Question

Among the following series of transition metal ions, the one where all metal ions have same 3d electronic configuration is

• A. $T{i}^{2+}$, $V^{3+}$, $C{r}^{4+}$, $M{n}^{5+}$
• B. $T{i}^{3+}$, $V^{2+}$, $C{r}^{3+}$, $M{n}^{4+}$
• C. $T{i}^{+}$, $V^{4+}$, $C{r}^{6+}$, $M{n}^{7+}$
• D. $T{i}^{4+}$, $V^{3+}$, $C{r}^{2+}$, $M{n}^{3+}$

Answer 1

Answer: (A)

$T{i}^{2+}$, $V^{3+}$, $C{r}^{4+}$, $M{n}^{5+}$

$T{i}^{+2},V^{3+},C{r}^{+4},M{n}^{+5}$ have $3d^2$ configurations

$Ti=\left[Ar\right]3d^{2}4s^2$

$T{i}^{+2}=\left[Ar\right]3d^2$

$V=\left[Ar\right]3d^{3}4s^2$

$V^{+3}=\left[Ar\right]3d^2$

$Cr=\left[Ar\right]3d^{5}4s^1$

$C{r}^{+4}=\left[Ar\right]3d^2$

$Mn=\left[Ar\right]3d^{5}4s^2$

$M{n}^{+5}=\left[Ar\right]3d^2$

Hence option A is correct