Co-ordination Compounds NEET 2008: Unpaired Electrons

Co-ordination compounds NEET 2008: which complex is most paramagnetic?

Option C is correct in the co-ordination compounds NEET 2008 question, with four unpaired electrons under the supplied solution’s spin assignments. The task is to identify the complex with the greatest number of unpaired electrons and hence the strongest paramagnetic behaviour among these choices:

Here, gly means coordinated glycinato, en is ethylenediamine, bpy is bipyridyl and OX is oxalate. The supplied atomic numbers are:

T i = 22, V = 23, F e = 26, C o = 27

The question bank tags this question as hard, with an expected solving time of 90 seconds. These are question-bank labels, not measured student performance. Start with ligand charges, not a guess about which metal is most magnetic.

How do you find each metal’s oxidation state and d-electron count?

Use this rule: metal oxidation state plus the sum of ligand charges equals the charge outside the brackets. It fixes the metal ion and its d-electron count before any pairing decision. The ligand-charge key is:

g l y : - 1, O H : - 1, O X : - 2

e n : 0, b p y : 0, N H_{3} : 0

When forming these cations, remove 4s electrons before 3d electrons. In each equation below, the unknown is the metal’s oxidation state.

A: glycinato and hydroxide each contribute negative charge. Ammonia contributes no charge.

x + 2 (- 1) + 2 (- 1) + 2 (0) = + 1

x - 4 = + 1; x = + 5

Neutral vanadium has the configuration:

V : [A r] 3 d^{3} 4 s^{2}

It loses two 4s and three 3d electrons:

V^{5 +} : [A r] 3 d^{0}

B: every ligand here is neutral. The metal therefore carries the entire complex charge.

x + 0 + 0 + 2 (0) = + 2; x = + 2

Neutral iron has the configuration:

F e : [A r] 3 d^{6} 4 s^{2}

It loses two 4s electrons, leaving all six 3d electrons:

F e^{2 +} : [A r] 3 d^{6}

C: each oxalate contributes two negative charges. Include both hydroxide charges too.

x + 2 (- 2) + 2 (- 1) = - 1

x - 6 = - 1; x = + 5

Neutral cobalt has the configuration:

C o : [A r] 3 d^{7} 4 s^{2}

It loses two 4s and three 3d electrons:

C o^{5 +} : [A r] 3 d^{4}

D: ammonia contributes no charge. Again, the metal carries the entire complex charge.

x + 6 (0) = + 3; x = + 3

Neutral titanium has the configuration:

T i : [A r] 3 d^{2} 4 s^{2}

It loses two 4s and one 3d electron:

T i^{3 +} : [A r] 3 d^{1}

Do not replace the calculated cobalt oxidation state with a more familiar one: the written complex charge requires removal of five electrons. Treating coordinated glycinato as neutral would also corrupt the vanadium calculation before electron pairing enters the solution.

How do the unpaired-electron counts establish option C?

The supplied solution assigns four unpaired electrons to C, one to D and none to A or B. Charge balance fixes the d counts, but pairing requires a separate ligand-field step. The assignments below follow the supplied solution rather than treating the d count as enough to decide spin.

A has no d electrons, so its unpaired-electron count is 0.
B is treated as low spin because the supplied solution uses the strong-field character of en and bpy. Its six d electrons occupy the lower set as three pairs:

d^{6} ⟶ t_{2 g}^{6} e_{g}^{0}, n = 0

C is treated as high spin with the O-donor ligands oxalate and hydroxide. Its four d electrons occupy separate orbitals:

d^{4} ⟶ t_{2 g}^{3} e_{g}^{1}, n = 4

D has one d electron, which must be unpaired:

d^{1}, n = 1

The compact audit below lists option / oxidation state / d count / unpaired-electron count:

+ 5 / d^{0} / 0

+ 2 / d^{6} / 0

+ 5 / d^{4} / 4

+ 3 / d^{1} / 1

Comparing unpaired-electron counts gives: $4 > 1 > 0$ $C > D > A = B$

Thus option C, [Co(OX)₂(OH)₂]⁻, matches the supplied official answer. The d count alone does not establish pairing: the same count can give different orbital occupations under different ligand fields. Do not turn this question’s high-spin treatment into a rule that every O-donor complex is always high spin.

