Question

Give reasons for the following:
(i) $Z{{n}^{2+}}$ salts are white but $C{{u}^{2+}}$ salts are blue in colour
(ii) Fluorine gives only one oxide but chlorine gives a series of oxides.

Answer 1

Hint:
(i) Here, to reflect a colour a compound must have incompletely filled $d$ orbital, that is $d$ orbital should have unpaired electrons. So that the $d-d$ transition can take place and gives rise to a coloured compound.
(ii) To be able to make multiple bonds with oxygen it is necessary for a halogen to have vacant $d$ orbital so that the halogen can accept electrons in the vacant $d$ orbital and expand its octet.

Complete answer:
(i) Zinc and copper are transition metals and the ions of the transition metals are generally coloured due to $d-d$transition of unpaired electrons that takes place within the d subshell.

But in case of Z{{n}^{2+}}$$$$3d is completely filled and they are no unpaired electrons available for transition, so the transitions take place from the different shell which is higher in energy and thus these transitions requires more energy than the energy provided by visible light, that is they absorb light from ultraviolet region and hence, they appear white. However, ion $C{{u}^{2+}}$ $3d$ has 9 electrons. The unpaired electron absorbs particular wavelengths in the visible region of light getting excited and the transmitted light shows the complementary colour which is 'blue'.

$Zn$ has electronic configuration = $3{{d}^{10}}4{{s}^{2}}$
$Z{{n}^{2+}}=3{{d}^{10}}$
$Cu=3{{d}^{10}}4{{s}^{1}}$
$C{{u}^{2+}}=3{{d}^{4}}$.
In case of $Z{{n}^{2+}}$ fully filled $d$ orbital is present therefore no $d-d$ transition can be possible in this case and it is colourless.
In the case of $C{{u}^{2+}}$ because of $d-d$ transition electrons emit light in the visible range and hence they are coloured compounds.

(ii) Fluorine has only two orbits. In 2nd orbit it has $2{{s}^{2}}2{{p}^{5}}$ configuration with no vacant '$d$' orbitals. So it cannot form more than one oxide that is $O{{F}_{2}}$​.
Chlorine, on the other hand, having configuration $3{{s}^{2}}3{{p}^{5}}3{{d}^{0}}$ has 3 orbits in its atoms and $3d$ is lying vacant in it. Therefore, it can show variable covalency and oxidation states and gives a series of oxides for example $C{{l}_{2}}O,Cl{{O}_{2}}$​.

Note: Numerous chlorine oxides are known for example di-chlorine oxide, $C{{l}_{2}}O$, and chlorine dioxide, $Cl{{O}_{2}}$, and two mixed anhydrides, di-chlorine hexoxide, $C{{l}_{2}}{{O}_{6}}$, and di-chlorine heptoxide, $C{{l}_{2}}{{O}_{7}}$, they are fairly stable under certain conditions.