Question

(A) Account for the following:
(i) $Mn$ shows the highest oxidation state of +7 with oxygen but with fluorine, it shows oxidation state of +4.
(ii) $C{r}^{2+}$ is a strong reducing agent.
(iii) $C{u}^{2+}$ salts are coloured, while $Z{n}^{2+}$ salts are white.
(B) Complete the following equations:
(i) $2Mn{O}_2+4KOH+O_2\stackrel{\Delta }{\to }$
(ii) $C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to$

Answer 1

(A)-(i)

Element Name and Symbol	Atomic Number	Common Oxidation States
Scandium (se)	21	+3
Titanium (Ti)	22	+4
Vanadium (V)	23	+2, +3, +4, +5
Chromium (Cr)	24	+2, +3, +6
Manganese (Mn)	25	+2, +3, +4, +6, +7
Iron (Fe)	26	+2, +3
Cobalt (Co)	27	+2, +3

So Manganese ($Mn$) shows highest number of oxidation state.

Element	Sc	Ti	V	Cr	Mn	Fe	Co
M.P. (${}^{o}C$)	1540	1668	1890	1875	1245	1537	1495

(ii) Scandium shows only +3 oxidation state.

Zn has electric cofiguration $=3d^{10}4s^2$

$Z{n}^{2+}=3d^{10}$

$Cu=3d^{5}4s^1$

$C{u}^{2+}=3d^6$

In case of $Zn$ fully filled d orbital is present therefore no $d-d$ transition can be possible in this case it is colorless.

In case of copper because of $d-d$ transition electrons emits light in the visible range and hence they are colored compounds.

(iii) $C{r}^{2+}$gets converted to $C{r}^{3+}$ as the +3 oxidation state has half filled $t_{2g}$ orbitals - thus it is a good reducing agent.

(B)

(i) $2Mn{O}_2+4KOH+O_2\to 2K_{2}Mn{O}_4+2H_{2}O$

(ii) $C{r}_{2}O{7}^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$