Question

۱۹ عنصر انتقالي (۸) العزم المغناطيسي لأيونه (A) أقل من العزم المغناطيسي لذرته. أي الاختيارات التالية صحيحا بالنسبة لهذا العنصر؟

• A. هو أحد فلزات العملة.
• B. له أعلى عزم مغناطيسي في سلسلته.
• C. له حالة تأكسد أعلى من رقم مجموعته.
• D. جميع مركباته بارامغناطيسية.

Answer 1

Answer: (B)

له أعلى عزم مغناطيسي في سلسلته.

Answer 2

The problem states that for a transition element (A), the magnetic moment of its ion (A²⁺) is less than the magnetic moment of its atom (A). We need to identify the correct statement about this element.

The magnetic moment ($\mu$) is directly related to the number of unpaired electrons (n) by the spin-only formula:

$$\mu = \sqrt{n(n+2)} \text{ B.M.}$$

For $\mu_{A^{2+}} < \mu_A$, it implies that the number of unpaired electrons in the A²⁺ ion ($n_{A^{2+}}$) must be less than the number of unpaired electrons in the neutral atom A ($n_A$).

Let's examine the electronic configurations and number of unpaired electrons for the 3d transition elements and their corresponding A²⁺ ions:

1. Scandium (Sc):

   • Atom (Sc): [Ar] 3d¹ 4s²; $n_A = 1$
   • Ion (Sc²⁺): [Ar] 3d¹; $n_{A^{2+}} = 1$
   • Here, $n_{A^{2+}} = n_A$.

2. Titanium (Ti):

   • Atom (Ti): [Ar] 3d² 4s²; $n_A = 2$
   • Ion (Ti²⁺): [Ar] 3d²; $n_{A^{2+}} = 2$
   • Here, $n_{A^{2+}} = n_A$.

3. Vanadium (V):

   • Atom (V): [Ar] 3d³ 4s²; $n_A = 3$
   • Ion (V²⁺): [Ar] 3d³; $n_{A^{2+}} = 3$
   • Here, $n_{A^{2+}} = n_A$.

4. Chromium (Cr): (Exception to Aufbau principle)

   • Atom (Cr): [Ar] 3d⁵ 4s¹; $n_A = 5 (\text{from } 3d) + 1 (\text{from } 4s) = 6$
   • Ion (Cr²⁺): [Ar] 3d⁴ (loses 1 electron from 4s and 1 from 3d); $n_{A^{2+}} = 4$
   • Here, $n_{A^{2+}} = 4 < n_A = 6$. This fits the given condition. So, element A is Chromium.

5. Manganese (Mn):

   • Atom (Mn): [Ar] 3d⁵ 4s²; $n_A = 5$
   • Ion (Mn²⁺): [Ar] 3d⁵; $n_{A^{2+}} = 5$
   • Here, $n_{A^{2+}} = n_A$.

6. Iron (Fe):

   • Atom (Fe): [Ar] 3d⁶ 4s²; $n_A = 4$
   • Ion (Fe²⁺): [Ar] 3d⁶; $n_{A^{2+}} = 4$
   • Here, $n_{A^{2+}} = n_A$.

7. Cobalt (Co):

   • Atom (Co): [Ar] 3d⁷ 4s²; $n_A = 3$
   • Ion (Co²⁺): [Ar] 3d⁷; $n_{A^{2+}} = 3$
   • Here, $n_{A^{2+}} = n_A$.

8. Nickel (Ni):

   • Atom (Ni): [Ar] 3d⁸ 4s²; $n_A = 2$
   • Ion (Ni²⁺): [Ar] 3d⁸; $n_{A^{2+}} = 2$
   • Here, $n_{A^{2+}} = n_A$.

9. Copper (Cu): (Exception to Aufbau principle)

   • Atom (Cu): [Ar] 3d¹⁰ 4s¹; $n_A = 1$
   • Ion (Cu²⁺): [Ar] 3d⁹ (loses 1 electron from 4s and 1 from 3d); $n_{A^{2+}} = 1$
   • Here, $n_{A^{2+}} = n_A$.

