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Question

For the Eu3+ ion (At No: 63),

A. the calculated and the observed magnetic moments are in agreement with each other.

B. the higher energy states 7F1 and 7F2 are populated and increase the observed magnetic moment.

C. the 4f orbital is more than half-filled.

D. the ground state term symbol is 7F0.

Of the above, the correct statements are

The correct answer is

B and D only

Europium Ion (Eu3+) Properties

The question asks about several properties of the Europium ion, specifically Eu3+, which has an atomic number of 63. We need to determine which of the given statements about Eu3+ are correct.

Eu3+ Electronic Configuration

First, let's find the electronic configuration of Eu3+.

  • The neutral Europium atom (Eu, Z=63) has the configuration: [Xe] $4f^7 6s^2$.
  • To form the Eu3+ ion, we remove 3 electrons. Electrons are removed from the outermost shell first. So, we remove the two $6s$ electrons and one $4f$ electron.
  • The electronic configuration of Eu3+ is therefore [Xe] $4f^6$.

Analysis of Statements

Let's evaluate each statement based on the $4f^6$ configuration.

Statement A: "the calculated and the observed magnetic moments are in agreement with each other."

  • For lanthanide ions, the magnetic moment arises from both spin and orbital angular momentum, coupled together. The magnetic moment is typically calculated using the total angular momentum quantum number J ($\mu_{eff} = g_J \sqrt{J(J+1)}$).
  • However, for ions like Eu3+ ($4f^6$) and Sm3+ ($4f^5$), the energy difference between the ground state and the first few excited states is relatively small, comparable to thermal energy at room temperature.
  • This means that excited states can be significantly populated at room temperature. The observed magnetic moment is an average over the magnetic moments of the ground state and the populated excited states, weighted by their Boltzmann distribution.
  • The magnetic moment calculated using only the ground state J value for Eu3+ ($^7F_0$, J=0, as shown in Statement D) would be zero. The observed magnetic moment for Eu3+ is non-zero (around 3.4-3.6 BM).
  • Therefore, the observed magnetic moment is generally *not* in agreement with the value calculated solely from the ground state. Statement A is likely incorrect.

Statement B: "the higher energy states 7F1 and 7F2 are populated and increase the observed magnetic moment."

  • As discussed above, the small energy gap between the ground state ($^7F_0$) and the first excited states ($^7F_1$, $^7F_2$, etc.) in Eu3+ allows these excited states to be populated at room temperature.
  • The $^7F_1$ (J=1) and $^7F_2$ (J=2) states have non-zero magnetic moments.
  • The population of these states contributes to the overall observed magnetic moment, making it higher than the theoretical value of zero expected from the $^7F_0$ ground state alone.
  • This population of excited states is the reason for the significant observed magnetic moment of Eu3+. Statement B is correct.

Statement C: "the 4f orbital is more than half-filled."

  • The $4f$ subshell can hold a maximum of 14 electrons.
  • Half-filled means having 7 electrons in the $4f$ subshell ($4f^7$).
  • The electronic configuration of Eu3+ is $4f^6$.
  • Since 6 is less than 7, the $4f$ orbital in Eu3+ is less than half-filled. Statement C is incorrect.

Statement D: "the ground state term symbol is 7F0."

  • To find the ground state term symbol for the $4f^6$ configuration, we use Hund's rules.
  • The $4f$ subshell has 7 orbitals ($m_l = +3, +2, +1, 0, -1, -2, -3$).
  • We have 6 electrons to place in these orbitals. According to Hund's first rule, we maximize the total spin (S). We place one electron in each of the first 6 orbitals with parallel spins (e.g., spin up).
  • Total spin angular momentum: $S = 6 \times (\frac{1}{2}) = 3$. The spin multiplicity is $2S+1 = 2(3)+1 = 7$.
  • According to Hund's second rule, we maximize the total orbital angular momentum (L). We sum the $m_l$ values for the occupied orbitals: $L = (+3) + (+2) + (+1) + (0) + (-1) + (-2) = 3$. An L value of 3 corresponds to an F term (L=0:S, L=1:P, L=2:D, L=3:F, etc.).
  • The term symbol is $^{2S+1}L$, which is $^7F$.
  • According to Hund's third rule, we determine the total angular momentum (J). For a subshell that is less than half-filled ($4f^6$ is less than $4f^7$), the ground state has the lowest possible J value, which is $|L-S|$.
  • $J = |L-S| = |3-3| = 0$.
  • Thus, the ground state term symbol is $^7F_0$. Statement D is correct.

Conclusion

Based on our analysis:

  • Statement A is incorrect.
  • Statement B is correct.
  • Statement C is incorrect.
  • Statement D is correct.

Therefore, the correct statements are B and D.

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Important Questions from Term Symbol

  1. The number of micro states corresponding to the atomic term symbol 4F is

  2. The possible terms arising from a p1d1 configuration are

  3. The ground state term symbol and the gj value for Pr3+ ion, respectively are (Atomic number of Pr is 59)

  4. For the d3 electron configuration, the ground state term symbol is

  5. The term symbol for the ground dinitrogen cation radical (\(N^+_2\)) is

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