401. Two square-planar platinum(II) entities have the same composition, \(\mathrm{[Pt(NH_3)_2Br_2]}\), but different ligand arrangements. A structural measurement shows that both have four donor attachments and no external bromide counter ions. Under an idealised bond-dipole model, equivalent opposing metal–ligand bond-dipole contributions cancel, and the cis and trans isomers have distinguishable net dipole moments. Sample P has a nonzero molecular dipole moment, whereas sample Q has a net dipole moment of zero. Which statement is correct?
ⓐ. P is trans and Q is cis, because a nonzero dipole moment requires opposite bromido ligands.
ⓑ. P is cis and Q is trans; both are neutral coordination entities with platinum in oxidation state \(+2\).
ⓒ. P and Q are linkage isomers, because bromide can attach through two different donor atoms.
ⓓ. Q must have coordination number \(2\) because the two bromido bond-dipole contributions cancel.
Correct Answer: P is cis and Q is trans; both are neutral coordination entities with platinum in oxidation state \(+2\).
Explanation: The composition contains two ammonia and two bromido ligands.
All four ligands are unidentate.
The coordination number is therefore four.
The geometry is specified as square planar.
Platinum has oxidation state \(+2\), while the two bromido ligands contribute a combined charge of \(-2\).
The two ammonia ligands are neutral.
Consequently, both complete coordination entities have charge zero.
Their formal charges and coordination numbers are identical.
However, their bond-dipole arrangements differ.
In the trans form, the two platinum–bromido bonds point in opposite directions.
The corresponding two platinum–ammine bonds also occupy opposite positions.
Under the stipulated idealised model, the opposing contributions cancel.
The trans form therefore has zero net molecular dipole moment.
In the cis form, the two bromido ligands occupy adjacent positions.
Their contributions do not cancel in the same way.
A nonzero resultant molecular dipole moment is therefore possible and is the specified distinguishing observation.
Accordingly, P is cis and Q is trans.
The observation concerns the spatial arrangements of identical ligand pairs rather than a change in bonding connectivity or oxidation state.
The two compounds are geometrical isomers, not linkage isomers.
A zero molecular dipole moment also does not imply that the metal–ligand bonds are absent or that coordination number is reduced.
402. Werner's classical cobalt(III) ammine–chloride compounds may be represented by the idealised series:
\[
\mathrm{[Co(NH_3)_{6-x}Cl_x]Cl_{3-x}}
\]
where \(x=0,1,2,3\). Each cobalt centre remains octahedral, and the coordination sphere remains intact during an immediate silver nitrate precipitation test.
The horizontal axis represents the number \(x\) of coordinated chlorido ligands, with the four permitted integer values 0, 1, 2 and 3. The vertical axis represents the moles of immediately precipitable AgCl obtained per mole of coordination compound. The data points lie on a straight line descending from \((0,3)\) to \((3,0)\).
Which statement correctly combines the graph with the geometrical-isomer possibilities?
ⓐ. At \(x=2\), two moles of AgCl are formed; at \(x=3\), three moles are formed, because all chloride ions satisfy the primary valence.
ⓑ. At \(x=2\), one mole of AgCl is formed and exactly one octahedral geometrical arrangement is possible.
ⓒ. At \(x=3\), no AgCl is formed and no geometrical isomerism is possible because the entity is neutral.
ⓓ. At \(x=2\), one mole of AgCl is formed and cis/trans forms are possible; at \(x=3\), no AgCl is formed and facial/meridional forms are possible.
Correct Answer: At \(x=2\), one mole of AgCl is formed and cis/trans forms are possible; at \(x=3\), no AgCl is formed and facial/meridional forms are possible.
Explanation: \(\textbf{Interpreting the precipitation graph:}\)
The total number of chloride ions per cobalt centre is three.
Of these, \(x\) are directly coordinated.
Therefore, the number of external chloride ions is:
\[
n_{\mathrm{external\ Cl}}=3-x
\]
Only these external chloride ions are immediately precipitated under the stated conditions.
Thus:
\[
n_{\mathrm{AgCl}}=3-x
\]
The graph consequently has slope \(-1\) and vertical intercept 3.
\(\textbf{Structure at }x=2\textbf{:}\)
Substituting:
\[
n_{\mathrm{AgCl}}=3-2
\]
\[
n_{\mathrm{AgCl}}=1\,\mathrm{mol}
\]
The corresponding coordination formula is:
\[
\mathrm{[Co(NH_3)_4Cl_2]Cl}
\]
The coordination entity contains four ammine and two chlorido ligands.
It is an octahedral \(\mathrm{MA_4B_2}\) arrangement.
The two coordinated chloride ligands can occupy adjacent or opposite positions.
Therefore, cis and trans geometrical forms are possible.
\(\textbf{Structure at }x=3\textbf{:}\)
Substituting:
\[
n_{\mathrm{AgCl}}=3-3
\]
\[
n_{\mathrm{AgCl}}=0
\]
The corresponding entity is:
\[
\mathrm{[Co(NH_3)_3Cl_3]}
\]
It is neutral and contains no external chloride counter ion.
However, it remains octahedral with ligand composition \(\mathrm{MA_3B_3}\).
This permits facial and meridional geometrical arrangements.
\(\textbf{What the graph cannot reveal:}\)
At a fixed \(x\), geometrical isomers have the same number of external chloride ions.
They therefore occupy the same precipitation point on the graph.
Consequently, the graph reveals how chloride is distributed between coordinated and external positions, but it does not distinguish geometrical isomers that share the same coordination-sphere composition.
The geometrical classification requires the additional spatial structure of the octahedron.
This explains why a neutral coordination entity can still exhibit geometrical isomerism.
403. Three cobalt(III) coordination salts have the same overall elemental composition:
P: \(\mathrm{[Co(NH_3)_5(NO_2)]Br_2}\), with nitrogen-bound nitrite.
Q: \(\mathrm{[Co(NH_3)_5(ONO)]Br_2}\), with oxygen-bound nitrite.
R: \(\mathrm{[Co(NH_3)_5Br](NO_2)Br}\), with coordinated bromido and one external nitrite ion.
Assume that all coordination spheres remain intact during dissolution and that an ion-specific analysis measures free bromide without interference from nitrite. Which statement correctly classifies the relationships and predicts the free-bromide results?
ⓐ. P and Q are geometrical isomers; P and R are linkage isomers; all three release two bromide ions.
ⓑ. P and Q are ionisation isomers; P and R are optical isomers; R releases no bromide.
ⓒ. P and Q are linkage isomers; P and R are ionisation isomers; P and Q each release two bromide ions, whereas R releases one.
ⓓ. All three are coordination isomers because their donor atoms are distributed differently between two metal centres.
Correct Answer: P and Q are linkage isomers; P and R are ionisation isomers; P and Q each release two bromide ions, whereas R releases one.
Explanation: The three compounds contain cobalt(III) and five neutral ammonia ligands.
Nitrite and bromido each carry charge \(-1\).
The coordinated anionic ligand therefore makes the complex cation's charge \(+2\) in each formula.
All three salts have identical complete elemental compositions.
However, their coordination connectivities and external anions are not identical.
