_3}$ . Given $K_{sp}(\mathrm{AgCl})=1.8\times10^{-10}$ and the formation constant $K_f$ for $[\mathrm{Ag(NH_3)_2}]^{+}$ is $1.6\times10^{7}$ , estimate by what factor the solubility of AgCl increases on addition of $0.10\ \mathrm{M}$ NH $_3$ at 298 K (assume $[\mathrm{NH_3}]\approx0.10\ \mathrm{M}$ ).

(A)

No significant change (factor $\approx 1$ )

(B)

Increases by a factor of $\approx 40$

(C)

Increases by a factor of $\approx 4.0\times10^{2}$

(D)

Increases by a factor of $\approx 4.0\times10^{4}$

Q6. Calculate the spin-only magnetic moment (in Bohr magneton, BM) for a high-spin octahedral complex of $Mn^{2+}$ (electronic configuration $d^5$ ).

(A)

$4.90\ \mathrm{BM}$

(B)

$3.87\ \mathrm{BM}$

(C)

$5.92\ \mathrm{BM}$

(D)

$1.73\ \mathrm{BM}$

Q7. Assume all the following octahedral aqua complexes are high-spin and compare their crystal field stabilization energies (CFSE) in terms of $\Delta_o$ (neglect pairing energy differences). Which complex has the largest (most negative) CFSE?

(A)

$[V(H_2O)_6]^{2+}$

(B)

$[Ti(H_2O)_6]^{3+}$

(C)

$[Cr(H_2O)_6]^{2+}$

(D)

$[Mn(H_2O)_6]^{3+}$

Q8. Among the following octahedral hexa-aqua complexes, which is expected to exhibit the strongest Jahn–Teller distortion (i.e., the largest difference between axial and equatorial M–O bond lengths)?

(A)

$[Ti(H_2O)_6]^{3+}\ (d^1)$

(B)

$[Cu(H_2O)_6]^{2+}\ (d^9)$

(C)

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

(D)

$[Cr(H_2O)_6]^{3+}\ (d^3)$

Q9. Consider isoelectronic ions $Fe^{2+}$ (3d $^6$ ) and $Ru^{2+}$ (4d $^6$ ) forming octahedral complexes with the same ligand of moderate field strength. Empirically $Ru^{2+}$ tends to give low-spin complexes while $Fe^{2+}$ often gives high-spin complexes. Which explanation best accounts for this difference?

(A)

4d orbitals are more contracted than 3d orbitals, so metal–ligand overlap and $\Delta_o$ are smaller for $Ru^{2+}$ .

(B)

The higher nuclear charge of Ru raises pairing energy so 4d metals favor high-spin states.

(C)

Stronger spin–orbit coupling in 4d metals forces low-spin configurations irrespective of $\Delta_o$ and pairing energy.

(D)

4d orbitals are more spatially extended than 3d, producing larger metal–ligand overlap (larger $\Delta_o$ ) and generally lower pairing energy $P$ , so electron pairing (low-spin) is energetically favored for $Ru^{2+}$ .

Q10. Among the lanthanoid series, which pair of elements is most likely to form stable $+2$ ions in aqueous solution? (Choose the pair whose $+2$ state is stabilized by a half-filled or completely filled 4f subshell.)

(A)

$Eu^{2+}$ and $Yb^{2+}$

(B)

$La^{2+}$ and $Ce^{2+}$

(C)

$Gd^{2+}$ and $Tb^{2+}$

(D)

$Sm^{2+}$ and $Pr^{2+}$

Q11. Calculate the spin-only magnetic moment (in Bohr magneton, BM) for the high-spin $Mn^{2+}$ ion. Use $\mu_{\text{so}}=\sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons.

(A)

$\approx 4.90\ \mathrm{BM}$

(B)

$\approx 1.73\ \mathrm{BM}$

(C)

$\approx 5.92\ \mathrm{BM}$

(D)

$\approx 3.87\ \mathrm{BM}$

Q12. For an octahedral $Fe^{2+}$ ( $d^6$ ) complex, ligand A produces $\Delta_o=12{,}000\ \mathrm{cm^{-1}}$ and ligand B produces $\Delta_o=20{,}000\ \mathrm{cm^{-1}}$ . The corresponding pairing energies are $P_A=160\ \mathrm{kJ\,mol^{-1}}$ and $P_B=150\ \mathrm{kJ\,mol^{-1}}$ . (Use $1\ \mathrm{cm^{-1}}=0.01196\ \mathrm{kJ\,mol^{-1}}$ .) Which complex will be low-spin?

(A)

Only the complex with ligand A will be low-spin.

(B)

Only the complex with ligand B will be low-spin.

(C)

Both complexes (with ligand A and B) will be low-spin.

(D)

Neither complex will be low-spin (both high-spin).

Q13. Consider the nickel(II) complexes $[Ni(CN)_4]^{2-}$ (square planar) and $[NiCl_4]^{2-}$ (tetrahedral). Assuming $CN^-$ is a strong-field ligand and $Cl^-$ a weak-field ligand, predict their magnetic behaviour.

(A)

Both complexes are diamagnetic (no unpaired electrons).

(B)

Both complexes are paramagnetic with two unpaired electrons each.

(C)

$[Ni(CN)_4]^{2-}$ is paramagnetic and $[NiCl_4]^{2-}$ is diamagnetic.

(D)

$[Ni(CN)_4]^{2-}$ is diamagnetic; $[NiCl_4]^{2-}$ is paramagnetic with two unpaired electrons.

Q14. Assertion (A): Zirconium ( $Zr$ ) and Hafnium ( $Hf$ ) have very similar ionic radii and exhibit nearly identical chemical behaviour. Reason (R): Lanthanoid contraction—caused by poor shielding of the increasing nuclear charge by 4f electrons across the lanthanoid series—reduces the atomic radius of $Hf$ , making it similar to that of $Zr$ .

(A)

Both A and R are true and R is the correct explanation of A.

(B)

Both A and R are true but R is NOT the correct explanation of A.

(C)

A is true but R is false.

(D)

A is false but R is true.

Q15. Which of the following best explains why actinoids (e.g., U, Np, Pu) exhibit a wider range of stable oxidation states than lanthanoids (e.g., Ce–Lu)?

(A)

Because actinoids have substantially lower first ionisation energies than lanthanoids.

(B)

Because the $5f$ orbitals in actinoids are more radially extended and energetically comparable to $6d$ and $7s$ orbitals, so $5f$ electrons participate in bonding and can be removed to give multiple oxidation states.

(C)

Because relativistic contraction of $4f$ orbitals in lanthanoids causes them to be highly delocalised, preventing variable oxidation states.

(D)

Because actinoids have completely filled $5f$ subshells which stabilise many oxidation states.

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