Spin Only Magnetic Moment Order of Transition Ions

❓ Question

The correct decreasing order of spin only magnetic moment values (BM) of:

$$Cu^+,\ Cu^{2+},\ Cr^{2+},\ Cr^{3+}$$

ions is:

1. $$$$

Cr^{2+}>Cr^{3+}>Cu^{2+}>Cu^+
]

2. $$$$

Cu^{2+}>Cu^+>Cr^{2+}>Cr^{3+}
]

3. $$$$

Cu^+>Cu^{2+}>Cr^{3+}>Cr^{2+}
]

4. $$$$

Cr^{3+}>Cr^{2+}>Cu^+>Cu^{2+}
]

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Magnetic moment
👉 Unpaired electrons
👉 Electronic configuration of ions.

Main idea:

$$\boxed{ \mu=\sqrt{n(n+2)} }$$

Where:

$$n=\text{number of unpaired electrons}$$

More unpaired electrons ⇒ larger magnetic moment.

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🔷 Step 1 — Configuration of $Cu^+$ 💯

Copper:

$$Cu=[Ar]\,3d^{10}4s^1$$

For:

$$Cu^+$$

remove one electron:

$$Cu^+=[Ar]\,3d^{10}$$

All electrons paired.

$$n=0$$

Thus:

$$\mu=0$$

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🔷 Step 2 — Configuration of $Cu^{2+}$

For:

$$Cu^{2+}$$

configuration becomes:

$$[Ar]\,3d^9$$

Unpaired electrons:

$$n=1$$

Magnetic moment:

$$\mu=\sqrt{1(1+2)}$$
$$=\sqrt3 \approx1.73\ BM$$

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🔷 Step 3 — Configuration of $Cr^{2+}$

Chromium:

$$Cr=[Ar]\,3d^54s^1$$

For:

$$Cr^{2+}$$

remove two electrons:

$$[Ar]\,3d^4$$

Unpaired electrons:

$$n=4$$

Magnetic moment:

$$\mu=\sqrt{4(4+2)}$$
$$=\sqrt{24} \approx4.9\ BM$$

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🔷 Step 4 — Configuration of $Cr^{3+}$

For:

$$Cr^{3+}$$

configuration:

$$[Ar]\,3d^3$$

Unpaired electrons:

$$n=3$$

Magnetic moment:

$$\mu=\sqrt{3(3+2)}$$
$$=\sqrt{15} \approx3.87\ BM$$

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🔷 Step 5 — Arrange in Decreasing Order

$$Cr^{2+}(4.9) > Cr^{3+}(3.87) > Cu^{2+}(1.73) > Cu^+(0)$$

Thus:

$$\boxed{ Cr^{2+}>Cr^{3+}>Cu^{2+}>Cu^+ }$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ $4s$ electron removal order bhool jaana

❌ Unpaired electrons galat count kar dena

❌ Magnetic moment formula confuse kar lena

Remember:

$$\boxed{ \mu=\sqrt{n(n+2)} }$$

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✅ Final Answer

$$\boxed{{\displaystyle C{r}^{2+}>C{r}^{3+}>C{u}^{2+}>C{u}^{+}}}$$

(Option 1)

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