Q34 / 150 · MHT CET 2026 (16 April Shift 1) · Chemistry
Identify the number of unpaired electrons present and geometry respectively of $[\text{Co(NH}_3)_6]^{3+}$ complex.
- A0, square planar
- B2, square planar
- C4, octahedral
- Correct answerD0, octahedral
Solution
The oxidation state of Co in $[\text{Co(NH}_3)_6]^{3+}$ is $+3$.
The electronic configuration of Co is $[\text{Ar}] 3d^7 4s^2$.
The electronic configuration of Co$^{3+}$ is $[\text{Ar}] 3d^6$.
Since $\text{NH}_3$ acts as a strong field ligand for Co$^{3+}$, pairing of electrons takes place in the $3d$ subshell.
All $6$ electrons pair up in the inner $d$-orbitals, leaving two $3d$ orbitals empty.
The number of unpaired electrons is $0$.
The complex undergoes $d^2sp^3$ hybridization, resulting in an octahedral geometry.
Answer: 0, octahedral
The electronic configuration of Co is $[\text{Ar}] 3d^7 4s^2$.
The electronic configuration of Co$^{3+}$ is $[\text{Ar}] 3d^6$.
Since $\text{NH}_3$ acts as a strong field ligand for Co$^{3+}$, pairing of electrons takes place in the $3d$ subshell.
All $6$ electrons pair up in the inner $d$-orbitals, leaving two $3d$ orbitals empty.
The number of unpaired electrons is $0$.
The complex undergoes $d^2sp^3$ hybridization, resulting in an octahedral geometry.
Answer: 0, octahedral
Hence, the correct answer is option: 'D'.