(a) How do you prepare:
(i) $K_{2} M n O_{4}$ from $M n O_{2}$ ?
(ii) ${N a}_{2} {C r}_{2} O_{7}$ from ${N a}_{2} C r O_{4}$ ?
(b) Account for the following:
(i) ${M n}^{2 +}$ is more stable than ${F e}^{2 +}$ towards oxidation to $+ 3$ state.
(ii) The enthalpy of atomization is lowest for $Z n$ in $3 d$ series of the transition elements.
(iii) Actinoid elements show wide range of oxidation states.

Open in App

Solution

a)
i) $K_{2} M n O_{4}$ can be prepared from pyrolusite $(M n O_{2})$ . The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as $K N O_{3}$ or $K C l O_{4}$ , to give $K_{2} M n O_{4}$ .

$2 M n O_{2} + 4 K O H + O_{2} Δ \to K_{2} M n O_{4} g r e e n + 2 H_{2} O$

ii) $N a_{2} C r_{2} O_{7}$ can be prepared from $N a_{2} C r O_{4}$ in the following way:
For the preparation of sodium dichromate, the yellow solution of sodium chromate is acidified with sulphuric acid to give a solution from which orange sodium dichromate, $N a_{2} C r_{2} O_{7} .2 H_{2} O$ can be crystallized.
Balanced equation for above reaction is as follows:

$2 N a_{2} C r O_{4} Y e l l o w + 2 H^{+} \to N a_{2} C r_{2} O_{7} + 2 N a^{+} O r a n g e + H_{2} O$

b)
i) Electronic configuration of $M n^{2 +}$ is $[A r]^{18} 3 d^{5}$
Electronic configuration of $F e^{2 +}$ is $[A r]^{18} 3 d^{6}$
It is known the half-filled and fully-filled orbitals are more stable. Therefore, Mn in +2 state has a stable $d^{5}$ configuration. Therefore, $M n^{2 +}$ shows resistance to oxidation to $M n^{3 +}$ . Also, $F e^{2 +}$ has $3 d^{6}$ configuration and by losing one electron, its configuration changes to a more stable $3 d^{5}$ configuration. Therefore, $F e^{2 +}$ gets oxidised to $F e^{3 +}$ easily.

ii) The extent of metallic bonding an element undergoes, decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: $3 d^{10} 4 s^{2}$ ), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.

iii) Actinides exhibit larger oxidation states because of very small energy gap between 5f, 6d and 7s sub-shells. The energies are calculated on the basis of (n+I) rule. The (n+I) values of the three orbitals are:
$5 f = 5 + 3 = 8$
$6 d = 6 + 2 = 8$
$7 s = 7 + 0 = 7$
Since all the values are almost the same, therefore all orbitals can be involved in bonding resulting in a larger oxidation number for actinoids.

Suggest Corrections

Similar questions

3^{r d}

M n^{2 +}

F e^{2 +}

S c (Z = 21)

Z n (Z = 30)

+ 3

E^{o}

M n^{3 +} / M n^{2 +}

C r^{3 +} / C r^{2 +}

Join BYJU'S Learning Program