In P, nitrite coordinates through nitrogen.
In Q, the same ligand coordinates through oxygen.
The donor-atom change occurs without exchanging a coordinated ligand with an external ion.
P and Q are therefore linkage isomers.
Both contain two bromide counter ions outside the coordination sphere.
Each releases two free bromide ions per formula unit under ideal dissociation.
R contains a coordinated bromido ligand in place of the coordinated nitrite.
One nitrite ion is now outside the coordination sphere.
The other external counter ion is bromide.
Thus, the relationship between P and R involves exchange of appropriate ions between coordinated and external positions.
This is ionisation isomerism.
R releases one free bromide ion per formula unit under the stated conditions.
The coordinated bromido ligand does not behave as free bromide in the specified intact-sphere analysis.
The three forms have the same cobalt oxidation state and coordination number.
The different free-bromide results therefore arise from the placement of bromide ions rather than from a change in cobalt's oxidation state.
The comparison distinguishes donor-atom substitution within an ambidentate ligand from exchange between coordination and counter-ion regions.
404. Two hypothetical octahedral complexes P and Q contain the same metal ion with configuration \(d^3\). Their octahedral splitting energies are \(12\,000\) and \(20\,000\,\mathrm{cm^{-1}}\), respectively. Assume that the appropriate electronic absorption corresponds to the elementary \(t_{2g}\rightarrow e_g\) excitation and that pairing-energy corrections are irrelevant for the ground-state configurations. Which statement correctly compares their crystal-field stabilisation energies, excitation energies and unpaired-electron counts?
ⓐ. Q gains \(8\,000\,\mathrm{cm^{-1}}\) of crystal-field stabilisation, its excitation energy rises by \(9\,600\,\mathrm{cm^{-1}}\), and its three electrons become paired.
ⓑ. Q gains \(9\,600\,\mathrm{cm^{-1}}\) of crystal-field stabilisation, its excitation energy rises by \(8\,000\,\mathrm{cm^{-1}}\), and both complexes retain three unpaired electrons.
ⓒ. Q gains \(8\,000\,\mathrm{cm^{-1}}\) of crystal-field stabilisation, its excitation energy rises by \(8\,000\,\mathrm{cm^{-1}}\), and both complexes retain three unpaired electrons.
ⓓ. Q gains \(9\,600\,\mathrm{cm^{-1}}\) of crystal-field stabilisation, but its excitation energy decreases because the lower \(t_{2g}\) set becomes more stable.
Correct Answer: Q gains \(9\,600\,\mathrm{cm^{-1}}\) of crystal-field stabilisation, its excitation energy rises by \(8\,000\,\mathrm{cm^{-1}}\), and both complexes retain three unpaired electrons.
Explanation: \(\textbf{Ground-state occupancy:}\)
A \(d^3\) ion in an octahedral field has configuration:
\[
t_{2g}^{3}e_g^{0}
\]
The three electrons occupy separate lower-energy orbitals.
This configuration is unchanged by the specified increase in splitting.
Both P and Q therefore contain three unpaired electrons.
\(\textbf{Crystal-field energy of P:}\)
Each \(t_{2g}\) electron contributes \(-0.4\Delta_o\).
Therefore:
\[
E_{\mathrm{CF},P}=3(-0.4\Delta_{o,P})
\]
\[
E_{\mathrm{CF},P}=-1.2(12\,000)
\]
\[
E_{\mathrm{CF},P}=-14\,400\,\mathrm{cm^{-1}}
\]
\(\textbf{Crystal-field energy of Q:}\)
Similarly:
\[
E_{\mathrm{CF},Q}=3(-0.4\Delta_{o,Q})
\]
\[
E_{\mathrm{CF},Q}=-1.2(20\,000)
\]
\[
E_{\mathrm{CF},Q}=-24\,000\,\mathrm{cm^{-1}}
\]
\(\textbf{Change in stabilisation:}\)
\[
E_{\mathrm{CF},Q}-E_{\mathrm{CF},P}=-24\,000-(-14\,400)
\]
\[
E_{\mathrm{CF},Q}-E_{\mathrm{CF},P}=-9\,600\,\mathrm{cm^{-1}}
\]
The crystal-field energy becomes more negative.
Q consequently has an additional stabilisation magnitude of \(9\,600\,\mathrm{cm^{-1}}\).
\(\textbf{Change in excitation energy:}\)
The specified elementary electronic excitation requires energy \(\Delta_o\).
Therefore:
\[
\Delta E_{\mathrm{excitation}}=\Delta_{o,Q}-\Delta_{o,P}
\]
\[
\Delta E_{\mathrm{excitation}}=20\,000-12\,000
\]
\[
\Delta E_{\mathrm{excitation}}=8\,000\,\mathrm{cm^{-1}}
\]
The excitation energy increases, even though the ground-state crystal-field energy becomes more negative.
This distinction arises because ground-state stabilisation and the energy gap between orbital sets are different quantities.
\(\textbf{Magnetic consistency:}\)
The ground-state configuration remains \(t_{2g}^{3}e_g^{0}\) in both complexes.
The three electrons do not pair merely because the splitting is larger.
Thus, the spin-only magnetic moments remain the same under the stated assumptions.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta|E_{\mathrm{CF}}|=9\,600\,\mathrm{cm^{-1}},\quad\Delta E_{\mathrm{excitation}}=8\,000\,\mathrm{cm^{-1}}}
\]
405. A hypothetical octahedral cobalt coordination entity undergoes the following coupled changes:
I. Cobalt is reduced from oxidation state \(+3\) to \(+2\).
II. One coordinated hydroxido ligand is protonated to form a coordinated aqua ligand.
III. The remaining five ammonia ligands stay coordinated.
The initial entity is \(\mathrm{[Co(NH_3)_5(OH)]^{2+}}\). Assume the initial cobalt(III) entity is low spin and the final cobalt(II) entity is high spin. Which statement correctly describes the product?
ⓐ. The coordination-entity charge decreases to \(+1\), and both complexes are diamagnetic.
ⓑ. The charge remains \(+2\), but coordination number decreases to five because hydroxido is replaced by water.
ⓒ. The charge increases to \(+3\), and cobalt retains the original \(d^6\) configuration.
ⓓ. The charge remains \(+2\), coordination number remains six, and the number of unpaired d electrons increases from zero to three.
Correct Answer: The charge remains \(+2\), coordination number remains six, and the number of unpaired d electrons increases from zero to three.
Explanation: \(\textbf{Initial charge and electronic structure:}\)
Cobalt initially has oxidation state \(+3\).
The five ammonia ligands are neutral, while hydroxido contributes charge \(-1\).
Therefore:
\[
q_i=(+3)+5(0)+(-1)=+2
\]
Cobalt(III) has a \(d^6\) configuration.
The specified low-spin octahedral arrangement is \(t_{2g}^{6}e_g^{0}\), containing no unpaired electrons.
\(\textbf{Coupled reduction and protonation:}\)
Reduction of cobalt(III) to cobalt(II) decreases the metal oxidation state by one.
Protonation simultaneously converts the anionic hydroxido ligand into neutral aqua.
