MCQ
The octahedral diamagnetic low spin complex among the following is
- A$\left[ NiCl _4\right]^{2-}$
- B$\left[ CoCl _6\right]^{3-}$
- C$\left[ CoF _6\right]^{3-}$
- ✓$\left[ Co \left( NH _3\right)_6\right]^{3+}$
(2) Paramagnetic, High Spin and Octahedral
(3) Paramagnetic, High Spin and Octahedral
(4) Diamagnetic, Low Spin and Octahedral
$\left[ Co \left( NH _3\right)_6\right]^{3+}, CN =6(\text { Octahedral })$
$NH _3= SFL$
$Co ^{+3}=[ Ar ] 3 d ^6$
Diamagnetic and Low spin complex
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Reason $(R)$ : Oxygen forms $\mathrm{p} \pi-\mathrm{p} \pi$ multiple bonds with itself and other elements having small size and high electronegativity like $\mathrm{C}, \mathrm{N}$, which is not possible for sulphur.
In the light of the above statements, choose the most appropriate answer from the options given below :
$C$ and $ D$ are