Question

Match each element with the full ground-state electron configuration of the monatomic ion it is most likely to form.

A   1s22s22p63s23p64s1 B   1s22s22p63s23p64s23d104p6 C   1s22s22p6 D   1s2 E   1s22s22p63s23p64s23d104p5 F   1s22s22p63s23p6 G   1s22s22p63s23p1

Br Na Li Al N K

Answer 1

1)
Number of electron in Br = 35
Since charge on Br is -1
So, we need to add 1 electrons more
Number of electrons to be arranged = 36
Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
Answer: Br —> B

2)
Number of electron in Na = 11
Electronic configuration is 1s2 2s2 2p6 3s1
Since charge on Na is +1
So, we need to remove 1 electrons
Electrons are always removed from highest number orbital
Removing 1 electron from 3s
Final electronic configuration is : 1s2 2s2 2p6
Answer: Na —> C

3)
Number of electron in Li = 3
Electronic configuration is 1s2 2s1
Since charge on Li 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
Answer: Li —> D

4)
Number of electron in Al = 13
Electronic configuration is 1s2 2s2 2p6 3s2 3p1
Since charge on Al is +3
So, we need to remove 3 electrons
Electrons are always removed from highest number orbital
Removing 1 electron from 3p
Removing 2 electron from 3s
Final electronic configuration is : 1s2 2s2 2p6
Answer: Al —> C

5)
Number of electron in N = 7
Since charge on N is -3
So, we need to add 3 electrons more
Number of electrons to be arranged = 10
Electronic configuration is 1s2 2s2 2p6
Answer: N—> C

6)
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: K —> F