Which of the following complexes formed by Cu²⁺ ions is most stable? (i) Cu²⁺ + 4NH₃ leftharpoons [Cu(NH₃)₄]²⁺

NCERT Exemplar · Q1

Cu^{2+}

(i)

Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}

\log K = 11.6

(ii)

Cu^{2+} + 4CN^{-} \rightleftharpoons [Cu(CN)_4]^{2-}

\log K = 27.3

(iii)

Cu^{2+} + 2en \rightleftharpoons [Cu(en)_2]^{2+}

\log K = 15.4

(iv)

Cu^{2+} + 4H_2O \rightleftharpoons [Cu(H_2O)_4]^{2+}

\log K = 8.9

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Concept understanding — Magnetic Moment Calculation

From a Paperclip to a Magnet: The Intuition

You already know that a magnet can pick up iron nails. But what is actually happening inside that nail when it gets near the magnet? And why does a plastic comb, rubbed on hair, pick up tiny bits of paper — but never iron filings?

The answer lies in magnetization — the process by which a material becomes magnetic.

Think of a piece of iron as a chaotic crowd of tiny compass needles. Each needle is an atomic magnetic moment (a tiny magnet, arising from the spin of electrons). In unmagnetized iron, these needles point in random directions. Their magnetic effects cancel out, so the iron as a whole shows no net magnetism.

Now bring a strong magnet close. Its magnetic field acts like a command: "Line up!" The tiny compass needles inside the iron start rotating, aligning themselves with the external field. The more they align, the stronger the iron's own magnetic field becomes. This alignment is magnetization.

Note

Magnetization is not the same as inducing a current. It is a purely magnetic reorientation of atomic dipoles inside a material.

The Precise Definition

Magnetization ( $\vec{M}$ ) is the net magnetic dipole moment per unit volume of a material. It tells you how strongly a material is magnetized — how many tiny atomic magnets are aligned, and in which direction.

If a material has $N$ atoms per unit volume, each with an average magnetic moment $\vec{\mu}_{\text{avg}}$ , then:

\vec{M} = N \vec{\mu}_{\text{avg}}

The SI unit of $\vec{M}$ is amperes per metre (A/m). Why? Because a magnetic dipole moment has units of A·m², and dividing by volume (m³) gives A/m.

\vec{M} = \frac{\text{total magnetic dipole moment}}{\text{volume}}

How Magnetization Connects to the Magnetic Field

When a material gets magnetized, it produces its own magnetic field. The total magnetic field $\vec{B}$ inside the material is the sum of:

The external applied field $\vec{H}$ (caused by free currents, like the current in a solenoid)
The material's response — the magnetization $\vec{M}$

The fundamental relation is:

\vec{B} = \mu_0 (\vec{H} + \vec{M})

where $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space.

Watch out

Do not confuse $\vec{H}$ (magnetic field intensity, or "magnetizing field") with $\vec{B}$ (magnetic flux density). $\vec{H}$ is what you apply; $\vec{M}$ is what the material does; $\vec{B}$ is the total field you measure.

The Three Kinds of Magnetic Materials

Not all materials respond the same way to an external field. The magnetization $\vec{M}$ is proportional to $\vec{H}$ for most materials (at least for small fields):

\vec{M} = \chi_m \vec{H}

where $\chi_m$ is the magnetic susceptibility — a dimensionless number that tells you how easily a material magnetizes.

Material Type	$\chi_m$	Behaviour	Example
Diamagnetic	Small and negative ( $\approx -10^{-5}$ )	Weakly repelled by a magnet; $\vec{M}$ opposes $\vec{H}$	Water, copper, bismuth
Paramagnetic	Small and positive ( $\approx 10^{-5}$ to $10^{-3}$ )	Weakly attracted; $\vec{M}$ aligns with $\vec{H}$	Aluminium, oxygen gas
Ferromagnetic	Large and positive ( $\gg 1$ )	Strongly attracted; $\vec{M}$ can be huge and persists even after $\vec{H}$ is removed	Iron, nickel, cobalt

Important

Ferromagnetic materials have spontaneous magnetization — their atomic moments align even without an external field, forming magnetic domains. Magnetization in these materials is not linear; it saturates and shows hysteresis.