Thus, one positive charge unit is lost through metal reduction, while one negative ligand charge is removed through protonation.
The resulting entity is \(\mathrm{[Co(NH_3)_5(H_2O)]^{2+}}\).
Its charge is:
\[
q_f=(+2)+5(0)+0
\]
\[
q_f=+2
\]
\(\textbf{Coordination and magnetic consequences:}\)
The hydroxido oxygen donor remains attached after protonation.
Five ammonia donor atoms and one aqua donor atom therefore maintain coordination number six.
Cobalt(II) has a \(d^7\) configuration.
The specified high-spin arrangement is \(t_{2g}^{5}e_g^{2}\), containing three unpaired electrons.
The initial entity is diamagnetic, while the final entity is paramagnetic.
The unchanged entity charge does not imply unchanged oxidation state, ligand charge or magnetic behaviour.
The charge preservation results specifically from the two coupled chemical changes.
\(\textbf{Final answer:}\)
\[
q_i=q_f=+2,\quad CN_i=CN_f=6,\quad n:0\rightarrow3
\]
406. In the elementary crystal-field model, ligands are treated as point charges or dipoles. A student argues that an anionic ligand must always produce greater splitting than a neutral ligand because an anion carries a negative electrical charge. Consider the following statements.
Statement I: The standard spectrochemical series places neutral CO among the strongest-field ligands.
Statement II: Fluorido, although anionic, lies below ammonia in the standard spectrochemical series.
Statement III: Formal ligand charge alone is sufficient to predict the complete spectrochemical series.
Statement IV: The elementary purely electrostatic treatment has limitations because actual metal–ligand bonding can contain contributions not captured by a point-charge description.
Which combination is correct?
ⓐ. Statements I, II and IV only
ⓑ. Statements I and III only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II, III and IV
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because carbon monoxide occupies the strong-field end of the standard spectrochemical series.
CO is conventionally treated as a neutral ligand in formal oxidation-state calculations.
Its strong-field behaviour therefore cannot be explained simply by assigning it a large negative formal charge.
Statement II is correct because fluorido is a weaker-field ligand than ammonia in the standard series.
Fluorido carries charge \(-1\), whereas ammonia is electrically neutral.
This provides another counterexample to the student's proposed charge-only ranking.
Statement III is incorrect because formal electrical charge does not uniquely determine the magnitude of crystal-field splitting.
The order of ligands depends on their chemical nature and interactions with the metal.
Consequently, comparing only neutral and negatively charged ligands cannot reproduce the full spectrochemical series.
Statement IV is correct because the elementary crystal-field model describes metal–ligand interactions primarily in electrostatic terms.
That model helps explain d-orbital splitting, electron occupancy and several magnetic and optical observations.
However, it does not completely represent the covalent contributions to actual metal–ligand bonding.
Its limitations become important when attempting to explain the detailed ligand-strength order.
The model should therefore be used within its stated assumptions rather than converted into an unconditional rule about ligand charge.
The correct combination is I, II and IV.
407. Two octahedral coordination entities contain the same divalent metal M:
P: \(\mathrm{[M(NH_3)_6]^{2+}}\)
Q: \(\mathrm{[M(en)_3]^{2+}}\)
Under the stated conditions, Q has the greater experimentally measured formation stability. A student concludes that this observation proves both that Q has the greater octahedral splitting and that every electronic transition of Q must occur at a shorter wavelength than the corresponding transition of P. Which evaluation is most appropriate?
ⓐ. The conclusion follows necessarily because the stability of a coordination entity and its crystal-field splitting are the same physical quantity.
ⓑ. The conclusion follows necessarily because three en molecules provide fewer donor atoms than six ammonia molecules.
ⓒ. The observed stability difference is consistent with the chelate effect, but stability alone does not quantitatively establish the splitting energies or complete absorption spectra.
ⓓ. The greater stability of Q proves that its metal oxidation state has increased from \(+2\) to \(+3\).
Correct Answer: The observed stability difference is consistent with the chelate effect, but stability alone does not quantitatively establish the splitting energies or complete absorption spectra.
Explanation: Both coordination entities contain the same metal ion in oxidation state \(+2\).
The six ammonia ligands in P are neutral and unidentate.
They supply six independent nitrogen donor attachments.
The three en ligands in Q are also neutral but didentate.
They likewise supply six nitrogen donor attachments.
Thus, both entities have coordination number six and charge \(+2\).
However, Q contains three chelating ligands rather than six independent unidentate ligands.
Its greater measured formation stability is consistent with the general chelate effect.
Formation stability describes the equilibrium favourability of producing a coordination entity under specified conditions.
Crystal-field splitting describes the energy separation between particular groups of metal d orbitals.
These are different chemical quantities.
Although changes in ligand identity can affect both properties, a measured stability ranking is not a numerical measurement of \(\Delta_o\).
Nor does a formation-stability measurement provide a complete electronic absorption spectrum.
Predicting a particular absorption wavelength requires information about the relevant electronic energy difference.
The student's argument therefore makes an unjustified inference from thermodynamic stability to electronic transition energy.
The observation supports a chelation-based stability comparison but not the complete optical conclusion.
408. Calcium(II) and magnesium(II) are the principal metal ions responsible for hardness in many water samples. Under suitable conditions, each can form a stable \(1:1\) complex with the hexadentate ligand \(\mathrm{EDTA^{4-}}\). Which statement correctly explains the coordination-chemistry principle behind using EDTA to bind these ions?
ⓐ. Each EDTA molecule supplies six independent ligand molecules, so six metal ions are trapped by one EDTA.
ⓑ. One EDTA ligand can attach through six donor atoms to one metal ion, lowering the availability of that free metal ion.
ⓒ. EDTA binds the metal ions by converting calcium(II) and magnesium(II) into metal atoms of oxidation state zero.
ⓓ. EDTA must precipitate both metal ions as insoluble carbonates before coordinate bonds can form.
Correct Answer: One EDTA ligand can attach through six donor atoms to one metal ion, lowering the availability of that free metal ion.
Explanation: \(\textbf{Chelating behaviour of EDTA:}\)
EDTA can bind a suitable metal centre through two nitrogen and four oxygen donor atoms.
When all six donor atoms coordinate, EDTA is hexadentate and forms a chelate with the metal.
One EDTA molecule is still one ligand, not six independent ligands.
\(\textbf{Effect on dissolved metal ions:}\)
In the stated \(1:1\) complexes, calcium(II) or magnesium(II) is retained within a metal–EDTA coordination entity.
Under conditions favouring complex formation, some metal ions that were freely available become ligand-bound.
This can reduce their availability for reactions requiring free \(\mathrm{Ca^{2+}}\) or \(\mathrm{Mg^{2+}}\).
Chelation does not require reduction to a neutral metal atom or precipitation as a carbonate.
\(\textbf{Conclusion:}\)
The analytical importance of EDTA arises from its multidentate coordination and the resulting change in free-ion availability.
