Question

Write the full electron configuration of the following in spdf notation.
A).Ag.
B) V3+.
C).Pd
D) W.
E) Cd2+.

Answer 1

Here are the full electron configurations in spdf notation for each of the specified elements and ions:

A) **Ag (Silver)**:
Atomic number = 47
Electron configuration:
\[ \text{[Kr]} 4d^{10} 5s^1 \]

B) **V³⁺ (Vanadium ion)**:
Atomic number = 23
First, we find the electron configuration for neutral Vanadium:
\[ \text{[Ar]} 4s^2 3d^3 \]
When it loses three electrons to form V³⁺, the electrons are removed from the 4s and 3d orbitals:
Electron configuration:
\[ \text{[Ar]} 3d^2 \]

C) **Pd (Palladium)**:
Atomic number = 46
Electron configuration:
\[ \text{[Kr]} 4d^{10} \]
(Note: Palladium has a filled d-subshell and does not have a 5s electron under this configuration.)

D) **W (Tungsten)**:
Atomic number = 74
Electron configuration:
\[ \text{[Rn]} 5f^{14} 6d^4 7s^2 \]

E) **Cd²⁺ (Cadmium ion)**:
Atomic number = 48
First, we find the electron configuration for neutral Cadmium:
\[ \text{[Kr]} 5s^2 4d^{10} \]
When it loses two electrons to form Cd²⁺, the electrons are removed from the 5s orbital:
Electron configuration:
\[ \text{[Kr]} 4d^{10} \]

These are the electron configurations you requested in spdf notation.