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What is the spin only magnetic moment value in Bohr magneton unit for $K_2Co(SCN)_4$?

$\begin{array}{1 1} 3.873\;\mu B \\ 3.982 \;\mu B \\ 2.25\;\mu B \\ 1.77\;\mu B\end{array}$

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Answer: 3.873 $\mu_B$
Cobolt in $[Co(SCN)_4]^{2-}$ is present as $Co^{2+}$. It involves $sp^3$ hybridization with three unpaired electrons.
Therefore, the spin only magnetic moment value $ = \sqrt{n(n+2)}\;\mu_B = \sqrt{3(3+2)}\;\mu_B = 3.873 \;\mu_B$
answered Jan 30, 2014 by sreemathi.v
edited Aug 5, 2014 by balaji.thirumalai
 

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