Chemistry - PERIODIC TABLE Question with Solution | TestHub

The question asks to identify the option that does NOT represent the correct order of the indicated properties.

Let's analyze each option:

A. $Sc>Ti>Cr>Ni$ (Atomic size): Across a period, atomic size generally decreases due to increasing nuclear charge. This order is correct.

B. $S{c}^{3+}>T{i}^{2+}>C{r}^{1+}>Mn$ (Ionic size): This order is incorrect. Ionic size generally decreases with increasing nuclear charge for isoelectronic species or species with similar electron configurations. $S{c}^{3+}$ has an electron configuration of $[Ar]$, $T{i}^{2+}$ is $[Ar]3d^2$, $C{r}^{1+}$ is $[Ar]3d^5$, and $Mn$ is $[Ar]3d^{5}4s^2$. Comparing $S{c}^{3+}$ and $T{i}^{2+}$, $S{c}^{3+}$ has a smaller nuclear charge than $T{i}^{2+}$ for the same number of electrons (isoelectronic with Ar), so $S{c}^{3+}$ should be larger than $T{i}^{2+}$ if they were isoelectronic. However, they are not isoelectronic. For transition metals, ionic size generally decreases across a period.

C. $S{c}^{3+}<T{i}^{2+}<C{r}^{1+}<Mn$ (Ionic size): This order is incorrect. This is the reverse of option B, and also incorrect.

D. $F{e}^{2+}>C{o}^{2+}>N{i}^{2+}>C{u}^{2+}$ (Number of unpaired electrons):

$F{e}^{2+}$: $[Ar]3d^6$, 4 unpaired electrons

$C{o}^{2+}$: $[Ar]3d^7$, 3 unpaired electrons

$N{i}^{2+}$: $[Ar]3d^8$, 2 unpaired electrons

$C{u}^{2+}$: $[Ar]3d^9$, 1 unpaired electron

This order is correct.

The question asks for the option that does NOT represent the correct order. Option B states $S{c}^{3+}>T{i}^{2+}>C{r}^{1+}>Mn$ for ionic size. This order is incorrect because ionic size generally decreases across a period for transition metal ions with similar charges, and also depends on the electron configuration. For example, $S{c}^{3+}$ (isoelectronic with Ar) is smaller than $T{i}^{2+}$ (which has $3d^2$ electrons).

The final answer is $\boxed{B}$