A Simple Example

Take a long iron rod placed inside a solenoid carrying current $I$ . The solenoid produces a uniform field $\vec{H}$ inside. The iron rod becomes magnetized: its atomic moments align, producing $\vec{M}$ in the same direction as $\vec{H}$ .

If $\chi_m$ for iron is about $5000$ , then $\vec{M} = 5000 \, \vec{H}$ . The total field inside the rod becomes:

\vec{B} = \mu_0 (\vec{H} + 5000 \vec{H}) = \mu_0 (5001) \vec{H}

That is why an iron core can amplify the magnetic field of a solenoid by thousands of times.

The Bottom Line

Magnetization is the measure of how much a material becomes magnetic when placed in an external field. It arises from the alignment of atomic magnetic dipoles. For linear materials, $\vec{M} = \chi_m \vec{H}$ . For ferromagnets, the response is nonlinear, strong, and can be permanent — that is how you get a bar magnet from a piece of iron.

Magnetic moment calculation using the spin-only formula is a numerically important topic in the NCERT/CBSE Class 12 Chemistry chapters on d-Block Elements and Coordination Compounds, and ‘spin only formula magnetic moment’ is a frequently searched important-question topic for board exams, JEE Main and NEET. Calculating the number of unpaired electrons correctly is a skill tested repeatedly in competitive-exam chemistry numericals.

Why this formula?

Magnetic Moment Calculation: Why the Formula Holds

Let's build this from first principles — understanding the why before the formula.

1. What is Magnetic Moment?

A magnetic moment ( $\vec{\mu}$ ) is a measure of the strength and orientation of a magnet or current loop. It tells us how strongly an object will interact with an external magnetic field.

The core idea: any moving charge creates a magnetic field. A loop of current is like a tiny bar magnet — its magnetic moment quantifies this.

2. The Fundamental Formula: Current Loop

The Setup

Consider a planar loop of wire carrying a steady current $I$ , enclosing an area $A$ .

Why $\mu = I A$ ?

Step 1: Force on a moving charge

A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences:

\vec{F} = q (\vec{v} \times \vec{B})

Step 2: Torque on a current loop

For a rectangular loop of sides $a$ and $b$ ( $A = ab$ ), placed in a uniform $\vec{B}$ :

Current $I$ means charge flows. On side of length $a$ , the force magnitude is $F = I a B$ (since $I = \frac{q}{t}$ and $v = \frac{a}{t}$ ).
These forces on opposite sides form a couple (equal, opposite, not collinear).
Torque $\tau = \text{force} \times \text{perpendicular distance} = (I a B) \times (b \sin\theta)$

Step 3: Recognize the pattern

\tau = I (ab) B \sin\theta = I A B \sin\theta

This looks exactly like:

\tau = \mu B \sin\theta

Comparing, we identify:

\boxed{\mu = I A}

Why this works: The torque on a current loop is proportional to the current and the area — this product naturally defines the magnetic moment.

3. For a Single Moving Charge (Orbital Magnetic Moment)

The Setup

An electron of charge $-e$ moves in a circular orbit of radius $r$ with speed $v$ .

Why $\mu = \frac{e v r}{2}$ ?

Step 1: Treat orbit as a current loop

Time for one revolution: $T = \frac{2\pi r}{v}$
Current (charge per unit time): $I = \frac{e}{T} = \frac{e v}{2\pi r}$

Step 2: Apply $\mu = I A$

Area of orbit: $A = \pi r^2$
So: $\mu = \left(\frac{e v}{2\pi r}\right) (\pi r^2) = \frac{e v r}{2}$

Step 3: Express in terms of angular momentum

Orbital angular momentum: $L = m v r$
Therefore: $\boxed{\mu = \frac{e}{2m} L}$

Why this matters: The magnetic moment is directly proportional to angular momentum. The factor $\frac{e}{2m}$ is called the gyromagnetic ratio — it links mechanics to magnetism.

4. For a Solenoid (Many Turns)

The Setup

A solenoid of $N$ turns, length $l$ , carrying current $I$ , cross-sectional area $A$ .

Why $\mu = N I A$ ?

