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Exercises · 3.5

Q.In terms of period and group where would you locate the element with Z=114Z = 114?

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Element 114 (flerovium) sits in Period 7, Group 14 — directly below lead in the carbon family, found by filling electrons through the 7th shell and counting four valence electrons in the p-block.

Why electron configuration determines position

The periodic table is organized by how electrons fill atomic orbitals. The period number tells you the highest principal quantum level (nn) occupied by electrons, while the group number reflects the valence electron count and the block (s, p, d, f) tells you which subshell is being filled last.

For superheavy elements like Z=114Z = 114, we follow the Aufbau principle through the entire sequence of orbital filling, then read off the period from the outermost shell and the group from the valence configuration.

Step-by-step location

  1. Write the full electron configuration for Z=114Z = 114

    Following the standard filling order (1s → 2s → 2p → 3s → 3p → 4s → 3d → … → 7p), we build up 114 electrons:

[Rn] 5f14 6d10 7s2 7p2[\text{Rn}] \, 5f^{14} \, 6d^{10} \, 7s^2 \, 7p^2

Here radon (Z=86Z = 86) accounts for the first 86 electrons, then we fill the 5f subshell (14 electrons), the 6d subshell (10 electrons), the 7s subshell (2 electrons), and finally place 2 electrons in 7p.

Check: 86+14+10+2+2=11486 + 14 + 10 + 2 + 2 = 114 ✓

  1. Identify the period

    The outermost electrons occupy the n=7n = 7 shell (7s and 7p orbitals). The period number equals the highest principal quantum number with electrons.

    Period = 7

  2. Identify the block

    The last electron enters a 7p orbital, so element 114 belongs to the p-block.

  3. Determine the group number

    For p-block elements, the group number is 10+(number of s and p valence electrons)10 + \text{(number of s and p valence electrons)}.

    Valence electrons: 7s2 7p27s^2 \, 7p^2 gives us 2+2=42 + 2 = 4 valence electrons. …

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