409. In an ideal octahedral crystal-field energy diagram, the five d orbitals split into a lower threefold-degenerate set and an upper twofold-degenerate set. The splitting is \(20\,000\,\mathrm{cm^{-1}}\), and the weighted barycentre of the five orbital energies is defined as zero. A student constructs a diagram with the lower set at \(-8\,000\,\mathrm{cm^{-1}}\) and the upper set at \(+8\,000\,\mathrm{cm^{-1}}\). Which statement correctly diagnoses the diagram and gives the appropriate upper-level energy?
ⓐ. The diagram is correct because the upper and lower shifts must be equal in magnitude.
ⓑ. The upper level should be \(+12\,000\,\mathrm{cm^{-1}}\) to preserve both \(\Delta_o\) and the barycentre.
ⓒ. The diagram is incorrect; the upper level should be \(+8\,000\,\mathrm{cm^{-1}}\), but the lower level should be \(-12\,000\,\mathrm{cm^{-1}}\).
ⓓ. The diagram is correct only for a \(d^5\) ion because five electrons occupy all five orbitals.
Correct Answer: The upper level should be \(+12\,000\,\mathrm{cm^{-1}}\) to preserve both \(\Delta_o\) and the barycentre.
Explanation: \(\textbf{Barycentre condition:}\)
The lower octahedral set contains three orbitals.
The upper set contains two orbitals.
Their weighted energy shifts must preserve the barycentre:
\[
3E_{\mathrm{lower}}+2E_{\mathrm{upper}}=0
\]
\(\textbf{Testing the proposed diagram:}\)
The student assigns:
\[
E_{\mathrm{lower}}=-8\,000\,\mathrm{cm^{-1}}
\]
\[
E_{\mathrm{upper}}=+8\,000\,\mathrm{cm^{-1}}
\]
The weighted energy sum would be:
\[
3(-8\,000)+2(+8\,000)
\]
\[
=-24\,000+16\,000
\]
\[
=-8\,000\,\mathrm{cm^{-1}}
\]
The barycentre would therefore not remain at zero.
The proposed energy separation is also only:
\[
8\,000-(-8\,000)=16\,000\,\mathrm{cm^{-1}}
\]
This contradicts the stated splitting of \(20\,000\,\mathrm{cm^{-1}}\).
\(\textbf{Constructing the correct upper level:}\)
Using the given lower level in the barycentre equation:
\[
3(-8\,000)+2E_{\mathrm{upper}}=0
\]
\[
2E_{\mathrm{upper}}=24\,000
\]
\[
E_{\mathrm{upper}}=12\,000\,\mathrm{cm^{-1}}
\]
\(\textbf{Splitting verification:}\)
\[
\Delta_o=E_{\mathrm{upper}}-E_{\mathrm{lower}}
\]
\[
\Delta_o=12\,000-(-8\,000)
\]
\[
\Delta_o=20\,000\,\mathrm{cm^{-1}}
\]
Both conditions are satisfied.
The upper and lower shifts are unequal because the two orbital sets have different degeneracies.
The three lower orbitals each undergo a shift of \(-0.4\Delta_o\), while the two upper orbitals each undergo a shift of \(+0.6\Delta_o\).
This is a property of the orbital-energy diagram and does not depend on a particular d-electron occupancy.
\(\textbf{Final answer:}\)
\[
\boxed{E_{\mathrm{lower}}=-8\,000\,\mathrm{cm^{-1}},\quad E_{\mathrm{upper}}=+12\,000\,\mathrm{cm^{-1}}}
\]
410. A solution initially contains \(0.080\,\mathrm{mol}\) of \(\mathrm{[Mg(EDTA)]^{2-}}\). Exactly \(0.060\,\mathrm{mol}\) of free calcium(II) ions is added. Under the specified conditions, the following exchange reaction proceeds quantitatively until one reactant is exhausted:
\[
\mathrm{[Mg(EDTA)]^{2-}+Ca^{2+}\rightarrow[Ca(EDTA)]^{2-}+Mg^{2+}}
\]
Assume no competing reactions and complete retention of EDTA in one-metal coordination entities. Which statement correctly gives the final amounts of calcium–EDTA, magnesium–EDTA and free magnesium(II)?
ⓐ. \(0.080,\ 0,\ 0.080\,\mathrm{mol}\), respectively
ⓑ. \(0.060,\ 0.020,\ 0.020\,\mathrm{mol}\), respectively
ⓒ. \(0.060,\ 0.020,\ 0.060\,\mathrm{mol}\), respectively
ⓓ. \(0.020,\ 0.060,\ 0.060\,\mathrm{mol}\), respectively
Correct Answer: \(0.060,\ 0.020,\ 0.060\,\mathrm{mol}\), respectively
Explanation: \(\textbf{Coordination-exchange stoichiometry:}\)
The stated reaction exchanges one coordinated magnesium ion for one calcium ion.
The metal-to-EDTA ratio remains \(1:1\).
No EDTA molecule is created or destroyed during the exchange.
\(\textbf{Identifying the limiting reactant:}\)
The initial amounts are:
\[
n_{\mathrm{MgEDTA}}=0.080\,\mathrm{mol}
\]
\[
n_{\mathrm{Ca^{2+}}}=0.060\,\mathrm{mol}
\]
The exchange reaction requires equal molar amounts of its two reactants.
Calcium(II) is therefore the limiting reactant.
\(\textbf{Calcium–EDTA formation:}\)
Every mole of calcium consumed forms one mole of calcium–EDTA.
Thus:
\[
n_{\mathrm{CaEDTA,formed}}=0.060\,\mathrm{mol}
\]
\(\textbf{Magnesium–EDTA remaining:}\)
The exchange consumes \(0.060\,\mathrm{mol}\) of the original magnesium complex.
Therefore:
\[
n_{\mathrm{MgEDTA,remaining}}=0.080-0.060
\]
\[
n_{\mathrm{MgEDTA,remaining}}=0.020\,\mathrm{mol}
\]
\(\textbf{Magnesium released:}\)
One mole of free magnesium(II) is released per mole of calcium–EDTA formed.
Therefore:
\[
n_{\mathrm{Mg^{2+},released}}=0.060\,\mathrm{mol}
\]
\(\textbf{Conservation checks:}\)
The total amount of coordinated EDTA remains:
\[
0.060+0.020=0.080\,\mathrm{mol}
\]
The total magnesium amount also remains:
\[
0.020+0.060=0.080\,\mathrm{mol}
\]
The incoming calcium is entirely accounted for in the calcium–EDTA complex.
Both coordinated entities carry charge \(-2\), because each divalent metal combines with \(\mathrm{EDTA^{4-}}\).
The exchange changes which metal is coordinated but does not change the formal charge or ordinary maximum denticity of EDTA.
The calculation relies on the explicitly supplied quantitative-exchange condition; it does not assume that every metal can displace every other metal from EDTA under arbitrary conditions.
\(\textbf{Final answer:}\)
\[
\boxed{n_{\mathrm{CaEDTA}}=0.060,\quad n_{\mathrm{MgEDTA}}=0.020,\quad n_{\mathrm{free\ Mg^{2+}}}=0.060\ \mathrm{mol}}
\]
411. Two cobalt(III) coordination salts are studied:
P: \(\mathrm{[Co(NH_3)_6]Cl_3}\)
Q: \(\mathrm{K_3[CoF_6]}\)
Assume complete ideal dissociation into counter ions and intact coordination entities. The cobalt(III) hexaammine entity is low spin, whereas the hexafluorido entity is high spin. Which statement is correct?