Each turn contributes $\mu_{\text{turn}} = I A$ . For $N$ identical turns in series:

\mu_{\text{total}} = N \cdot (I A) = N I A

If the solenoid has $n = N/l$ turns per unit length:

\mu = (n l) I A

Key insight: The magnetic moment adds linearly for multiple turns because each turn's torque contribution adds.

5. Summary Table of Key Results

System	Formula	Why
Single current loop	$\mu = I A$	Torque on loop $\propto I A$
Orbiting electron	$\mu = \frac{e}{2m} L$	Current from orbital motion
Solenoid	$\mu = N I A$	Sum of individual loop moments
General definition	$\vec{\mu} = \frac{1}{2} \int \vec{r} \times \vec{J} \, dV$	For continuous current distributions

6. Exam-Relevant Takeaway

Always remember:

Magnetic moment always involves current × area (or equivalent)
For particles, it's charge-to-mass ratio × angular momentum
Direction: given by right-hand rule (curl fingers along current, thumb points along $\vec{\mu}$ )

The formula isn't arbitrary — it emerges naturally from the torque a current loop experiences in a magnetic field.

The stability of a complex is directly measured by its formation constant $K$ ; the larger the $\log K$ , the more stable the complex. Here, the complex with $\log K = 27.3$ is the most stable.

The question asks which complex is most stable. In coordination chemistry, the stability of a complex is quantified by its formation constant (also called stability constant) $K$ . The reaction given is the formation of the complex from the metal ion and ligands. A larger $K$ means the equilibrium lies further to the right — the complex is more stable and less likely to dissociate.

The values are given as $\log K$ , so we compare these directly. No conversion is needed: the highest $\log K$ corresponds to the highest $K$ , hence the most stable complex.

Let’s go through each option:

Option (i): $Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$ , $\log K = 11.6$

This is a moderately stable complex. Ammonia is a good ligand, but not exceptionally strong for copper(II).
Option (ii): $Cu^{2+} + 4CN^{-} \rightleftharpoons [Cu(CN)_4]^{2-}$ , $\log K = 27.3$

Cyanide ion is a very strong ligand (high field strength, forms strong $\sigma$ and $\pi$ bonds). The $\log K$ is dramatically higher than the others — over 10 orders of magnitude larger in $K$ than the next closest.
Option (iii): $Cu^{2+} + 2en \rightleftharpoons [Cu(en)_2]^{2+}$ , $\log K = 15.4$

Ethylenediamine (en) is a bidentate ligand, which gives a chelate effect — this usually increases stability compared to monodentate ligands like NH $_3$ . Indeed, $\log K = 15.4$ is higher than for NH $_3$ (11.6), but still far below CN $^-$ .
Option (iv): $Cu^{2+} + 4H_2O \rightleftharpoons [Cu(H_2O)_4]^{2+}$ , $\log K = 8.9$

Water is a weak ligand. This is the least stable complex here.

Watch out

A common mistake is to think that chelating ligands (like en) always form the most stable complexes. While the chelate effect does enhance stability, the intrinsic ligand strength matters more. Here, CN $^-$ is such a powerful ligand that it overcomes the chelate advantage.

Tip

You don’t need to calculate $K$ from $\log K$ — just compare the $\log K$ values directly. The largest $\log K$ means the largest $K$ , hence the most stable complex.

✓Final answer

The most stable complex is formed with cyanide ions, option (ii) .

Common Mistakes & How to Avoid Them

Mistake 1: Confusing $\log K$ with $K$

The error: Students compare $\log K$ values directly and think the largest $\log K$ means the least stable complex.

Why it's wrong:

A higher $\log K$ means a larger equilibrium constant $K$ , which indicates greater stability of the complex.

How to avoid:

Remember:

$\log K \uparrow \implies K \uparrow \implies$ more stable complex
For this question: $27.3 > 15.4 > 11.6 > 8.9$ , so option (ii) is most stable.

Mistake 2: Forgetting that $\log K$ is directly proportional to stability

The error: Some students think a lower $\log K$ means the reaction "goes more to completion" — this is backwards.

