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Among the following series of transition metals ions,the one where all metal ions have $3d^2$ electronic configuration is

$\begin{array}{1 1}(a)\;Ti^{3+},V^{2+},Cr^{3+},Mn^{4+}\\(b)\;Ti^{+},V^{4+},Cr^{6+},Mn^{7+}\\(c)\;Ti^{4+},V^{3+},Cr^{2+},Mn^{3+}\\(d)\;Ti^{2+},V^{3+},Cr^{4+},Mn^{5+}\end{array}$

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$22Ti:3d^2,4s^2$
$24Cr:3d^5,4s^1$
$23V:3d^3,4s^2$
$25Mn:3d^5,4s^2$
Hence (d) is the correct option.
answered Jan 13, 2014 by sreemathi.v
 

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