ⓐ. Both yield four ideal ionic species, but their complex ions differ magnetically; equal ion counts need not mean equal conductivity.
ⓑ. P produces four ionic species, while Q produces seven because its six coordinated fluoride ligands dissociate separately.
ⓒ. Both salts are diamagnetic because cobalt remains in oxidation state \(+3\).
ⓓ. Both have equal measured conductivity and equal spin-only magnetic moments because their coordination numbers are six.
Correct Answer: Both yield four ideal ionic species, but their complex ions differ magnetically; equal ion counts need not mean equal conductivity.
Explanation: P contains the hexaamminecobalt(III) cation and three external chloride counter ions.
Its ideal dissociation produces one complex cation and three chloride ions.
Therefore, four ionic species arise per formula unit.
Q contains three potassium counter ions and the hexafluoridocobaltate(III) anion.
Its ideal dissociation produces three potassium ions and one intact complex anion.
Thus, Q also produces four ionic species per formula unit.
However, the coordinated fluoride ligands are not counted as separate free ions under the stated conditions.
Both cobalt centres have oxidation state \(+3\) and configuration \(d^6\).
The low-spin hexaamminecobalt(III) entity has occupancy \(t_{2g}^{6}e_g^{0}\).
It is diamagnetic because all its d electrons are paired.
The high-spin hexafluorido entity has occupancy \(t_{2g}^{4}e_g^{2}\).
It contains four unpaired d electrons and is paramagnetic.
Therefore, equal ideal ionic-species counts do not imply equal magnetic behaviour.
Nor do they prove identical measured electrical conductivities.
Actual conductivity also depends on the concentrations, charges and mobilities of the ions under the measurement conditions.
The two salts also contain complex ions of opposite electrical charges, although both are neutral as complete compounds.
The comparison distinguishes ion-count bookkeeping, electrical transport and magnetic electronic structure.
412. An aqueous solution of \(\mathrm{KMnO_4}\) is intensely purple, whereas a solution containing \(\mathrm{[Cr(H_2O)_6]^{3+}}\) is coloured because of electronic transitions associated with its chromium d orbitals. Which statement correctly distinguishes the electronic origins of these colours?
ⓐ. Both colours must arise exclusively from ordinary metal-centred d–d transitions because manganese and chromium are d-block elements.
ⓑ. Manganese in permanganate is \(+2\), with configuration \(d^5\), while chromium(III) is \(d^0\).
ⓒ. Both ions have \(d^{10}\) configurations, so neither can absorb visible radiation.
ⓓ. Permanganate contains \(d^0\) Mn(VII), so its colour is not ordinary d–d absorption; \(d^3\) chromium(III) can undergo d–d transitions.
Correct Answer: Permanganate contains \(d^0\) Mn(VII), so its colour is not ordinary d–d absorption; \(d^3\) chromium(III) can undergo d–d transitions.
Explanation: Permanganate is the anion \(\mathrm{MnO_4^-}\).
Each oxygen is assigned oxidation state \(-2\) in ordinary oxidation-state bookkeeping.
Therefore, manganese has oxidation state \(+7\).
Manganese(VII) has a \(d^0\) configuration.
An ordinary metal-centred d–d transition requires a d electron in an occupied initial state.
Such an electron is absent from the manganese(VII) d subshell.
Therefore, the intense purple colour of permanganate cannot be attributed to an ordinary manganese-centred d–d excitation.
Its visible absorption is associated with charge-transfer electronic transitions involving the oxygen-containing ligand environment.
Chromium(III), by contrast, has configuration \(3d^3\).
In an octahedral aqua environment, its electrons occupy the lower \(t_{2g}\) orbitals.
Suitable absorption can promote electronic excitation involving the higher d-orbital set.
This provides an ordinary crystal-field explanation for relevant absorption features of the chromium(III) entity.
The two substances therefore illustrate different possible origins of colour.
The presence of a transition-metal element does not guarantee a partially filled d subshell in every oxidation state.
Nor does the absence of an ordinary d–d transition imply that a compound must be colourless.
The distinction must be established from metal oxidation state, electron configuration and the available electronic transitions.
413. Two coordination-compound solutions, P and Q, are placed in separate transparent cells. Their idealised visible-light absorption properties are supplied below:
| Solution | Radiation absorbed completely | Radiation transmitted completely |
|---|
| P | Red | Green and blue |
| Q | Green | Red and blue |
White light containing red, green and blue components of equal initial intensity is directed first through P and then through Q. A second experiment reverses the order of the cells. Assume negligible reflection, scattering, emission and chemical interaction between solutions. Which statement correctly predicts the transmitted light?
ⓐ. The first arrangement transmits only green light, while the reversed arrangement transmits only red light.
ⓑ. Both arrangements transmit only blue light; changing the order does not restore a colour component already absorbed.
ⓒ. The first arrangement transmits only red light, while the reversed arrangement transmits only green light.
ⓓ. Neither arrangement transmits any light because two differently coloured coordination compounds are placed in sequence.
Correct Answer: Both arrangements transmit only blue light; changing the order does not restore a colour component already absorbed.
Explanation: The question specifies how each solution interacts with the three incident colour components.
P absorbs red light completely.
It transmits both green and blue components.
Therefore, after passing through P, the original white light no longer contains its red component.
The remaining beam contains green and blue radiation.
This beam then enters Q.
Q absorbs green light completely and transmits blue light.
Consequently, only blue light emerges from the sequence P followed by Q.
In the reversed arrangement, white light first enters Q.
The green component is removed, leaving red and blue radiation.
This remaining beam enters P.
P removes the red component and transmits blue.
The resulting transmitted light is again blue.
The order of the cells changes which colour is removed first but not which components survive both ideal filters.
Absorption cannot recreate a previously removed component under the stated absence of emission or chemical change.
This example demonstrates why observed colour must be interpreted through the actual absorption properties and incident spectrum rather than through the informal addition of colour names.
It does not require assigning any particular oxidation state or ligand field to P or Q.
414. Two coordination entities P and Q are compared experimentally. P has greater measured thermodynamic formation stability, but Q has a larger octahedral crystal-field splitting. A separate substitution experiment shows that P undergoes a selected ligand-exchange reaction more rapidly than Q under identical conditions. Consider the following statements.
Statement I: Greater formation stability necessarily implies greater crystal-field splitting.
Statement II: Thermodynamic formation stability and the rate of ligand substitution describe different properties.
Statement III: A larger crystal-field splitting does not, by itself, uniquely determine the rate of every ligand-exchange reaction.
Statement IV: The three supplied observations are inherently contradictory because stability, colour-related splitting and substitution rate must have the same ordering.
Which combination is correct?
ⓐ. Statements I and II only
ⓑ. Statements I, III and IV only
ⓒ. Statements II and III only
ⓓ. Statements II, III and IV only
Correct Answer: Statements II and III only
Explanation: Statement I is incorrect because formation stability and crystal-field splitting are not identical physical quantities.