Why it's wrong:

The equilibrium constant $K$ for complex formation is:

K = \frac{[\text{complex}]}{[\text{Cu}^{2+}][\text{ligand}]^n}

A larger $K$ means the equilibrium lies far to the right — more complex formed, hence more stable.

How to avoid:

Write the expression for $K$ and reason:

Big $K$ → products favoured → stable complex
Small $K$ → reactants favoured → unstable complex

Mistake 3: Ignoring the denticity of ligands

The error: Students compare $\log K$ values without considering that en (ethylenediamine) is bidentate, while NH $_3$ , CN $^-$ , and H $_2$ O are monodentate.

Why it matters:

A bidentate ligand like en forms a chelate ring, which gives extra stability (chelate effect). Even though $\log K$ for en (15.4) is less than for CN $^-$ (27.3), the chelate effect is already included in the given $\log K$ value.

How to avoid:

The $\log K$ values already account for denticity — compare them directly.
Do not try to "adjust" the values manually.

Mistake 4: Overthinking magnetic moment or geometry

The error: Students try to use magnetic moment or crystal field theory to decide stability.

Why it's wrong:

The question gives experimental $\log K$ values — these are the direct measure of stability. Magnetic moment tells you about unpaired electrons, not thermodynamic stability.

How to avoid:

When $\log K$ (or $K$ ) is given, use it directly.
Save magnetic moment reasoning for questions about geometry, spin state, or colour.

Mistake 5: Misreading the question as "least stable"

The error: Students accidentally pick the smallest $\log K$ (option iv) because they read "most stable" as "least stable".

How to avoid:

Circle the word "most" or "least" in the question.
Double-check: largest $\log K$ = most stable.

Final Answer

Most stable complex: Option (ii) $[\text{Cu(CN)}_4]^{2-}$ with $\log K = 27.3$

Quick check:

(ii) $\log K = 27.3$ → largest → most stable ✓
(iv) $\log K = 8.9$ → smallest → least stable

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04194 marksMCQ
$^{2+}$
›Reveal solutionSolution
[!TLDR]

Using $\mu=\sqrt{n(n+2)}$ , Ni²⁺ has 2 unpaired electrons ( $\mu\approx2.83$ BM), so its listed value of 4.73 BM is the incorrect match.

Concept

The spin-only magnetic moment (from the NCERT/CBSE-aligned d-block chapter KEAM follows) is $\mu = \sqrt{n(n+2)}$ Bohr Magnetons, where $n$ is the count of unpaired $d$ electrons. First find each ion's $d^x$ configuration, then $n$ , then $\mu$ .

Solution

Evaluate each ion (remove electrons from 4s first, then 3d):

(A) Ni²⁺ (Z=28) $\to 3d^8$ : 2 unpaired, $\mu=\sqrt{2\cdot4}=\sqrt{8}=2.83$ BM. Listed 4.73 → wrong.

(B) Ti²⁺ (Z=22) $\to 3d^2$ : 2 unpaired, $\mu=\sqrt{8}=2.83\approx2.84$ BM ✓

(C) Mn²⁺ (Z=25) $\to 3d^5$ : 5 unpaired, $\mu=\sqrt{5\cdot7}=\sqrt{35}=5.92$ BM ✓

(D) Fe²⁺ (Z=26) $\to 3d^6$ : 4 unpaired, $\mu=\sqrt{4\cdot6}=\sqrt{24}=4.90$ BM ✓

(E) Co²⁺ (Z=27) $\to 3d^7$ : 3 unpaired, $\mu=\sqrt{3\cdot5}=\sqrt{15}=3.87$ BM ✓

Only Ni²⁺ fails: its correct value is 2.83 BM, not 4.73 BM.

[!ANSWER]

(A) Ni²⁺ (Z=28) — the mismatch; its spin-only moment is ≈ 2.83 BM.
KEAM 2026Set pha-2026-0418F4 marksMCQ
$Ti^{2+}$
›Reveal solutionSolution
$CH_3CH_2CH(C_2H_5)CH_2CH(CH_3)CH_2CH_3$ is 3-ethyl-5-methylheptane.

Reasoning

Select the longest continuous chain: it contains 7 carbons (heptane). Numbering to give the lowest locants:

an ethyl group at C-3,

a methyl group at C-5.

