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Question

An element forms an ion in the +IV oxidation state with electronic configuration : $4d^4 \ 5d^{10}$. Which group and period of the Modern Periodic Table does the element belong to?

The correct answer is

p-block element from Period - 6 and Group - 14

To determine the group and period of the element in the Modern Periodic Table, we need to analyze the given electronic configuration of the ion: 4d^4 \ 5d^{10}. This configuration is typical of elements in the p-block, particularly in the heavy p-block elements.

  1. First, consider the configuration 5d^{10}. The fact that there is a fully filled 5d subshell indicates the preceding 5d block is complete.
  2. The 4d^4 configuration suggests that the element has four electrons beyond the filled 4d orbitals, occupying 5p orbitals.
  3. This places the element in Group 14, as elements in this group typically have four electrons added to the p-orbitals.
  4. Given the configuration of 5d and 4d, and knowing that the 5d corresponds to the 6th period, this element belongs to Period 6 of the period table.

Therefore, based on the electronic configuration and the periodic properties, the element is a p-block element from Period 6 and Group 14. This matches with option: p-block element from Period - 6 and Group - 14.

To further ensure the understanding, it is critical to grasp that in the periodic table, the filling order and the sequence of the periods and groups adhere to the Aufbau principle, where electrons occupy the lowest available energy orbital. Additionally, Period 6 involves both the filling of the 4f, 5d, and 6p orbitals, aligning with the given configuration.

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