401. An octahedral \(d^6\) complex containing \(\mathrm{H_2O}\) is converted into an analogous complex containing \(\mathrm{CN^-}\), with the metal oxidation state unchanged. The most likely change is:
ⓐ. \(\Delta_o\) increases and a low-spin arrangement becomes more favourable
ⓑ. \(\Delta_o\) decreases and a high-spin arrangement becomes less favourable
ⓒ. the \(d\)-electron count changes from \(d^6\) to \(d^5\)
ⓓ. the complex must become tetrahedral because cyanido is monodentate
Correct Answer: \(\Delta_o\) increases and a low-spin arrangement becomes more favourable
Explanation: Cyanido lies much farther toward the strong-field end than aqua. Replacing aqua by cyanido therefore generally increases \(\Delta_o\). When \(\Delta_o\) becomes larger than the pairing energy, electrons pair in the lower \(\mathrm{t_{2g}}\) orbitals. For a \(d^6\) ion, this favours \(\mathrm{t_{2g}^6e_g^0}\). The metal oxidation state is unchanged, so the \(d\)-electron count remains \(d^6\).
402. Complex \(P\) is an octahedral \(d^5\) chlorido complex, while \(Q\) is the analogous cyanido complex. Which comparison is most plausible?
ⓐ. Both must have five unpaired electrons because their metal ions are identical
ⓑ. \(P\) is more likely high spin, while \(Q\) is more likely low spin
ⓒ. \(P\) must be low spin, while \(Q\) must be high spin
ⓓ. Both must be diamagnetic because each complex is octahedral
Correct Answer: \(P\) is more likely high spin, while \(Q\) is more likely low spin
Explanation: Chlorido is a weak-field ligand and commonly produces a relatively small \(\Delta_o\). The \(d^5\) electrons then occupy \(\mathrm{t_{2g}^3e_g^2}\), giving five unpaired electrons. Cyanido is a strong-field ligand and can favour \(\mathrm{t_{2g}^5e_g^0}\). That low-spin arrangement contains one unpaired electron. The identical metal ion can therefore display different spin states because the ligand field has changed.
403. A graph has ligand position from weak field to strong field on the horizontal axis and octahedral splitting \(\Delta_o\) on the vertical axis. Which interpretation is appropriate?
ⓐ. The graph must fall because stronger ligands produce smaller splitting
ⓑ. The graph represents an exact universal numerical law independent of the metal
ⓒ. Every adjacent pair of ligands must be separated by the same increase in \(\Delta_o\)
ⓓ. A general rise, with exact values depending on the metal environment
Correct Answer: A general rise, with exact values depending on the metal environment
Explanation: Moving toward the strong-field end generally corresponds to increasing splitting. The spectrochemical series is primarily a qualitative empirical order. It does not state that equal horizontal spacing produces equal numerical changes in \(\Delta_o\). The actual splitting also depends on metal identity, oxidation state, bond distance, and geometry. The graph should therefore be interpreted as a trend rather than a universal calibration curve.
404. An octahedral \(d^6\) ion forms two complexes. For complex \(P\), \(\Delta_o=12{,}000\,\mathrm{cm^{-1}}\); for complex \(Q\), \(\Delta_o=24{,}000\,\mathrm{cm^{-1}}\). The pairing energy is \(18{,}000\,\mathrm{cm^{-1}}\) in both. Which conclusion is correct?
ⓐ. Both complexes are low spin and diamagnetic
ⓑ. Both complexes are high spin with four unpaired electrons
ⓒ. \(P\) is high spin with four unpaired electrons, while \(Q\) is low spin and diamagnetic
ⓓ. \(P\) is low spin and diamagnetic, while \(Q\) is high spin with four unpaired electrons
Correct Answer: \(P\) is high spin with four unpaired electrons, while \(Q\) is low spin and diamagnetic
Explanation: For complex \(P\):
\[
\Delta_o=12{,}000\,\mathrm{cm^{-1}}
\]
\[
P=18{,}000\,\mathrm{cm^{-1}}
\]
Therefore:
\[
\Delta_o\lt P
\]
The high-spin \(d^6\) configuration is:
\[
\mathrm{t_{2g}^4e_g^2}
\]
It contains:
\[
4\ \text{unpaired electrons}
\]
For complex \(Q\):
\[
\Delta_o=24{,}000\,\mathrm{cm^{-1}}
\]
Hence:
\[
\Delta_o\gt P
\]
The low-spin configuration is:
\[
\mathrm{t_{2g}^6e_g^0}
\]
It contains no unpaired electrons and is diamagnetic.
