JEE Main 2013 — D And F Block Elements Question with Solution

(1)

V = 3d34s2	V2+ = 3d3	= 3 unpaired electron
Cr = 3d54s1	Cr2+ = 3d4	= 4 unpaired electron
Mn = 3d54s2	Mn2+ = 3d5	= 5 unpaired electron
Fe = 3d64s2	Fe2+ = 3d6	= 4 unpaired electron

hence the correct order of paramagnetic behaviour

$$V^{2+}<Cr{}^{2+}=F{e}^{2+}<M{n}^{2+}$$

(2)

For the same oxidation state, the-ionic radii generally $$dc$$-creases as the atomic number increases in a particular transition series.

Hence the order is

$$M{n}^{++}>F{e}^{++}>C{o}^{++}>N{i}^{++}$$

(3)

In solution, the stability of the compounds depends upon electrode potentials, $$SEP$$ of the transitions metal ions are given as

$$C{o}^{3+}/Co=+1.97,F{e}^{3+}/Fe=+0.77;$$

$$C{r}^{3+}/C{r}^{2+}=-0.41,S{c}^{3+}$$ is highly stable as it does not show $$+2$$ $$O.$$ $$S.$$

(4)

$$Sc-\left(+2\right),\left(+3\right)$$

$$Ti-\left(+2\right),\left(+3\right),\left(+4\right)$$

$$Cr-\left(+1\right),\left(+2\right),\left(+3\right),\left(+4\right),\left(+5\right),\left(+6\right)$$

$$Mn-\left(+2\right),\left(+3\right),\left(+4\right),\left(+5\right),\left(+6\right),\left(+7\right)$$

$$i.e.Sc<Ti<Cr=Mn$$

$$E_a=53598.6J/mol$$

$$=53.6kJ/mol.$$