Either direction gives locants {3,5}; alphabetically 'ethyl' is cited first and takes the lower number (3). Name: 3-ethyl-5-methylheptane.

(Options naming a 'hexane' parent are wrong because the longest chain has 7 carbons.)

✓Final answer
$The correct option is (D) .$
KEAM 2026Set pha-2026-0419F4 marksMCQ
$Ti^{2+}$
›Reveal solutionSolution

Diamagnetic means no unpaired electrons. $Sc^{3+}$ is $3d^0$ and $Zn^{2+}$ is $3d^{10}$ — both have all electrons paired.

$Sc^{3+}$ : [Ar] $3d^0$ (0 unpaired) and $Zn^{2+}$ : [Ar] $3d^{10}$ (0 unpaired) are both diamagnetic. The other options ( $Ti^{2+}$ , $Mn^{2+}$ , $Ni^{2+}$ , $V^{2+}$ , $Cr^{2+}$ , $Co^{2+}$ ) all have unpaired d-electrons and are paramagnetic.

✓Final answer
$The correct option is (E) .$
KEAM 2025Set eng-2025-04254 marksMCQ
$Q. The magnetic moment of a divalent ion in aqueous solution is 3.87 BM. The number of unpaired electrons present in it is (A) 4 (B) 5 (C) 3 (D) 2 (E) 1$
›Reveal solutionSolution

The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM. Setting $\mu = 3.87$ gives $n(n+2) = 15$ , so $n = 3$ unpaired electrons.

The spin-only formula relates magnetic moment to the number of unpaired electrons:
$\mu = \sqrt{n(n+2)} \text{ BM}$
Squaring the given moment:
$n(n+2) = (3.87)^2 \approx 15$
Solving, $n = 3$ (since $3 \times 5 = 15$ ), giving $\mu = \sqrt{15} = 3.87$ BM. Hence the divalent ion has 3 unpaired electrons.

✓Final answer
$The correct option is (C) .$
KEAM 2025Set eng-2025-04264 marksMCQ
$Sc^{3+}$
›Reveal solutionSolution
Magnetic moment rises with unpaired electrons; $Mn^{2+}$ ( $d^5$ ) has 5 unpaired e⁻, the maximum here.

Spin-only magnetic moment $\mu=\sqrt{n(n+2)}$ BM, so more unpaired electrons ( $n$ ) means a larger moment.

$Sc^{3+}$ : $d^0$ , $n=0$ .

$Ti^{3+}$ : $d^1$ , $n=1$ .

$Cr^{2+}$ : $d^4$ , $n=4$ .

$Fe^{2+}$ : $d^6$ , $n=4$ .

$Mn^{2+}$ : $d^5$ , $n=5$ , giving $\mu=\sqrt{5\times7}=\sqrt{35}=5.92$ BM.

$Mn^{2+}$ has the most unpaired electrons and hence the highest magnetic moment.

✓Final answer
$The correct option is (E) .$
KEAM 2025Set eng-2025-04274 marksMCQ
$Fe^{2+}$
›Reveal solutionSolution
Magnetic moment rises with the number of unpaired electrons. $Mn^{2+}$ ( $d^5$ ) has 5 unpaired electrons, more than any other ion listed, so it has the highest moment.

Reasoning

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}\,BM$ , where $n$ = number of unpaired electrons.

Electronic configurations (high spin):

(A) $Fe^{2+}$ : $d^6$ → 4 unpaired

(B) $Mn^{2+}$ : $d^5$ → 5 unpaired

(C) $Ni^{2+}$ : $d^8$ → 2 unpaired

(D) $Co^{2+}$ : $d^7$ → 3 unpaired

(E) $Cr^{2+}$ : $d^4$ → 4 unpaired

$Mn^{2+}$ has the most unpaired electrons (5):
$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\,BM$
which is the highest of all the ions.

✓Final answer
$The correct option is (B) .$
KEAM 2025Set eng-2025-04294 marksMCQ
$Zn^{2+}$
›Reveal solutionSolution
$Zn^{2+} = 3d^{10}$ (fully filled, no unpaired electrons) is diamagnetic; the others have unpaired d-electrons and are paramagnetic.