The ligand in \(Q\) is therefore expected to lie farther toward the strong-field end than the ligand in \(P\).
405. The CFSE of low-spin \(\mathrm{[Fe(CN)_6]^{3-}}\), excluding pairing energy, is:
ⓐ. \(-2.0\Delta_o\)
ⓑ. \(-1.2\Delta_o\)
ⓒ. \(-0.8\Delta_o\)
ⓓ. \(0\)
Correct Answer: \(-2.0\Delta_o\)
Explanation: The low-spin \(d^5\) configuration is:
\[
\mathrm{t_{2g}^5e_g^0}
\]
Each of the five lower-level electrons contributes:
\[
-0.4\Delta_o
\]
Thus:
\[
\mathrm{CFSE}=5(-0.4\Delta_o)
\]
\[
\mathrm{CFSE}=-2.0\Delta_o
\]
No electron occupies the destabilised \(\mathrm{e_g}\) level.
The two pairing-energy costs are considered separately from this CFSE.
406. Assertion: \(\mathrm{[Fe(CN)_6]^{3-}}\) is paramagnetic even though it is a low-spin complex.
Reason: Its \(\mathrm{t_{2g}^5}\) configuration contains one singly occupied orbital.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Low spin means that the number of unpaired electrons is minimised, not necessarily reduced to zero. Five electrons placed in three \(\mathrm{t_{2g}}\) orbitals produce two pairs and one single electron. The single electron makes the complex paramagnetic. Its expected spin-only moment is approximately \(1.73\,\mathrm{BM}\). The Reason therefore directly explains the magnetic property stated in the Assertion.
407. A measured spin-only moment near \(1.73\,\mathrm{BM}\) is obtained for an octahedral iron(III) complex. Which ligand and configuration best fit the observation?
ⓐ. \(\mathrm{F^-}\) and \(\mathrm{t_{2g}^3e_g^2}\)
ⓑ. \(\mathrm{CN^-}\) and \(\mathrm{t_{2g}^5e_g^0}\)
ⓒ. \(\mathrm{Cl^-}\) and \(\mathrm{t_{2g}^3e_g^2}\)
ⓓ. \(\mathrm{F^-}\) and \(\mathrm{t_{2g}^4e_g^1}\)
Correct Answer: \(\mathrm{CN^-}\) and \(\mathrm{t_{2g}^5e_g^0}\)
Explanation: A magnetic moment of approximately \(1.73\,\mathrm{BM}\) corresponds to one unpaired electron. Low-spin \(d^5\) has the configuration \(\mathrm{t_{2g}^5e_g^0}\) and contains one unpaired electron. Cyanido is sufficiently strong field to favour this arrangement in the standard example. Fluorido and chlorido are weak-field ligands and would more commonly give five unpaired electrons. The magnetic measurement therefore supports a cyanido low-spin complex.
408. For a \(d^5\) octahedral complex, the low-spin total-energy expression relative to the high-spin arrangement is \(-2.0\Delta_o+2P\). Low spin is favoured when:
ⓐ. \(\Delta_o\gt P\)
ⓑ. \(\Delta_o\lt P\)
ⓒ. \(\Delta_o=0\)
ⓓ. \(2P=0\) only
Correct Answer: \(\Delta_o\gt P\)
Explanation: High-spin \(d^5\) has zero CFSE and no paired orbitals. Low-spin \(d^5\) gains \(2.0\Delta_o\) of crystal-field stabilisation. It also creates two electron pairs, costing \(2P\). Low spin is favoured when the stabilisation exceeds the pairing cost:
\[
2\Delta_o\gt 2P
\]
Dividing by \(2\) gives \(\Delta_o\gt P\).
