Question

Find sum of unpaired electrons in :-

Atomic Nitrogen ; K$_3$[Fe(CN)$_6$] ; unstable S$_2$ molecule ; BaO$_2$ ; Nitric oxide ; Ozone ; Dioxygen ; [Ti(H$_2$O)$_6$]Cl$_3$ ; [MnCl$_6$]$^{3-}$ ; [Mn(CN)$_6$]$^{3-}$

Answer 1

16 unpaired electrons

Answer 2

We'll determine the number of unpaired electrons species‐wise and then add them up.

Step‐by‐Step:

1. Atomic Nitrogen, N:
   Configuration: 1s² 2s² 2p³ → Three electrons in 2p (each unpaired).
   Unpaired electrons = 3

2. K₃[Fe(CN)₆]:
   In $[Fe(CN)_6]^{3-}$, Fe is in +3 oxidation state ⇒ $d^5$. Cyanide is a strong‐field ligand so the complex is low‐spin.
   Configuration: $t_{2g}^5$ → One orbital must pair two electrons, leaving the other two orbitals singly occupied.
   Unpaired electrons = 1

3. Unstable S₂ molecule:
   MO theory (analogous to O₂) gives two electrons in degenerate $\pi^*$ orbitals.
   Unpaired electrons = 2

4. BaO₂ (Barium peroxide):
   The O₂²⁻ (peroxide ion) has two extra electrons compared to O₂ filling the $\pi^*$ orbitals. This leads to all electrons being paired.
   Unpaired electrons = 0

5. Nitric Oxide, NO:
   MO theory shows one unpaired electron.
   Unpaired electrons = 1

6. Ozone, O₃:
   Resonance forms yield a structure with all electrons paired.
   Unpaired electrons = 0

7. Dioxygen, O₂:
   Standard MO diagram for O₂ shows two unpaired electrons in the $\pi^*$ orbitals.
   Unpaired electrons = 2

8. [Ti(H₂O)₆]Cl₃:
   In $[Ti(H_2O)_6]^{3+}$, Ti is in +3 oxidation state. Ti (atomic no. 22) loses three electrons so that $d^1$ remains.
   Unpaired electrons = 1

9. [MnCl₆]³⁻:
   For $[MnCl_6]^{3-}$: 6 Cl⁻ (each –1) gives Mn oxidation state +3: $d^4$. Chloride is a weak-field ligand → high-spin configuration.
   High spin $d^4$: three electrons in $t_{2g}$ and one electron in $e_g$ (all unpaired).
   Unpaired electrons = 4

10. [Mn(CN)₆]³⁻:
    Here, Mn is also +3 ($d^4$), but CN⁻ is a strong-field ligand → low-spin configuration.
    For low-spin $d^4$: electrons fill $t_{2g}$ orbitals as: one electron in each orbital (3 unpaired) and the 4th electron pairs in one, leaving 2 unpaired electrons overall.
    Unpaired electrons = 2

Total Sum:

$$3 + 1 + 2 + 0 + 1 + 0 + 2 + 1 + 4 + 2 = 16$$

Each species’ unpaired electrons were determined via electronic configuration (using MO theory for diatomic molecules and ligand field theory for complexes). Their total is 16.