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Which has maximum number of unpaired electron :

$(a)\;Mg^{2+}\qquad(b)\;Ti^{3+}\qquad(c)\;V^{3+}\qquad(d)\;Fe^{2+}$

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$Fe^{2+}$ has four unpaired electron($1s^2,2s^22p^6,3s^23p^63d^6)$
Hence (d) is the correct answer.
answered Jan 15, 2014 by sreemathi.v
 

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