Among the elements $\mathrm{B}, \mathrm{Al}, \mathrm{C}$ and Si ,
(a) which element has the highest first ionisation enthalpy?
(b) which element has the most metallic character?
Justify your answer in each case.
The placing of elements are as
| Period | Group-13 | Group-14 |
|---|---|---|
| 2nd Period | Boron | Carbon |
| 3rd Period | Aluminium | Silicon |
(a) Ionisation enthalpy increases along a period (as we move from left to right in a period) with decrease in atomic size and decreases down the group with increase in atomic size. Hence, carbon has the highest first ionisation enthalpy.
(b) Metallic character decreases across a period but increases on moving down the group. Hence, aluminium has the most metallic character.
Write four characteristic properties of $p$-block elements.
The four important characteristic properties of p-block elements are the following
(a) $p$-Block elements include both metals and non-metals but the number of non-metals is much higher than that of metals. Further, the metallic character increases from top to bottom within a group and non-metallic character increases from left to right along a period in this block.
(b) Their ionisation enthalpies are relatively higher as compared to s-block elements.
(c) They mostly form covalent compounds.
(d) Some of them show more than one (variable) oxidation states in their compounds. Their oxidising character increases from left to right in a period and reducing character increases from top to bottom in a group.
Choose the correct order of atomic radii of fluorine and neon (in pm) out of the options given below and justify your answer.
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.