Question

write the electronic configuration of Cu²⁺, H^-, Fe³⁺, Mn^​2+, Cr³⁺
also, find the ion having
1. maximum number of protons
2. maximum number of neutrons
3. maximum number of unpaired electrons.

Answer 1

Dear student


Electronic configuration of Cu²⁺, H^-, Fe³⁺, Mn²⁺, Cr³⁺
Cu²⁺ = 3d⁹
H = 1s²2s¹
​Fe³⁺ =3d⁵
Mn²⁺ = 3d⁵
​Cr³⁺ =3d³

1.No.of protons = no.of electrons.= atomic number
In case of ions, the protons remain same as the element, the number of electrons change

So, no.of protons in Cu²⁺ =29
No.of protons in H^- = 1
No.of protons in Fe³⁺ = 26
No.of protons in Mn²⁺ = 25
No.of protons in Cr =24

Cu²⁺ has maximum number of protons


2.
Number of neutrons =mass number - no.of protons
no.of neutrons in Cu²⁺ = 63-29 =34
No.of neutrons in H^- = 2-1= 1
No.of neutrons in Fe³⁺ =56- 26 = 30
No.of neutrons in Mn²⁺= 55-25 = 30
no.of neutrons in Cr³⁺= 52-24 = 28
Thus Cu has maximum number of neutrons


3. Cu²⁺ has 1 unpaired electron
H^- has 1 unpaired electron
Fe³⁺ has 5 unpaired electrons
Mn²⁺ has 5 unpaired electrons
Cr³⁺ has 3 unpaired electrons.

Thus Fe³⁺ and Mn²⁺ have maximum unpaired electrons

Regards