Formation stability concerns the equilibrium favourability of forming a coordination entity under specified conditions.
Crystal-field splitting concerns the energy separation between metal d-orbital groups in a particular ligand environment.
A complex may therefore be more stable in one thermodynamic comparison without necessarily possessing the greater splitting energy.
Statement II is correct because equilibrium stability and reaction rate answer different chemical questions.
The former describes relative equilibrium favourability.
The latter describes how quickly a particular transformation occurs.
Statement III is correct because knowing the d-orbital splitting alone does not specify the complete mechanism or activation-energy barrier for ligand substitution.
The elementary crystal-field model is useful for understanding electronic structure, magnetic behaviour and relevant electronic absorption.
It does not provide a complete kinetic prediction for every possible coordination reaction.
Statement IV is incorrect because the experimental orderings need not be identical.
There is no general requirement that the more stable complex must absorb at the shorter wavelength or react more slowly in every substitution.
The supplied observations may therefore coexist without logical contradiction.
Interpreting them correctly requires distinguishing thermodynamic, spectroscopic and kinetic evidence.
The comparison demonstrates a limitation of reducing all coordination-compound properties to one ligand-field parameter.
415. In a hypothetical metallurgical recovery experiment, a solution contains \(0.120\,\mathrm{mol}\) of \(\mathrm{[Au(CN)_2]^-}\) and \(0.080\,\mathrm{mol}\) of \(\mathrm{[Ag(CN)_2]^-}\). Exactly \(0.075\,\mathrm{mol}\) of zinc is added. Assume zinc can reduce either metal(I) complex according to the \(1:2\) stoichiometry:
\[
\mathrm{Zn+2[M(CN)_2]^-\rightarrow[Zn(CN)_4]^{2-}+2M}
\]
where M is Au or Ag. All zinc is consumed, no side reactions occur, and no information about preferential reduction is supplied. Which conclusion is justified?
ⓐ. Exactly \(0.120\,\mathrm{mol}\) of gold and \(0.030\,\mathrm{mol}\) of silver must be recovered because gold is more valuable.
ⓑ. Exactly \(0.080\,\mathrm{mol}\) of silver and \(0.070\,\mathrm{mol}\) of gold must be recovered because silver is initially present in the smaller amount.
ⓒ. All gold must be recovered before silver reduction begins because zinc reacts with the more abundant coordination entity first.
ⓓ. A total of \(0.150\,\mathrm{mol}\) of metal is recovered, but the individual gold and silver amounts cannot be uniquely determined from the supplied information.
Correct Answer: A total of \(0.150\,\mathrm{mol}\) of metal is recovered, but the individual gold and silver amounts cannot be uniquely determined from the supplied information.
Explanation: \(\textbf{Available metal-complex amount:}\)
The solution initially contains:
\[
n_{\mathrm{Au(I)}}=0.120\,\mathrm{mol}
\]
\[
n_{\mathrm{Ag(I)}}=0.080\,\mathrm{mol}
\]
The combined amount of reducible metal(I) coordination entities is:
\[
n_{\mathrm{M(I),total}}=0.120+0.080
\]
\[
n_{\mathrm{M(I),total}}=0.200\,\mathrm{mol}
\]
\(\textbf{Zinc reduction capacity:}\)
The supplied reaction shows that one mole of zinc reduces two moles of metal(I) complex.
Therefore:
\[
n_{\mathrm{M,recovered}}=2n_{\mathrm{Zn}}
\]
Substituting:
\[
n_{\mathrm{M,recovered}}=2(0.075)
\]
\[
n_{\mathrm{M,recovered}}=0.150\,\mathrm{mol}
\]
The zinc amount is insufficient to recover the entire \(0.200\,\mathrm{mol}\) of dissolved metals.
\(\textbf{Individual-metal uncertainty:}\)
Let \(x\) be the amount of gold recovered and \(y\) the amount of silver recovered.
The zinc balance gives:
\[
x+y=0.150
\]
The initial inventories impose:
\[
0\leq x\leq0.120
\]
\[
0\leq y\leq0.080
\]
These conditions permit more than one solution.
For example:
\[
x=0.120,\quad y=0.030
\]
also satisfies the balances, as does:
\[
x=0.070,\quad y=0.080
\]
The experiment does not state that one metal is preferentially reduced.
Therefore, neither particular allocation can be selected uniquely.
\(\textbf{What is determined:}\)
The total metallic amount recovered is fixed by zinc stoichiometry.
The distribution of that amount between gold and silver requires additional chemical or experimental information.
This is an example in which a valid overall reaction balance does not establish a unique product composition for competing metal-ion reactions.
\(\textbf{Final answer:}\)
\[
\boxed{n_{\mathrm{Au,recovered}}+n_{\mathrm{Ag,recovered}}=0.150\,\mathrm{mol}}
\]
416. An aqueous iron(II) coordination entity initially has six coordinated cyanido ligands. Under a hypothetical controlled substitution, exactly three cyanido ligands are replaced by three water molecules:
\[
\mathrm{[Fe(CN)_6]^{4-}\rightarrow[Fe(CN)_3(H_2O)_3]^-}
\]
Assume that both entities are octahedral, iron remains in oxidation state \(+2\), the initial splitting is \(30\,000\,\mathrm{cm^{-1}}\), the final splitting is \(12\,000\,\mathrm{cm^{-1}}\), and the pairing energy is \(18\,000\,\mathrm{cm^{-1}}\) in both environments. Which statement correctly predicts the electronic and ionic consequences?
ⓐ. The initial entity is high spin and the final entity is low spin; both require four potassium counter ions.
ⓑ. Initially low spin and diamagnetic; finally high spin with four unpaired electrons, requiring one rather than four potassium ions.
ⓒ. Both entities are low spin because iron remains in oxidation state \(+2\), and both complex ions retain charge \(-4\).
ⓓ. The final entity is diamagnetic because three coordinated cyanido ligands remain, and its coordination number falls to three.
Correct Answer: Initially low spin and diamagnetic; finally high spin with four unpaired electrons, requiring one rather than four potassium ions.
Explanation: \(\textbf{Initial iron-ion configuration:}\)
Iron(II) has configuration \(d^6\).
The initial complex contains six unidentate cyanido ligands, so it is six-coordinate.
Its specified splitting is:
\[
\Delta_{o,i}=30\,000\,\mathrm{cm^{-1}}
\]
The pairing energy is:
\[
P_{\mathrm{pair}}=18\,000\,\mathrm{cm^{-1}}
\]
Since:
\[
\Delta_{o,i}\gt P_{\mathrm{pair}}
\]
the low-spin arrangement is favoured.
Its occupancy is \(t_{2g}^{6}e_g^{0}\), with no unpaired electrons.
\(\textbf{Effect of ligand substitution on charge:}\)
Each cyanido ligand contributes charge \(-1\).
The starting entity charge is:
\[
q_i=(+2)+6(-1)
\]
\[
q_i=-4
\]
In the product, three cyanido ligands remain, while three neutral aqua ligands are coordinated.