Magnetic behaviour depends on unpaired electrons. The $d$ -electron counts of the $M^{2+}$ ions are:

$Zn^{2+}$ : $[Ar]3d^{10}$ — 0 unpaired electrons → diamagnetic.

$Ni^{2+}$ : $3d^8$ — 2 unpaired.

$Co^{2+}$ : $3d^7$ — 3 unpaired.

$Cu^{2+}$ : $3d^9$ — 1 unpaired.

$Mn^{2+}$ : $3d^5$ — 5 unpaired.

Only $Zn^{2+}$ , with a completely filled $3d^{10}$ subshell, has all electrons paired and is diamagnetic.

✓Final answer
$The correct option is (A) .$
KEAM 2025Set pha-2025-0429F4 marksMCQ
$Q. What is the magnetic moment of divalent ion with three unpaired electrons? (A) 2.84 BM (B) 5.92 BM (C) 3.87 BM (D) 4.90 BM (E) 1.73 BM$
›Reveal solutionSolution

The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM. For 3 unpaired electrons, $\mu = \sqrt{15} \approx 3.87$ BM.

Reasoning

The spin-only formula gives the magnetic moment from the number of unpaired electrons $n$ :
$\mu = \sqrt{n(n+2)}\ \text{BM}$
With $n = 3$ :
$\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87\ \text{BM}$
(The '+2' charge is not needed; only the unpaired-electron count matters.)

✓Final answer
$The correct option is (C) .$
KEAM 2023Set eng-2023-P1-A14 marksMCQ
$[MnBr_4]^{2-}, [FeF_6]^{4-}$
›Reveal solutionSolution
[MnBr4]2- is tetrahedral (sp3, μ=5.9), [FeF6]4- octahedral (sp3d2, μ=4.9), and [Ni(CN)4]2- square planar (dsp2, μ=0).

Concept and Intuition

Geometry and magnetic moment depend on the metal d-count, ligand field strength, and coordination number. Weak-field ligands give high-spin outer-orbital complexes; strong-field CN- pairs electrons giving diamagnetic dsp2 square-planar Ni2+.

Step-by-Step Solution

[MnBr4]2-: Mn2+ is d5; weak-field Br-, CN 4 → tetrahedral, sp3, 5 unpaired → μ = √(5·7) ≈ 5.9 BM.

[FeF6]4-: Fe2+ is d6; weak-field F-, CN 6 → octahedral high-spin, sp3d2, 4 unpaired → μ ≈ 4.9 BM.

[Ni(CN)4]2-: Ni2+ is d8; strong-field CN-, CN 4 → square planar, dsp2, 0 unpaired → μ = 0.

Matches option B: tetrahedral, octahedral, square planar; sp3, sp3d2, dsp2; 5.9, 4.9, 0.

Common Mistakes

Assigning dsp2 to weak-field tetrahedral/octahedral complexes; only strong-field CN- forces the low-spin square-planar Ni case.

✓Final answer
$The correct option is (B) — Tetrahedral, octahedral, square planar; sp3, sp3d2, dsp2; 5.9, 4.9, 0.$

ANSWER: B
KEAM 2021Set eng-2021-P1-A14 marksMCQ
$Z = 24$
›Reveal solutionSolution
The magnetic moment of $Cr^{3+}$ is 3.87 BM.

Concept and Intuition

The spin-only magnetic moment depends on the number of unpaired electrons: $\mu = \sqrt{n(n+2)}$ BM.

Step-by-Step Solution

$Z=24$ is Cr, configuration $[Ar]3d^5 4s^1$ .

$Cr^{3+}$ : remove three electrons → $[Ar]3d^3$ .

$3d^3$ has $n=3$ unpaired electrons.

$\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87$ BM.

Common Mistakes

Counting electrons from $4s$ first but miscounting unpaired d-electrons for $Cr^{3+}$ .

✓Final answer
$The correct option is (A) — 3.87 BM.$

ANSWER: A

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Q.Which of the following complexes formed by Cu2+Cu^{2+}Cu2+ ions is most stable?

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Q.Which of the following complexes formed by $Cu^{2+}$ ions is most stable?

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