Why does the even-electron shortcut incorrectly select D?

The shortcut “an even number of d electrons means diamagnetic” is invalid. It wrongly assigns zero unpaired electrons to both B and C, then retains one for D and selects it. The error lies in placing the electrons, not in the final comparison.

Applied mechanically, the wrong reasoning is:

A has no d electrons, so assign zero unpaired electrons.
B has six d electrons, so assume three pairs.
C has four d electrons, so assume two pairs.
D has one d electron, so select D.

Electron parity does not determine orbital occupancy. Ligand-field splitting and the energy cost of pairing must be considered. High-spin C is the counterexample: its even total of four d electrons still gives four unpaired electrons.

Use this checking sequence rather than the even-electron shortcut:

Ligand charges → oxidation state → d count → spin assignment → unpaired electrons.

Can you solve two related magnetic-properties questions?

Use charge balance and ligand strength for the first comparison, then the established unpaired-electron counts for the magnetic moments. These are original practice questions, not additional verified PYQs. Attempt each before reading its answer.

Original practice question 1: Which has more unpaired electrons?

[F e F_{6}]^{4 -} or [F e (C N)_{6}]^{4 -}

Fluoride and cyanide each carry one negative charge. Fluoride is weak field; cyanide is strong field.

For both complexes:

x + 6 (- 1) = - 4; x = + 2

F e^{2 +} : [A r] 3 d^{6}

The fluoride complex is high spin:

t_{2 g}^{4} e_{g}^{2}, n = 4

The cyanide complex is low spin:

t_{2 g}^{6} e_{g}^{0}, n = 0

The fluoride complex has more unpaired electrons. Identical metal oxidation states and d counts do not guarantee identical magnetism.

Original practice question 2: Using the worked PYQ’s unpaired-electron counts, calculate the spin-only magnetic moments of C and D. In the formula below, the electron-count variable denotes unpaired electrons:

μ = \sqrt{n (n + 2)} B M

For C:

μ_{C} = \sqrt{4 (4 + 2)} = \sqrt{24} \approx 4.90 B M

For D:

μ_{D} = \sqrt{1 (1 + 2)} = \sqrt{3} \approx 1.73 B M

These are calculated spin-only values, not experimental measurements or magnetic moments supplied by the question bank. Calculating them was unnecessary for the original MCQ because comparing unpaired-electron counts already settled the answer.

Apply the five-step sequence to your next set: Coordination Compounds Practice Questions NEET: 8 Free MCQs.

Frequently asked questions

Which complex is most paramagnetic in the NEET 2008 question?

Option C, [Co(OX)₂(OH)₂]⁻, is most paramagnetic under the supplied solution’s spin assignments, with four unpaired electrons. D has one unpaired electron, while A and B have none. The resulting order is C > D > A = B.

How do you find the oxidation state of cobalt in [Co(OX)₂(OH)₂]⁻?

Each oxalate ligand carries a −2 charge, and each hydroxide carries a −1 charge. Charge balance gives x + 2(−2) + 2(−1) = −1, so cobalt is in the +5 oxidation state. Removing two 4s electrons and three 3d electrons from neutral cobalt gives a d⁴ configuration.

Does an even number of d electrons mean a complex is diamagnetic?

No. Diamagnetism requires all electrons to be paired; an even d-electron count alone does not establish pairing. Under the supplied high-spin assignment, option C has four d electrons and all four are unpaired.

How do you calculate the spin-only magnetic moments of options C and D?

Use μ = √[n(n + 2)] BM, where n is the number of unpaired electrons. With the supplied counts, C has n = 4 and μ ≈ 4.90 BM, while D has n = 1 and μ ≈ 1.73 BM. These are calculated spin-only values, not experimental measurements.

Co-ordination Compounds NEET 2008: Paramagnetism