Question

Which of the following arrangements does not represent the correct order of property stated against it ?

• A. $V^{2+}<C{r}^{2+}<M{n}^{2+}<F{e}^{2+}:$ paramagnetic behaviour
• B. $N{i}^{2+}<C{o}^{2+}<F{e}^{2+}<M{n}^{2+}:$ Ionic size
• C. $C{o}^{3+}<F{e}^{3+}<C{r}^{3+}<S{e}^{3+}:$ Stability in aqueous solution
• D. $Sc<Ti<Cr<Mn:$number of oxidation states

Answer 1

The correct option is A $V^{2+}<C{r}^{2+}<M{n}^{2+}<F{e}^{2+}:$ paramagnetic behaviour

Answer: a
(a)
$V^{2+}=3$ unpaired electrons
$C{r}^{2+}=4$ unpaired electrons
$M{n}^{2+}=5$ unpaired electrons
$F{e}^{2+}=4$ unpaired electrons

Hence, the order of paramagnetic behaviour should be
$V^{2+}<C{r}^{2+}=F{e}^{2+}<M{n}^{2+}$
(a) is incorrect.

(b) Ionic size decreases from left to right along a period till $N{i}^{2+}$ for $3d$ series.
So, the given order for ionic size is correct.

(c) (As per data from NCERT)
Standard reduction potentials
$C{o}^{3+}/C{o}^{2+}=1.97V;$
$F{e}^{3+}/F{e}^{2+}=0.77V;$
$C{r}^{3+}/C{r}^{2+}=-0.41V$

$S{c}^{3+}$ is highly stable , attains a noble gas configuration. So it does not show +2 oxidation state.
Higher the SRP, lesser is the stability of $+3$ in aqueous medium since the tendency is to reduce.
So the given order is correct.

(d) The Oxidation states increases as we go from group 3 to group 7 in the same period.
$Sc$ has +3 oxidation states.
$Ti$ has +2,+3,+4 oxidation states.
$Cr$ has +2,+3,+4,+5,+6 oxidation states.
$Mn$ has +2,+3,+4,+5,+6,+7 oxidation states.
The given order is correct.