Among the following series of transition metal ions, the one where all metal ions have $3d^2$ electronic configuration is :-
A
$Ti^+, V^{4+}, Cr^{6+}, Mn^{7+}$
B
$Ti^{4+}, V^{3+}, Cr^{2+}, Mn^{3+}$
C
$Ti^{2+}, V^{3+}, Cr^{4+}, Mn^{5+}$
D
$Ti^{3+}, V^{2+}, Cr^{3+}, Mn^{4+}$
The answer is C.
Solution
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 $Ti^{2+}, V^{3+}, Cr^{4+}, Mn^{5+}$.
Conceptd electron count of ions
The d- and f-Block Elements in past papers
76 questions from this chapter have appeared across 27 exam years.
