Chapter 40, Problem 031 Consider the elements selenium ( Z = 34), bromine ( Z =

Question

Chapter 40, Problem 031

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

1s 2s 2p 3s 3p 3d 4s 4p . . . .

(a) What are the highest occupied subshell for selenium and (b) the number of electrons in it (c) What are the highest occupied subshell for bromine and (d) the number of electrons in it (e) What are the highest occupied subshell for krypton and (f) the number of electrons in it?

Explanation / Answer

For Selenoum ( Z= 34) , electronic configuration will be:

1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p4 ----------------------------------1

The first 18 electrons are in same orbitals as that of argon ,So we can write symbol [Ar] to represent those 18 electrons.

[Ar] 3d10,4s2,4p4   ( rewrite eq.1)

a) from this configuration, Highest filled subshell for Selenium = 4p

b) the no of e- in it = 4

For Bromine ( z= 35), its electronic configuration is:

[Ar] 3d10, 4s2, 4p5

c) highest filled subshell is = 4p

d) no of e- in it = 5

For Krypton ( Z= 36) , its electronic configuration is:

[Ar] 3d10, 4s2, 4p6

e) highest filled subshell is = 4p

f) no of e in it = 6

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