10. Zinc (Zn):

    • Atom (Zn): [Ar] 3d¹⁰ 4s²; $n_A = 0$
    • Ion (Zn²⁺): [Ar] 3d¹⁰; $n_{A^{2+}} = 0$
    • Here, $n_{A^{2+}} = n_A$.

From the analysis, only Chromium (Cr) satisfies the condition. Therefore, element A is Chromium.

Now, let's evaluate the given options for Chromium:

• ① هو أحد فلزات العملة. (It is one of the coinage metals.)
  Coinage metals are Group 11 elements (Cu, Ag, Au). Chromium is in Group 6. So, this statement is incorrect.

• ② له أعلى عزم مغناطيسي في سلسلته. (It has the highest magnetic moment in its series.)
  The number of unpaired electrons in the atomic state of 3d series elements are:
  Sc (1), Ti (2), V (3), Cr (6), Mn (5), Fe (4), Co (3), Ni (2), Cu (1), Zn (0).
  Chromium (Cr) has 6 unpaired electrons in its atomic state, which is the highest among the 3d series elements. Therefore, it has the highest magnetic moment. This statement is correct.

• ③ له حالة تأكسد أعلى من رقم مجموعته. (It has an oxidation state higher than its group number.)
  Chromium is in Group 6. Its maximum oxidation state is +6 (e.g., in CrO₃, K₂Cr₂O₇). This is equal to its group number, not higher. So, this statement is incorrect.

• ④ جميع مركباته بارامغناطيسية. (All its compounds are paramagnetic.)
  Paramagnetic compounds have unpaired electrons. Diamagnetic compounds have all electrons paired. Chromium forms compounds in various oxidation states. For example, in Cr⁶⁺ compounds (like CrO₃ or K₂Cr₂O₇), the electronic configuration of Cr⁶⁺ is [Ar] 3d⁰, meaning it has zero unpaired electrons. Such compounds are diamagnetic. Therefore, not all compounds of Chromium are paramagnetic. This statement is incorrect.

Based on the analysis, only option ② is correct.

The final answer is $\boxed{②}$

Solution:

1. Identify the condition: The magnetic moment of A²⁺ is less than that of atom A. This implies the number of unpaired electrons ($n$) in A²⁺ is less than in A ($n_{A^{2+}} < n_A$).
2. Test 3d elements:
   • For most 3d elements (Sc, Ti, V, Mn, Fe, Co, Ni), A²⁺ is formed by losing 2 electrons from the 4s orbital, leaving the 3d configuration unchanged from the atom (or changing it slightly but not reducing unpaired electrons, like Fe where 3d⁶ remains 3d⁶). Thus, $n_{A^{2+}} = n_A$.
   • For Copper (Cu, 3d¹⁰4s¹), Cu²⁺ is 3d⁹. $n_A = 1$, $n_{A^{2+}} = 1$. So, $n_{A^{2+}} = n_A$.
   • For Chromium (Cr, 3d⁵4s¹), $n_A = 5+1 = 6$. When Cr forms Cr²⁺, it loses the 4s electron and one 3d electron, resulting in [Ar] 3d⁴. So, $n_{A^{2+}} = 4$.
   • Since $4 < 6$, Chromium is the element that satisfies the condition.
3. Evaluate options for Chromium (Cr):
   • ① Coinage metal: Incorrect. Cr is Group 6, coinage metals are Group 11.
   • ② Highest magnetic moment in its series: Correct. Cr atom (3d⁵4s¹) has 6 unpaired electrons, which is the maximum for any atom in the 3d series.
   • ③ Oxidation state higher than group number: Incorrect. Cr is Group 6, its max O.S. is +6 (equal to group number).
   • ④ All compounds paramagnetic: Incorrect. Cr⁶⁺ compounds (e.g., CrO₃) are diamagnetic (d⁰).

Answer: The correct option is ②.