Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration.
Oxidation state of an element depends upon the electrons present in the outermost shell or eight minus the number of valence shell electrons (outermost shell electrons). e.g.,
Alkali metals (Group 1 elements) General valence shell electronic configuration — $n s^1$; Oxidation state $=+1$.
Alkaline earth metals (Group 2 elements) General valence shell electronic configuration $-n s^2 ;$ Oxidation state $=+2$.
Alkali metals and alkaline earth metals belong to s-block elements and elements of group 13 to group 18 are known as $p$-block elements.
Group 13 elements General valence shell electronic configuration - $n s^2 n p^1$; Oxidation states $=+3$ and +1 .
Group 14 elements General valence shell electronic configuration $-n s^2 n p^2$; Oxidation states $=+4$ and +2 .
Group 15 elements General valence shell electronic configuration $-n s^2 n p^3$; Oxidation states $=-3,+3$ and +5 . Nitrogen shows $+1,+2,+4$ oxidation states also.
Group 16 elements General valence shell electronic configuration $-n s^2 n p^4$; Oxidation states $=-2,+2,+4$ and +6.
Group 17 elements General valence shell electronic configuration $-n s^2 n p^5$; Oxidation states $=-1 . \mathrm{Cl}, \mathrm{Br}$ and I also show $+1,+3,+5$ and +7 oxidation states.
Group 18 elements General valence shell configuration $-n s^2 n p^6$. Oxidation state $=$ zero.
Transition elements or d-block elements General electronic configuration $-(n-1) d^{1-10} n s^{1-2}$. These elements show variable oxidation states due to involvement of not only $n s$ electrons but $d$ or $f$-electrons (inner-transition elements) as well. Their most common oxidation states are +2 and +3 .
Nitrogen has positive electron gain enthalpy whereas oxygen has negative. However, oxygen has lower ionisation enthalpy than nitrogen. Explain.
Electronic configuration of ${ }_7 \mathrm{~N}=1 s^2, 2 s^2, 2 p_x^1, 2 p_y^1, 2 p_z^1$. Nitrogen has stable configuration because $p$-orbital is half-filled. Therefore, addition of extra electron to any of the $p$-orbital requires energy.
Electronic configuration of ${ }_8 \mathrm{O}=1 s^2, 2 s^2, 2 p_x^2, 2 p_y^1, 2 p_z^1$. Oxygen has $2 p^4$ electrons, so process of adding an electron to the $p$-orbital is exothermic.
Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 p$-orbital, oxygen acquires stable configuration, i.e., $2 p^3$. On the other hand, in case of nitrogen it is not easy to remove one of the three $2 p$-electrons due to its stable configuration.
First member of each group of representative elements (i.e., $s$ and $p$-block elements) shows anomalous behaviour. Illustrate with two examples.
First member of each group of representative elements (i.e., $s$ - and p-block elements) shows anomalous behaviour due to (i) small size (ii) high ionisation enthalpy (iii) high electronegativity and (iv) absence of $d$ - orbitals. e.g., in s-block elements, lithium shows anomalous behaviour from rest of the alkali metals.
(a) Compounds of lithium have significant covalent character. While compounds of other alkali metals are predominantly ionic.
(b) Lithium reacts with nitrogen to form lithium nitride while other alkali metals do not form nitrides.
In p-block elements, first member of each group has four orbitals, one $2 s$ - orbital and three $2 p$-orbitals in their valence shell. So, these elements show a maximum covalency of four while other members of the same group or different group show a maximum covalency beyond four due to availability of vacant $d$ - orbitals.
p-block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with water.
In p-block, when we move from left to right in a period, the acidic character of the oxides increases due to increase in electronegativity. e.g.,
(i) 2nd period $\mathrm{B}_2 \mathrm{O}_3<\mathrm{CO}_2<\mathrm{N}_2 \mathrm{O}_3$ acidic nature increases.
(ii) 3rd period $\mathrm{Al}_2 \mathrm{O}_3<\mathrm{SiO}_2<\mathrm{P}_4 \mathrm{O}_{10}<\mathrm{SO}_3<\mathrm{Cl}_2 \mathrm{O}_7$ acidic character increases.
