Ag⁺ ion
[Kr] 4d¹⁰. 0 unpaired electrons, diamagnetic.
Element or ion
Type a symbol, a name or an atomic number, add a charge such as Fe3+ or Cl-, or tap an element in the table. Orbitals fill from m = −l up, spin +½ first.
Ag+Silver ion, 46 electrons
[Kr] 4d10
In full: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰. Ag loses 5s¹
Ag loses 5s¹
- Blockd block
- Period5
- Group11
- Outer shell18
- Unpaired0
- MagneticDia
Last electron 4d
- n4shell
- l2d subshell
- m+2orbital
- s−½spin down
Step by step
- Silver has 47 electrons: [Kr] 4d¹⁰ 5s¹.
- The ion has 1 fewer, 46 electrons. Remove them from the highest n first.
- Ag loses 5s¹.
- Ag⁺ ion: [Kr] 4d¹⁰, in full 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰.
Ag⁺ ion: the numbers
| Atomic number | 47 |
|---|---|
| Electrons | 46 |
| Period | 5 |
| Group | 11 |
| Block | d block |
| Electrons in the outer shell | 18 |
| Unpaired electrons | 0 |
| Magnetic | diamagnetic |
| Last electron | 4d: n = 4, l = 2, m = +2, s = −1/2 |
Ag⁺ ion: quick answers
What is the electronic configuration of Ag⁺?
Ag⁺ has 46 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰, or [Kr] 4d¹⁰. Ag loses 5s¹.
How many unpaired electrons are there in Ag⁺?
Ag⁺ has 0 unpaired electrons, so it is diamagnetic.
Why does Ag lose its outer electrons first?
Electrons leave from the highest n first. The 4s level fills before 3d, but once 3d holds electrons the 4s electrons are the outermost, so they are lost first.