Question

Write ground-state electron configurations for the ions Bek*, O* and Fe2+ Which do you expect will be paramagnetic due to the presence of unpaired electrons? (Express your answer as a series of orbitals. For example, the electron configuration of Li would be entered in complete form as 152 2st or in condensed form as [He]2s.) ion electron configuration paramagnetic Bet Submit Answer 5 question attempts remaining

Answer 1

1)
a)
Number of electron in Be = 4
Electronic configuration is 1s2 2s2
Since charge on Be is +1
So, we need to remove 1 electrons
Electrons are always removed from highest number orbital
Removing 1 electron from 2s
Final electronic configuration is : 1s2 2s1
Answer: 1s2 2s1

b)
It has 1 unpaired electron in 2s orbital.
So, it is paramagnetic
Answer: yes

2)
a)
Number of electron in K = 19
Electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s1
Since charge on K is +1
So, we need to remove 1 electrons
Electrons are always removed from highest number orbital
Removing 1 electron from 4s
Final electronic configuration is : 1s2 2s2 2p6 3s2 3p6
Answer: 1s2 2s2 2p6 3s2 3p6

b)
All electrons are paired.
So, it is not paramagnetic
Answer: No

3)
a)
Number of electron in O = 8
Since charge on O is -2
So, we need to add 2 electrons more
Number of electrons to be arranged = 10
Electronic configuration is 1s2 2s2 2p6
Answer: 1s2 2s2 2p6

b)
All electrons are paired.
So, it is not paramagnetic
Answer: No

4)
a)
Number of electron in Fe = 26
Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d6 4s2
Since charge on Fe is +2
So, we need to remove 2 electrons
Electrons are always removed from highest number orbital
Removing 2 electron from 4s
Final electronic configuration is : 1s2 2s2 2p6 3s2 3p6 3d6
Answer: 1s2 2s2 2p6 3s2 3p6 3d6

b)
d orbitals has unpaired electrons.
So, it is paramagnetic
Answer: yes