Question

Number of paramagnetic ions among the following d- and f-block metal ions is _______.} \[ \text{Mn}^{2+}, \, \text{Cu}^{2+}, \, \text{Zn}^{2+}, \, \text{Yb}^{2+}, \, \text{Sc}^{3+}, \, \text{La}^{3+}, \, \text{Gd}^{3+}, \, \text{Lu}^{3+}, \, \text{Ti}^{4+}, \, \text{Ce}^{4+} \] \text{(Atomic number of Mn = 25, Cu = 29, Zn = 30, Yb = 70, Sc = 21, La = 57, Gd = 64, Lu = 71, Ti = 22, Ce = 58)}

Answer 1

Correct Answer: 5

Step 1: Understanding paramagnetism.
Paramagnetic ions have unpaired electrons in their d- or f-orbitals. If all the electrons are paired, the ion is diamagnetic. We need to determine how many of the given ions have unpaired electrons.
Step 2: Electron configuration of each ion.
1. Mn\(^{2+}\):
Atomic number of Mn = 25. The electron configuration of Mn is \( [Ar] 3d^5 4s^2 \). For Mn\(^{2+}\), the electron configuration becomes \( [Ar] 3d^5 \). Since there are 5 unpaired electrons, Mn\(^{2+}\) is paramagnetic.
2. Cu\(^{2+}\):
Atomic number of Cu = 29. The electron configuration of Cu is \( [Ar] 3d^{10} 4s^1 \). For Cu\(^{2+}\), the electron configuration becomes \( [Ar] 3d^9 \). Since there is 1 unpaired electron, Cu\(^{2+}\) is paramagnetic.
3. Zn\(^{2+}\):
Atomic number of Zn = 30. The electron configuration of Zn is \( [Ar] 3d^{10} 4s^2 \). For Zn\(^{2+}\), the electron configuration becomes \( [Ar] 3d^{10} \). Since all electrons are paired, Zn\(^{2+}\) is diamagnetic.
4. Yb\(^{2+}\):
Atomic number of Yb = 70. The electron configuration of Yb is \( [Xe] 4f^{14} 6s^2 \). For Yb\(^{2+}\), the electron configuration becomes \( [Xe] 4f^{13} \). Since there is 1 unpaired electron, Yb\(^{2+}\) is paramagnetic.
5. Sc\(^{3+}\):
Atomic number of Sc = 21. The electron configuration of Sc is \( [Ar] 3d^1 4s^2 \). For Sc\(^{3+}\), the electron configuration becomes \( [Ar] 3d^0 \). Since there are no unpaired electrons, Sc\(^{3+}\) is diamagnetic.
6. La\(^{3+}\):
Atomic number of La = 57. The electron configuration of La is \( [Xe] 4f^1 5d^0 6s^2 \). For La\(^{3+}\), the electron configuration becomes \( [Xe] 4f^0 5d^0 6s^0 \). Since there are no unpaired electrons, La\(^{3+}\) is diamagnetic.
7. Gd\(^{3+}\):
Atomic number of Gd = 64. The electron configuration of Gd is \( [Xe] 4f^7 5d^1 6s^2 \). For Gd\(^{3+}\), the electron configuration becomes \( [Xe] 4f^7 \). Since there are 7 unpaired electrons, Gd\(^{3+}\) is paramagnetic.
8. Lu\(^{3+}\):
Atomic number of Lu = 71. The electron configuration of Lu is \( [Xe] 4f^{14} 5d^1 6s^2 \). For Lu\(^{3+}\), the electron configuration becomes \( [Xe] 4f^{13} \). Since there is 1 unpaired electron, Lu\(^{3+}\) is paramagnetic.
9. Ti\(^{4+}\):
Atomic number of Ti = 22. The electron configuration of Ti is \( [Ar] 3d^2 4s^2 \). For Ti\(^{4+}\), the electron configuration becomes \( [Ar] 3d^0 \). Since there are no unpaired electrons, Ti\(^{4+}\) is diamagnetic.
10. Ce\(^{4+}\):
Atomic number of Ce = 58. The electron configuration of Ce is \( [Xe] 4f^1 5d^0 6s^2 \). For Ce\(^{4+}\), the electron configuration becomes \( [Xe] 4f^0 5d^0 6s^0 \). Since there are no unpaired electrons, Ce\(^{4+}\) is diamagnetic.
Step 3: Count the number of paramagnetic ions.
The paramagnetic ions are:
- Mn\(^{2+}\)
- Cu\(^{2+}\)
- Yb\(^{2+}\)
- Gd\(^{3+}\)
- Lu\(^{3+}\)
Thus, the number of paramagnetic ions is 5.
Answer: \boxed{5}