409. An iron(III) complex has \(\Delta_o=25{,}000\,\mathrm{cm^{-1}}\) and \(P=19{,}000\,\mathrm{cm^{-1}}\). Compared with the high-spin \(d^5\) state, the low-spin state is:
ⓐ. higher by \(12{,}000\,\mathrm{cm^{-1}}\)
ⓑ. lower by \(25{,}000\,\mathrm{cm^{-1}}\)
ⓒ. lower by \(12{,}000\,\mathrm{cm^{-1}}\)
ⓓ. equal in energy
Correct Answer: lower by \(12{,}000\,\mathrm{cm^{-1}}\)
Explanation: The high-spin \(d^5\) state has:
\[
E_{\mathrm{HS}}=0
\]
The low-spin \(d^5\) expression is:
\[
E_{\mathrm{LS}}=-2.0\Delta_o+2P
\]
Substituting the values:
\[
E_{\mathrm{LS}}=-2(25{,}000)+2(19{,}000)
\]
\[
E_{\mathrm{LS}}=-50{,}000+38{,}000
\]
\[
E_{\mathrm{LS}}=-12{,}000\,\mathrm{cm^{-1}}
\]
The negative result means that the low-spin state lies below the high-spin state.
It is lower by:
\[
12{,}000\,\mathrm{cm^{-1}}
\]
410. Study the standard crystal-field predictions.
| Row | Complex | Configuration | Unpaired electrons |
|---|
| P | \(\mathrm{[FeF_6]^{3-}}\) | \(\mathrm{t_{2g}^3e_g^2}\) | \(5\) |
| Q | \(\mathrm{[Fe(CN)_6]^{3-}}\) | \(\mathrm{t_{2g}^5e_g^0}\) | \(1\) |
| R | \(\mathrm{[FeF_6]^{3-}}\) | High spin | \(1\) |
| S | \(\mathrm{[Fe(CN)_6]^{3-}}\) | Low spin | \(1\) |
The inconsistent row is:
ⓐ. P
ⓑ. Q
ⓒ. S
ⓓ. R
Correct Answer: R
Explanation: Fluorido produces a weak field and gives high-spin \(d^5\). Its configuration is \(\mathrm{t_{2g}^3e_g^2}\), with five unpaired electrons. Cyanido produces low-spin \(\mathrm{t_{2g}^5e_g^0}\), with one unpaired electron. Rows P, Q, and S are therefore consistent. Row R incorrectly assigns the low-spin unpaired-electron count to the fluorido complex.
411. Replacing all six \(\mathrm{F^-}\) ligands in high-spin \(\mathrm{[FeF_6]^{3-}}\) by \(\mathrm{CN^-}\), without changing the oxidation state or geometry, most directly causes:
ⓐ. the iron ion to change from \(d^5\) to \(d^6\)
ⓑ. the coordination number to decrease from \(6\) to \(4\)
ⓒ. larger \(\Delta_o\) and fewer unpaired electrons
ⓓ. the \(\mathrm{t_{2g}}\) level to move above the \(\mathrm{e_g}\) level
Correct Answer: larger \(\Delta_o\) and fewer unpaired electrons
Explanation: Cyanido lies much farther toward the strong-field end of the spectrochemical series than fluorido. Ligand substitution therefore increases the octahedral splitting. The metal remains iron(III), so its \(d^5\) electron count is unchanged. Greater splitting favours pairing in the lower \(\mathrm{t_{2g}}\) orbitals. The number of unpaired electrons decreases from five to one.
412. The tetrahedral crystal-field configuration of \(\mathrm{Ni^{2+}}\) in \(\mathrm{[NiCl_4]^{2-}}\) is:
ⓐ. \(\mathrm{e^2t_2^6}\)
ⓑ. \(\mathrm{e^3t_2^5}\)
ⓒ. \(\mathrm{e^4t_2^4}\)
ⓓ. \(\mathrm{e^4t_2^2}\)
Correct Answer: \(\mathrm{e^4t_2^4}\)
Explanation: Nickel(II) has a \(d^8\) configuration. In a tetrahedral field, the lower \(\mathrm{e}\) set can hold four electrons. The remaining four electrons occupy the upper \(\mathrm{t_2}\) set. The resulting configuration is \(\mathrm{e^4t_2^4}\). This arrangement leaves two electrons unpaired in the three \(\mathrm{t_2}\) orbitals.
