Co-ordination Compounds NEET 2002: One Unpaired Electron

Co-ordination compounds NEET 2002: which complex has exactly one unpaired electron?

Option C, ferricyanide, is correct: its low-spin iron(III) centre has exactly one unpaired electron. The co-ordination compounds NEET 2002 question asks you to use chromium’s atomic number, 24, and iron’s, 26, to identify the complex meeting that exact condition.

The options are:

A) & [C r (C O)_{6}] B) & [F e (C O)_{5}] C) & [F e (C N)_{6}]^{3 -} D) & [C r (N H_{3})_{6}]^{3 +}

Here, “spin of one electron” means one unpaired electron, not simply a paramagnetic complex. Option C has the low-spin configuration: $d^{5}$

Option D is also paramagnetic, but it has three unpaired electrons. No spatial figure is needed: this problem requires electron counting and orbital occupancy, not interpretation of a physical arrangement.

How do you eliminate the two carbonyl complexes?

Both carbonyl complexes are diamagnetic, so neither qualifies. Follow the official solution’s sequence: determine oxidation state, establish the metal electron count, account for the ligand field, then count unpaired electrons. Do not stop at the isolated metal atom’s configuration.

For A, carbon monoxide is neutral. Chromium’s oxidation state is therefore zero:

x + 6 (0) = 0 \Rightarrow x = 0

Neutral chromium has six valence electrons:

C r = [A r] 3 d^{5} 4 s^{1}

In the official explanation, the strong-field carbonyl environment produces pairing. The resulting complex has zero unpaired electrons and is diamagnetic.

For B, carbon monoxide is again neutral. Iron’s oxidation state is also zero:

x + 5 (0) = 0 \Rightarrow x = 0

Neutral iron starts with:

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

The strong-field iron carbonyl complex has all electrons paired and is diamagnetic. These isolated-atom configurations are starting information, not the finished electron arrangements in the complexes.

Do not impose octahedral splitting on option B: it is five-coordinate. The supplied solution eliminates it through carbonyl pairing and diamagnetism.

Why does ferricyanide have exactly one unpaired electron?

Its five d electrons form two pairs and leave one electron unpaired in the lower octahedral orbital set. Strong-field pairing reduces the unpaired count, but it cannot pair an odd number of electrons completely.

Each cyanide ligand contributes a charge of minus one. The oxidation-state calculation gives iron(III):

x + 6 (- 1) = - 3 \Rightarrow x = + 3

Start from neutral iron, then remove the two 4s electrons before removing one 3d electron:

F e = [A r] 3 d^{6} 4 s^{2} ⟶ {F e}^{3 +} = [A r] 3 d^{5}

Cyanide is a strong-field ligand and gives low-spin iron(III) in this octahedral complex. Its configuration is:

t_{2 g}^{5} e_{g}^{0}

The orbital occupancy is:

t_{2 g} = (↑ ↓) (↑ ↓) (↑), e_{g} = () ()

This gives two electron pairs and one unpaired electron, so option C meets the exact requirement. “Strong field” does not mean “always diamagnetic”.

Why is the chromium ammine complex not the answer?

Option D has three unpaired electrons, not one. Its chromium(III) centre has three d electrons, which occupy the three lower octahedral orbitals singly. There is no competing high-spin versus low-spin choice for an ordinary octahedral complex with this electron count.

Ammonia is neutral. Chromium is therefore in the oxidation state plus three:

x + 6 (0) = + 3 \Rightarrow x = + 3

Remove chromium’s 4s electron first, followed by two 3d electrons:

C r = [A r] 3 d^{5} 4 s^{1} ⟶ {C r}^{3 +} = [A r] 3 d^{3}

The octahedral configuration and occupancy are:

t_{2 g}^{3} e_{g}^{0}

t_{2 g} = (↑) (↑) (↑), e_{g} = () ()

The three lower orbitals have equal energy, so the electrons occupy them singly before pairing. Option D is paramagnetic, but that alone is insufficient.

Why does stopping at “paramagnetic” give the wrong answer?