Therefore:
\[
q_f=(+2)+3(-1)+3(0)
\]
\[
q_f=-1
\]
\(\textbf{Final spin-state decision:}\)
The final splitting is:
\[
\Delta_{o,f}=12\,000\,\mathrm{cm^{-1}}
\]
This is smaller than the pairing energy:
\[
\Delta_{o,f}\lt P_{\mathrm{pair}}
\]
The high-spin \(d^6\) configuration is therefore favoured:
\[
t_{2g}^{4}e_g^{2}
\]
It contains four unpaired electrons.
\(\textbf{Counter-ion and coordination checks:}\)
The original potassium salt requires four potassium ions per coordination anion.
The substituted entity carries charge \(-1\), so its neutral potassium salt requires only one potassium ion.
All six coordinated ligands remain unidentate.
The coordination number consequently remains six.
Thus, ligand substitution changes spin state and overall entity charge without changing iron oxidation state or coordination number.
\(\textbf{Final answer:}\)
\[
n_i=0,\quad n_f=4,\quad q:-4\rightarrow-1,\quad N_{\mathrm{K^+}}:4\rightarrow1
\]
417. Disodium EDTA is commonly represented in solution using the ligand-acid form \(\mathrm{H_2EDTA^{2-}}\). In a hypothetical quantitative complexation experiment, a divalent metal ion M reacts according to:
\[
\mathrm{M^{2+}+H_2EDTA^{2-}\rightarrow[M(EDTA)]^{2-}+2H^+}
\]
Assume the EDTA ligand becomes fully deprotonated during complexation and coordinates through all six donor atoms. If \(0.0250\,\mathrm{mol}\) of metal ions is completely complexed, which statement correctly gives the amount of hydrogen ions produced, the coordination number and the formal charge of the resulting complex?
ⓐ. \(0.0250\,\mathrm{mol}\) hydrogen ions; coordination number six; complex charge zero
ⓑ. \(0.0500\,\mathrm{mol}\) hydrogen ions; coordination number two; complex charge \(-2\)
ⓒ. \(0.0500\,\mathrm{mol}\) hydrogen ions; coordination number six; complex charge \(-2\)
ⓓ. \(0.100\,\mathrm{mol}\) hydrogen ions; coordination number six; complex charge \(+2\)
Correct Answer: \(0.0500\,\mathrm{mol}\) hydrogen ions; coordination number six; complex charge \(-2\)
Explanation: \(\textbf{Distinguishing the protonated and coordinated ligand forms:}\)
The reactant contains \(\mathrm{H_2EDTA^{2-}}\).
The product contains fully deprotonated \(\mathrm{EDTA^{4-}}\) coordinated to M.
The reaction explicitly releases two hydrogen ions for every EDTA ligand that becomes incorporated.
\(\textbf{Metal–ligand combining ratio:}\)
The balanced equation shows:
\[
1\,\mathrm{mol\ M^{2+}}\longleftrightarrow1\,\mathrm{mol\ EDTA}
\]
The amount of metal complexed is:
\[
n_{\mathrm{M^{2+}}}=0.0250\,\mathrm{mol}
\]
Therefore:
\[
n_{\mathrm{EDTA,incorporated}}=0.0250\,\mathrm{mol}
\]
\(\textbf{Hydrogen-ion production:}\)
Each mole of complex formed releases two moles of hydrogen ions.
Hence:
\[
n_{\mathrm{H^+}}=2n_{\mathrm{complex}}
\]
\[
n_{\mathrm{H^+}}=2(0.0250)
\]
\[
n_{\mathrm{H^+}}=0.0500\,\mathrm{mol}
\]
\(\textbf{Coordination-number determination:}\)
The fully coordinated EDTA ligand uses two nitrogen and four oxygen donor atoms.
Thus, it supplies six metal–donor attachments.
The coordination number is six.
\(\textbf{Charge conservation:}\)
The metal contributes charge \(+2\), while coordinated EDTA contributes \(-4\).
Therefore:
\[
q_{\mathrm{complex}}=(+2)+(-4)
\]
\[
q_{\mathrm{complex}}=-2
\]
The reactant side has net charge zero.
On the product side, the coordination anion contributes charge \(-2\) and the two released hydrogen ions contribute \(+2\).
The reaction is consequently charge-balanced.
The appearance of hydrogen ions reflects ligand deprotonation, not oxidation of M.
\(\textbf{Final answer:}\)
\[
\boxed{n_{\mathrm{H^+}}=0.0500\,\mathrm{mol},\quad CN=6,\quad q_{\mathrm{complex}}=-2}
\]
418. A \(0.100\,\mathrm{mol}\) sample of \(\mathrm{[Co(NH_3)_5Cl]Cl_2}\) is dissolved in water. Initially, its intact coordination sphere is preserved. Excess ammonia is then added, and a quantitative ligand substitution occurs:
\[
\mathrm{[Co(NH_3)_5Cl]^{2+}+NH_3\rightarrow[Co(NH_3)_6]^{3+}+Cl^-}
\]
Assume complete ideal dissociation of the salts and that excess silver nitrate precipitates only freely available chloride. Which statement correctly compares the ionic-species amount and precipitable chloride before and after the substitution?
ⓐ. The amount of ionic species remains \(0.300\,\mathrm{mol}\), while precipitable chloride increases from \(0.200\) to \(0.300\,\mathrm{mol}\).
ⓑ. The amount of ionic species increases from \(0.300\) to \(0.400\,\mathrm{mol}\), and precipitable chloride increases from \(0.200\) to \(0.300\,\mathrm{mol}\).
ⓒ. The amount of ionic species increases from \(0.300\) to \(0.400\,\mathrm{mol}\), but precipitable chloride remains \(0.200\,\mathrm{mol}\).
ⓓ. The amount of ionic species decreases from \(0.400\) to \(0.300\,\mathrm{mol}\), while precipitable chloride decreases from \(0.300\) to \(0.200\,\mathrm{mol}\).
Correct Answer: The amount of ionic species increases from \(0.300\) to \(0.400\,\mathrm{mol}\), and precipitable chloride increases from \(0.200\) to \(0.300\,\mathrm{mol}\).
Explanation: \(\textbf{Initial dissociation:}\)
The starting salt contains one coordination cation and two external chloride ions.
Its ideal dissociation is:
\[
\mathrm{[Co(NH_3)_5Cl]Cl_2\rightarrow[Co(NH_3)_5Cl]^{2+}+2Cl^-}
\]
Each formula unit produces three ionic species.
For \(0.100\,\mathrm{mol}\) of salt:
\[
n_{\mathrm{ions},i}=3(0.100)
\]
\[
n_{\mathrm{ions},i}=0.300\,\mathrm{mol}
\]
Only the two external chloride ions are initially precipitable.
Therefore:
\[
n_{\mathrm{Cl^-,free},i}=2(0.100)
\]
\[
n_{\mathrm{Cl^-,free},i}=0.200\,\mathrm{mol}
\]
\(\textbf{Quantitative ligand substitution:}\)
One coordinated chloride ligand is replaced by ammonia.
The cobalt oxidation state remains \(+3\).