413. A four-coordinate nickel(II) complex has a spin-only magnetic moment close to \(2.83\,\mathrm{BM}\). Which identification is most plausible?
ⓐ. Square-planar \(\mathrm{[Ni(CN)_4]^{2-}}\)
ⓑ. Tetrahedral \(\mathrm{[NiCl_4]^{2-}}\)
ⓒ. Diamagnetic \(\mathrm{[Ni(CN)_4]^{2-}}\) with two electron pairs
ⓓ. Octahedral \(\mathrm{[NiCl_6]^{4-}}\)
Correct Answer: Tetrahedral \(\mathrm{[NiCl_4]^{2-}}\)
Explanation: A moment of approximately \(2.83\,\mathrm{BM}\) corresponds to two unpaired electrons. Tetrahedral \(d^8\) has configuration \(\mathrm{e^4t_2^4}\) and contains two unpaired electrons. The weak-field chlorido ligand favours this geometry in the standard nickel example. Square-planar \(\mathrm{[Ni(CN)_4]^{2-}}\) is diamagnetic. The magnetic result therefore identifies the tetrahedral chlorido complex.
414. If \(\Delta_t=9{,}000\,\mathrm{cm^{-1}}\) for tetrahedral \(\mathrm{[NiCl_4]^{2-}}\), its CFSE is:
ⓐ. \(-7{,}200\,\mathrm{cm^{-1}}\)
ⓑ. \(-5{,}400\,\mathrm{cm^{-1}}\)
ⓒ. \(-9{,}000\,\mathrm{cm^{-1}}\)
ⓓ. \(-3{,}600\,\mathrm{cm^{-1}}\)
Correct Answer: \(-7{,}200\,\mathrm{cm^{-1}}\)
Explanation: The tetrahedral \(d^8\) configuration is:
\[
\mathrm{e^4t_2^4}
\]
The lower-set contribution is:
\[
4(-0.6\Delta_t)=-2.4\Delta_t
\]
The upper-set contribution is:
\[
4(+0.4\Delta_t)=+1.6\Delta_t
\]
Therefore:
\[
\mathrm{CFSE}=-2.4\Delta_t+1.6\Delta_t
\]
\[
\mathrm{CFSE}=-0.8\Delta_t
\]
Substituting \(\Delta_t=9{,}000\,\mathrm{cm^{-1}}\):
\[
\mathrm{CFSE}=-0.8(9{,}000)
\]
\[
\mathrm{CFSE}=-7{,}200\,\mathrm{cm^{-1}}
\]
415. Assertion: \(\mathrm{[Ni(CN)_4]^{2-}}\) is diamagnetic, whereas \(\mathrm{[NiCl_4]^{2-}}\) is paramagnetic.
Reason: Strong-field cyanido favours a paired square-planar \(d^8\) arrangement, while weak-field chlorido favours a tetrahedral arrangement with two unpaired electrons.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Both complexes contain \(\mathrm{Ni^{2+}}\) with a \(d^8\) configuration. The cyanido complex adopts the standard paired square-planar arrangement and has no unpaired electrons. The chlorido complex adopts tetrahedral geometry and retains two unpaired electrons. Their magnetic properties therefore follow from the ligand-dependent geometries and electron arrangements. The Reason fully explains the contrast stated in the Assertion.
416. A weak-field ligand in a four-coordinate nickel(II) complex is replaced by a strong-field ligand. The product becomes diamagnetic. Which change is most consistent with the standard treatment?
