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The d- and f-Block Elements
Electronic Configurations of Transition Metal Ions
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Concepts tested
- d electron count of ions
All Questions
2004 AIPMT 1 question
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Among the following series of transition metal ions, the one where all metal ions have $3d^2$ electronic configuration is :-For first row transition metal ions, the 4s electrons are lost first. The number of d electrons = (group number of the element) $-$ (charge on the ion), i.e. total valence electrons minus the charge.
Valence electrons ($3d + 4s$): Ti = 4, V = 5, Cr = 6, Mn = 7.
For a $3d^2$ ion: Ti must be $Ti^{2+}$ (4 – 2 = 2), V must be $V^{3+}$ (5 – 3 = 2), Cr must be $Cr^{4+}$ (6 – 4 = 2), Mn must be $Mn^{5+}$ (7 – 5 = 2).
Checking the other series: $Ti^{4+}$ is $3d^0$, $V^{2+}$ is $3d^3$, $Cr^{6+}$ is $3d^0$, so those series are not all $3d^2$.
So the series with all ions $3d^2$ is $Ti^{2+}, V^{3+}, Cr^{4+}, Mn^{5+}$.