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Q4.1(v):
Write down the electronic configuration of: (v) \(Co^{2+}\)

Solution :

Step 1: Identifying the Atomic Number and Neutral Configuration

The element Cobalt ($Co$) possesses an atomic number ($Z$) of $27$. [Per the Aufbau Principle, electrons are filled into orbitals in order of increasing energy]. The electronic configuration for a neutral Cobalt atom in its ground state is:

$Co (Z=27): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7$

Alternatively, using the nearest noble gas (Argon, $Ar$, $Z=18$) core notation:

$Co: [Ar] 3d^7 4s^2$

Step 2: Determining the Configuration of the Ion

To derive the electronic configuration of the cation $Co^{2+}$, we must remove $2$ electrons from the neutral atom. [Per the shell ionization rule, electrons are removed from the orbital with the highest principal quantum number ($n$) first].

  • In $Co$, the $4s$ orbital ($n=4$) is higher in energy/further from the nucleus than the $3d$ orbital ($n=3$).
  • Therefore, electrons are removed from the $4s^2$ subshell before the $3d$ subshell.

Step 3: Calculating Final Electronic Configuration

Subtracting $2$ electrons from the $4s$ subshell:

$Co^{2+} = [Ar] 3d^7 4s^{2-2}$

$Co^{2+} = [Ar] 3d^7$

Visual Representation of Orbital Occupancy

3d Orbitals of Co²⁺: 3d⁷ configuration (3 unpaired electrons)

Final Solution: The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$ or $1s^2 2s^2 2p^6 3s^2 3p^6 3d^7$.


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