Consider the elements selenium `(Z=34)`, bromine `(Z=35)`, and krypton `(Z=36)`. In their part of the periodic table, the subshells of the electronic states are filled in the sequence.
`1s2s2p3s3p3d4s4p……`
What are (a) the highest occupied subshell for selenium and (b) the number of electrons in it, (c ) the highest occupied subshell for bromine and (d) the number of electrons in it, and (e) the highest occupied subshell for krypton and (f) the number of electrons in it?

Explanation

To solve the problem, we will analyze the electronic configurations of selenium, bromine, and krypton step by step.

### Step 1: Determine the electronic configuration of selenium (Z=34)
- The atomic number of selenium is 34, which means it has 34 electrons.
- The electronic configuration follows the order of filling subshells:
- 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴.
- Therefore, the electronic configuration of selenium is: **[Ar] 4s² 4p⁴**.

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