ⓐ. Square planar to tetrahedral, with the number of unpaired electrons increasing
ⓑ. Tetrahedral to square planar, with the \(d^8\) electrons becoming paired
ⓒ. Octahedral to tetrahedral, with nickel changing to \(d^6\)
ⓓ. Tetrahedral to octahedral, without any change in coordination number
Correct Answer: Tetrahedral to square planar, with the \(d^8\) electrons becoming paired
Explanation: Nickel(II) has a \(d^8\) configuration. A weak-field four-coordinate complex commonly adopts tetrahedral geometry and retains two unpaired electrons. A sufficiently strong-field ligand can favour pairing of the \(d\) electrons. The paired arrangement supports square-planar bonding and gives no unpaired electrons. The change in magnetic behaviour therefore agrees with a tetrahedral-to-square-planar conversion.
417. Which properties are the same for \(\mathrm{[NiCl_4]^{2-}}\) and \(\mathrm{[Ni(CN)_4]^{2-}}\)?
ⓐ. oxidation state, ligand charge, and colour
ⓑ. \(d\)-electron count, formula mass, and spin state
ⓒ. coordination number, crystal shape, and molar mass
ⓓ. oxidation state, \(d^n\), and coordination number
Correct Answer: oxidation state, \(d^n\), and coordination number
Explanation: In both complexes, four singly charged anionic ligands surround nickel. Charge balance gives nickel the oxidation state \(+2\) in each case. Nickel(II) therefore has a \(d^8\) configuration in both entities. Each complex also has coordination number \(4\). Their ligand-field strengths, geometries, and magnetic properties differ even though these three features remain unchanged.
418. A four-coordinate complex of \(\mathrm{Ni^{2+}}\) has a measured magnetic moment of approximately \(0\,\mathrm{BM}\). The most likely description is:
ⓐ. Square planar, paired \(d^8\), and diamagnetic
ⓑ. Tetrahedral, high-spin \(d^8\), and diamagnetic
ⓒ. Square planar, with two unpaired electrons
ⓓ. Tetrahedral, with four unpaired electrons
Correct Answer: Square planar, paired \(d^8\), and diamagnetic
Explanation: A magnetic moment of \(0\,\mathrm{BM}\) indicates that no unpaired electrons are present. Nickel(II) is a \(d^8\) ion. In the standard four-coordinate comparison, complete pairing is associated with square-planar geometry. A tetrahedral \(d^8\) arrangement retains two unpaired electrons and is paramagnetic. The magnetic observation therefore strongly supports the square-planar assignment.
419. A student calculates the CFSE of \(\mathrm{[Ni(CN)_4]^{2-}}\) by using the tetrahedral levels \(\mathrm{e}\) and \(\mathrm{t_2}\). Which evaluation is correct?
ⓐ. The method is valid because every four-coordinate complex is tetrahedral
ⓑ. The method is valid because nickel(II) always has configuration \(\mathrm{e^4t_2^4}\)
ⓒ. The tetrahedral scheme is invalid for a square-planar complex
ⓓ. The method is invalid only because cyanido is a neutral ligand
Correct Answer: The tetrahedral scheme is invalid for a square-planar complex
Explanation: The labels \(\mathrm{e}\) and \(\mathrm{t_2}\) describe tetrahedral splitting. Strong-field cyanido produces a square-planar nickel(II) complex in the standard example. Square-planar geometry has a different pattern of \(d\)-orbital energies. A tetrahedral CFSE expression cannot therefore be transferred directly to \(\mathrm{[Ni(CN)_4]^{2-}}\). Cyanido is an anionic ligand, so the final option also gives an incorrect reason.
420. A \(d\)-electron transition occurs when the energy of the absorbed photon satisfies:
ⓐ. \(h\nu\lt 0\)
ⓑ. \(h\nu=0\)
ⓒ. \(h\nu\gt P\) in every complex
ⓓ. \(h\nu=\Delta E\)
Correct Answer: \(h\nu=\Delta E\)
Explanation: A photon carries energy \(h\nu\), where \(h\) is Planck’s constant and \(\nu\) is frequency. For electronic excitation, this energy must match the separation between the initial and final levels. The relation is:
\[
h\nu=\Delta E
\]
The electron then absorbs the photon and moves to the higher-energy orbital set. A photon with an unsuitable energy is not absorbed through that particular transition.