On moving down the group, acidic character decreases and basic character increases. e.g.,
(a) Nature of oxides of 13 group elements
$$ \underset{\text { Weakly acidic }}{\mathrm{B}_2 \mathrm{O}_3} \underbrace{\mathrm{Al}_2 \mathrm{O}_3 \quad \mathrm{Ga}_2 \mathrm{O}_3}_{\text {Amphoteric }} \underset{\text { Basic }}{\mathrm{In}_2 \mathrm{O}_3} \underset{\text { Strongly basic}}{\mathrm{Tl}_2 \mathrm{O}} $$
(b) Nature of oxides of 15 group elements
$\underset{\text { Strongly acidic}}{\mathrm{N}_2 \mathrm{O}_5} ~\underset{\text { Moderately acidic}}{\mathrm{P}_4 \mathrm{O}_{10}} ~\underset{\text { Amphoteric}}{\mathrm{As}_4 \mathrm{O}_{10}}~ \underset{\text { Amphoteric}}{\mathrm{Sb}_4 \mathrm{O}_{10}} ~\underset{\text { Basic}}{\mathrm{Bi}_2 \mathrm{O}_3}$
Among the oxides of same element, higher the oxidation state of the element, stronger is the acid. e.g., $\mathrm{SO}_3$ is a stronger acid than $\mathrm{SO}_2$.
$\mathrm{B}_2 \mathrm{O}_3$ is weakly acidic and on dissolution in water, it forms orthoboric acid. Orthoboric acid does not act as a protonic acid (it does not ionise) but acts as a weak Lewis acid.
$$\underset{\text { Boron trioxide }}{\mathrm{B}_2 \mathrm{O}_3}+3 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { Orthoboric acid }}{2 \mathrm{H}_3 \mathrm{BO}_3}$$
$$\mathrm{B}(\mathrm{OH})_3+\mathrm{H}-\mathrm{OH} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}^{+}$$
$\mathrm{Al}_2 \mathrm{O}_3$ is amphoteric in nature. It is insoluble in water but dissolves in alkalies and reacts with acids.
$$\underset{\begin{array}{c} \text { Aluminium } \\ \text { trioxide } \end{array}}{\mathrm{Al}_2 \mathrm{O}_3}+2 \mathrm{NaOH} \xrightarrow{\Delta} \underset{\begin{array}{c} \text { Sodium meta } \\ \text { aluminate } \end{array}}{2 \mathrm{NaAlO}_2}+\mathrm{H}_2 \mathrm{O} \longleftarrow \mathrm{Al}_2 \mathrm{O}_3+6 \mathrm{HCl} \xrightarrow{\Delta} \underset{\begin{array}{c} \text { Aluminium } \\ \text { chloride } \end{array}}{2 \mathrm{AlCl}_3}+3 \mathrm{H}_2 \mathrm{O}$$
$$\begin{aligned} &\mathrm{Tl}_2 \mathrm{O} \text { is as basic as } \mathrm{NaOH} \text { due to its lower oxidation state (+1). }\\ &\mathrm{Tl}_2 \mathrm{O}+2 \mathrm{HCl} \longrightarrow 2 \mathrm{TICl}+\mathrm{H}_2 \mathrm{O} \end{aligned}$$
P$$_{4}$$O$$_{10}$$ on reaction with water gives orthophosphoric acid
$$\underset{\begin{array}{c} \text { Phosphorus } \\ \text { pentoxide } \end{array}}{\mathrm{P}_4 \mathrm{O}_{10}}+6 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\begin{array}{c} \text { Orthophosphoric } \\ \text { acid } \end{array}}{4 \mathrm{H}_3 \mathrm{PO}_4}$$
$$\begin{aligned} &\mathrm{Cl}_2 \mathrm{O}_7 \text { is strongly acidic in nature and on dissolution in water, it gives perchloric acid. }\\ &\underset{\substack{\text { Dichlorine heptoxide }}}{\mathrm{Cl}_2 \mathrm{O}_7}+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Perchloric acid }}{2 \mathrm{HClO}_4} \end{aligned}$$
How would you explain the fact that the first ionisation enthalpy of sodium is lower than that of magnesium but its second ionisation enthalpy is higher than that of magnesium?
First ionisation enthalpy of sodium $\left(\mathrm{Na}=1 s^2, 2 s^2, 2 p^6, 3 s^1\right)$ is lower than that of magnesium $\left(\mathrm{Mg}=1 s^2, 2 s^2 2 p^6, 3 s^2\right)$ because the electron to be removed in both the cases is from 3 s -orbital but the nuclear charge is lower in Na than that of magnesium. After the removal of first electron $\mathrm{Na}^{+}$acquires inert gas $(\mathrm{Ne})$ configuration $\left(\mathrm{Na}^{+}=1 s^2, 2 s^2, 2 p^6\right)$ and hence, removal of second electron from sodium is difficult. While in case of magnesium, after the removal of first electron, the electronic configuration of $\mathrm{Mg}^{+}$is $1 s^2, 2 s^2, 2 p^6, 3 s^1$. In this case $3 s^1$ electron is easy to remove in comparison to remove an electron from inert gas configuration. Therefore, $\mathrm{IE}_2$ of Na is higher than that of magnesium.