Paramagnetism establishes at least one unpaired electron; the question requires exactly one. Stopping at the broader property leaves both C and D in consideration. The missing test is the unpaired-electron count.

The faulty reasoning is:

Chromium(III) has three d electrons.
These electrons are unpaired in the octahedral complex.
Therefore, choose D.

The first two statements are correct. The selection is not, because it never tests the required count.

C and D are both paramagnetic, with 1 and 3 unpaired electrons respectively. Write an unpaired-electron count beside every option before selecting the answer. Select C because its count is exactly one, not merely because it is paramagnetic.

How do you apply the same method to related questions?

Keep the sequence fixed: oxidation state, metal electron count, ligand field, then unpaired-electron count. These are original related practice questions, not additional verified PYQs. They test a change in oxidation state, a change in ligand strength and an unchanged three-electron octahedral arrangement.

Question 1: Is the iron cyanide complex with a charge of minus four paramagnetic?

No. It is diamagnetic, because its low-spin iron(II) centre has six d electrons, all paired. Compared with option C, the changed overall charge changes the oxidation state and therefore the d-electron count.

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

x - 6 = - 4 \Rightarrow x = + 2

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

Strong-field cyanide gives the low-spin configuration:

t_{2 g}^{6} e_{g}^{0}

t_{2 g} = (↑ ↓) (↑ ↓) (↑ ↓)

There are zero unpaired electrons. Changing oxidation state has changed the d count and the magnetic result.

Question 2: How many unpaired electrons are in the iron fluoride complex with a charge of minus three?

Five electrons are unpaired. Iron has the same oxidation state as in option C, but weak-field fluoride gives a high-spin arrangement instead of cyanide’s low-spin arrangement. $[F e F_{6}]^{3 -}$

x - 6 = - 3 \Rightarrow x = + 3

{F e}^{3 +} = [A r] 3 d^{5}

The high-spin configuration and occupancy are:

t_{2 g}^{3} e_{g}^{2}

t_{2 g} = (↑) (↑) (↑), e_{g} = (↑) (↑)

Each of the five d orbitals contains one electron. Changing ligand strength can therefore change the spin state without changing the metal’s d-electron count.

Question 3: How many unpaired electrons are in the chromium aqua complex with a charge of plus three?

Three electrons are unpaired, just as in option D. Replacing ammonia with water does not change chromium’s oxidation state or its three-electron octahedral occupancy.

[C r (H_{2} O)_{6}]^{3 +}

Water is neutral, giving:

x + 6 (0) = + 3 \Rightarrow x = + 3

{C r}^{3 +} = [A r] 3 d^{3} \Rightarrow t_{2 g}^{3} e_{g}^{0}

t_{2 g} = (↑) (↑) (↑)

Ordinary octahedral chromium(III) retains three unpaired electrons here. Cover the solutions and redo all three questions, writing oxidation state, d count, orbital occupancy and unpaired count before checking your answer.

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For a worked example of the same idea, see Photoelectric Effect NEET 2026: Three-Cell Solution.

Frequently asked questions

Which complex has exactly one unpaired electron in the NEET 2002 question?

Option C, ferricyanide, [Fe(CN)6]3−, has exactly one unpaired electron. Its iron(III) centre is d5, and strong-field cyanide produces a low-spin octahedral arrangement with two pairs and one unpaired electron.

Why is [Cr(NH3)6]3+ not the correct answer?

Chromium(III) has a d3 configuration, with three electrons occupying the three lower octahedral orbitals singly. The complex therefore has three unpaired electrons, whereas the question requires exactly one.

Why are Cr(CO)6 and Fe(CO)5 diamagnetic?

Both carbonyl complexes have all their electrons paired in the strong-field carbonyl environment. Their zero unpaired-electron count eliminates both options; isolated metal-atom configurations are not the final electron arrangements in these complexes.

Does a strong-field ligand always make a complex diamagnetic?

No: strong-field pairing cannot completely pair an odd number of d electrons. Ferricyanide is low-spin d5 and retains one unpaired electron, while the low-spin d6 complex [Fe(CN)6]4− has all electrons paired and is diamagnetic.