Replacing an anionic ligand with a neutral ligand increases the coordination-cation charge from \(+2\) to \(+3\).
The departing chloride becomes freely available in solution.
\(\textbf{Final ionic composition:}\)
After complete conversion, the cobalt-containing coordination cation is \(\mathrm{[Co(NH_3)_6]^{3+}}\).
It is accompanied by three free chloride ions per cobalt centre.
Thus, the final dissolved ionic composition corresponds to the ideal dissociation of \(\mathrm{[Co(NH_3)_6]Cl_3}\).
The total ionic-species amount becomes:
\[
n_{\mathrm{ions},f}=4(0.100)
\]
\[
n_{\mathrm{ions},f}=0.400\,\mathrm{mol}
\]
The free chloride amount becomes:
\[
n_{\mathrm{Cl^-,free},f}=3(0.100)
\]
\[
n_{\mathrm{Cl^-,free},f}=0.300\,\mathrm{mol}
\]
\(\textbf{Coordination and redox checks:}\)
Both cobalt entities contain six directly attached donor atoms.
The coordination number therefore remains six.
Cobalt remains in oxidation state \(+3\), so the substitution is not a redox reaction.
The change in free chloride and ionic-species amounts arises solely from movement of a chloride ligand out of the coordination sphere.
\(\textbf{Final answer:}\)
\[
\boxed{n_{\mathrm{ions}}:0.300\rightarrow0.400\,\mathrm{mol},\quad n_{\mathrm{Cl^-,free}}:0.200\rightarrow0.300\,\mathrm{mol}}
\]
419. A hypothetical octahedral chromium(III) coordination entity undergoes complete deprotonation of all its coordinated aqua ligands:
\[
\mathrm{[Cr(H_2O)_6]^{3+}\rightarrow[Cr(OH)_6]^{3-}+6H^+}
\]
Assume that all six chromium–oxygen donor attachments remain intact, chromium retains oxidation state \(+3\), and the ordinary octahedral \(d^3\) occupancy applies before and after the change. Which statement correctly describes the consequences?
ⓐ. The charge shifts from \(+3\) to \(-3\) while coordination number six and three unpaired electrons remain; absorption can change.
ⓑ. The coordination-entity charge remains \(+3\), because protonation or deprotonation does not affect a ligand's formal charge.
ⓒ. Chromium is reduced to oxidation state \(-3\), and the entity must become diamagnetic because hydroxido is an anionic ligand.
ⓓ. The coordination number decreases from six to zero because deprotonated water can no longer donate an oxygen electron pair.
Correct Answer: The charge shifts from \(+3\) to \(-3\) while coordination number six and three unpaired electrons remain; absorption can change.
Explanation: \(\textbf{Initial coordination environment:}\)
The starting entity contains chromium(III) and six neutral water ligands.
Its coordination-entity charge is \(+3\).
Each aqua ligand supplies one oxygen donor atom.
Therefore, the initial coordination number is six.
\(\textbf{Effect of complete deprotonation:}\)
Each coordinated water molecule loses one hydrogen ion and becomes a coordinated hydroxido ligand.
The six oxygen donor attachments are explicitly retained.
The coordination number consequently remains six.
However, every new hydroxido ligand carries charge \(-1\).
\(\textbf{Final charge calculation:}\)
Chromium retains oxidation state \(+3\).
The six hydroxido ligands contribute combined charge \(-6\).
Thus:
\[
q_f=(+3)+6(-1)
\]
\[
q_f=-3
\]
The coordination-entity charge therefore changes by six negative charge units.
The six released hydrogen ions account for the charge difference.
\(\textbf{Electronic and magnetic behaviour:}\)
Chromium(III) has configuration \(3d^3\).
The specified octahedral ground-state occupancy is \(t_{2g}^{3}e_g^{0}\).
The three d electrons remain singly occupied.
Consequently, both entities contain three unpaired electrons and remain paramagnetic.
\(\textbf{Optical qualification:}\)
Aqua and hydroxido are chemically different ligands.
Changing the coordinated ligand identity can alter the crystal-field splitting.
The positions of suitable electronic absorption bands may therefore change.
However, the reaction does not establish a particular new observed colour without additional spectral information.
The complete change demonstrates that ligand protonation state, coordination-entity charge and optical properties may vary while the central-metal oxidation state, coordination number and ordinary spin count remain fixed.
\(\textbf{Final answer:}\)
\[
\boxed{q:+3\rightarrow-3,\quad CN=6,\quad n_{\mathrm{unpaired}}=3}
\]
420. Three observations about coordination compounds are presented:
I. \(\mathrm{[Co(NH_3)_4Cl_2]Cl}\) gives one mole of AgCl per mole of compound under conditions preserving the coordination sphere, and its coordinated chlorido ligands can occupy cis or trans positions.
II. \(\mathrm{[Co(NH_3)_6]^{3+}}\) is diamagnetic, whereas \(\mathrm{[CoF_6]^{3-}}\) is paramagnetic.
III. Two coordination compounds containing the same metal in the same oxidation state absorb visible light at different wavelengths because their ligand environments differ.
Which statement most appropriately identifies the explanatory roles of Werner's theory, the valence-bond approach and crystal-field theory?
ⓐ. Werner's theory explains all three observations quantitatively because primary and secondary valences determine electron pairing and absorption wavelengths.
ⓑ. The valence-bond approach explains chloride ionisation but cannot describe any magnetic difference between complexes.
ⓒ. Crystal-field theory is required to distinguish coordinated chloride from external chloride, while Werner's theory explains electronic absorption.
ⓓ. Werner's theory addresses the ionisation and geometrical evidence in I; electronic-bonding models address II, while crystal-field splitting provides an explanation for the optical difference in III.
Correct Answer: Werner's theory addresses the ionisation and geometrical evidence in I; electronic-bonding models address II, while crystal-field splitting provides an explanation for the optical difference in III.
Explanation: Observation I concerns the distribution of chloride ions between coordinated and external positions.
Werner's theory distinguishes primary and secondary valences and provides a structural explanation for this distribution.
In the stated cobalt compound, one chloride ion is external and is available for immediate precipitation.
The other two chloride ions are directly coordinated and occupy positions associated with the secondary valence.
The directional nature of these coordination positions also permits cis and trans arrangements.
Thus, Werner's model addresses both the ionisation evidence and the geometrical distinction.
Observation II requires information about metal d-electron configurations and electron pairing.
Cobalt(III) has a \(d^6\) configuration in both entities.
The hexaammine entity is low spin, whereas the hexafluorido entity is high spin in the standard NCERT examples.
The valence-bond approach qualitatively represents their differing orbital arrangements and magnetic properties.
Crystal-field theory explains this difference using ligand-field splitting and pairing considerations.
Observation III concerns the energies of electronic excitations.
These energies depend on the relevant differences between electronic energy levels.
Crystal-field theory provides an explanation for changes in suitable d-orbital absorption energies when ligand environments change.
Werner's valence postulates do not independently calculate such absorption wavelengths.
The elementary valence-bond approach also does not satisfactorily explain the colours of coordination compounds.
The three observations therefore illustrate the complementary explanatory roles and limitations of successive coordination-bonding models.
