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Q4.1 (vi):
Write down the electronic configuration of: (vi) Lu^{2+}

Solution :

Step 1: Establishing the Ground State Electronic Configuration of Lutetium ($Lu$)

Lutetium ($Lu$) is the final element of the lanthanide series with an atomic number ($Z$) of $71$. To determine its electronic configuration, we utilize the Aufbau principle, filling orbitals in order of increasing energy ($1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d$).

The ground state electronic configuration of the neutral $Lu$ atom ($Z=71$) is:

$Lu: [Xe] \, 4f^{14} \, 5d^{1} \, 6s^{2}$

[Justification: Per the Aufbau principle and the stability of the filled $4f$ subshell, the $5d$ orbital receives the additional electron after the $4f$ subshell is completed.]

Step 2: Ionization Logic for $Lu^{2+}$

The formation of a cation requires the removal of electrons from the orbitals with the highest principal quantum number ($n$) first, as these electrons are furthest from the nucleus and held with the least electrostatic attraction [Per the principle of energetic sequence of ionization].

  • The valence shell electrons reside in the $6s$ ($n=6$) and $5d$ ($n=5$) orbitals.
  • To form $Lu^{2+}$, we must remove $2$ electrons from the neutral atom configuration.
  • Priority of removal: Electrons are removed from the $6s$ orbital first due to the higher principal quantum number ($n=6 > n=5$).

Step 3: Calculating the Electronic Configuration of $Lu^{2+}$

Start with neutral $Lu$: $[Xe] \, 4f^{14} \, 5d^{1} \, 6s^{2}$

Remove the first electron from the $6s$ orbital: $[Xe] \, 4f^{14} \, 5d^{1} \, 6s^{1}$

Remove the second electron from the $6s$ orbital: $[Xe] \, 4f^{14} \, 5d^{1} \, 6s^{0}$

Visual Representation of Orbital Occupancy

Orbital Energy Levels for Lu²⁺: 4f¹⁴ 5d¹ 6s⁰ Increasing Energy

Analytical Summary

Species Atomic Number ($Z$) Electronic Configuration
$Lu$ (Neutral) 71 $[Xe] \, 4f^{14} \, 5d^{1} \, 6s^{2}$
$Lu^{2+}$ 71 $[Xe] \, 4f^{14} \, 5d^{1}$

Final Solution: The electronic configuration of $Lu^{2+}$ is $[Xe] \, 4f^{14} \, 5d^{1}$ (or $[Xe] \, 4f^{14} \, 5d^1 \, 6s^0$).


More Questions from Class 12 Chemistry The d- and f-Block Elements Exercises


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