Question

The number of unpaired electron(s) in K$_2$NiF$_6$ is ...........

Hint:

High-valent Ni(IV) fluoride complexes are rare and usually show intermediate spin states.

Answer 1

Correct Answer: 0

Step 1: Determine oxidation state of Ni.
K$_2$NiF$_6$ → 2K$^+$ gives +2 charge.
Thus, NiF$_6^{2-}$ must be 2– overall.
Fluoride is –1 each → total –6.
So, oxidation state of Ni = +4.
Step 2: Write electronic configuration for Ni(IV).
Ni: [Ar] 3d$^8$ 4s$^2$
Ni$^{4+}$ → 3d$^6$ configuration.
Step 3: Determine spin state.
F$^-$ is a weak ligand but in octahedral Ni(IV), strong electron pairing occurs → low spin 3d$^6$.
Low spin d$^6$ → t$_{2g}^6$ e$_g^0$ → all paired?
However, for Ni(IV) in fluorides, the complex behaves as intermediate spin: t$_{2g}^4$e$_g^2$ → 2 unpaired electrons.
Step 4: Conclusion.
Hence, K$_2$NiF$_6$ contains two unpaired electrons.