Question

The number of unpaired electrons in $\text{Ce}^{4+}$, $\text{Pr}^{4+}$, $\text{Nd}^{4+}$ and $\text{Tb}^{4+}$ are p, q, r and s respectively. The value of (r + s - p - q) is :

• A. 8
• B. 7
• C. 6
• D. 9

Answer 1

Answer: (A)

8

Answer 2

To determine the number of unpaired electrons for each ion, we first find their electronic configurations.

1. $\text{Ce}^{4+}$: Cerium (Ce) has atomic number 58. Its electronic configuration is $[Xe] 4f^1 5d^1 6s^2$. For $\text{Ce}^{4+}$, we remove 4 electrons: 2 from $6s$, 1 from $5d$, and 1 from $4f$. The electronic configuration of $\text{Ce}^{4+}$ is $[Xe]$. Number of unpaired electrons (p) = 0.

2. $\text{Pr}^{4+}$: Praseodymium (Pr) has atomic number 59. Its electronic configuration is $[Xe] 4f^3 6s^2$. For $\text{Pr}^{4+}$, we remove 4 electrons: 2 from $6s$, and 2 from $4f$. The electronic configuration of $\text{Pr}^{4+}$ is $[Xe] 4f^1$. Number of unpaired electrons (q) = 1.

3. $\text{Nd}^{4+}$: Neodymium (Nd) has atomic number 60. Its electronic configuration is $[Xe] 4f^4 6s^2$. For $\text{Nd}^{4+}$, we remove 4 electrons: 2 from $6s$, and 2 from $4f$. The electronic configuration of $\text{Nd}^{4+}$ is $[Xe] 4f^2$. According to Hund's rule, these two electrons will occupy different $f$ orbitals with parallel spins. Number of unpaired electrons (r) = 2.

4. $\text{Tb}^{4+}$: Terbium (Tb) has atomic number 65. Its electronic configuration is $[Xe] 4f^9 6s^2$. For $\text{Tb}^{4+}$, we remove 4 electrons: 2 from $6s$, and 2 from $4f$. The electronic configuration of $\text{Tb}^{4+}$ is $[Xe] 4f^7$. According to Hund's rule, with 7 electrons in the 7 $f$ orbitals, all 7 electrons will be unpaired. Number of unpaired electrons (s) = 7.

Now, we calculate the value of (r + s - p - q): (r + s - p - q) = (2 + 7 - 0 - 1) = 9 - 1 = 8.

The value of (r + s - p - q) is 8.