Q.Write the general outer electronic configuration of the Group 15 elements and list the range of oxidation states they show. Why can nitrogen show oxidation states all the way from $-3$ to $+5$?

Acidic strength of group-15 pentoxides decreases down the group; N2O5 (top of the group) is the most acidic.

All the oxides listed are +5 oxides of group-15 elements: N2O5, P2O5, As2O5, Sb2O5. Going down the group the elements become larger and more metallic, so their oxides become progressively less acidic (more basic/amphoteric):

N2O5 > P2O5 > As2O5 > Sb2O5 (decreasing acidic strength)

Nitrogen, being the smallest and most electronegative, forms the most strongly acidic oxide (N2O5 is the anhydride of the strong acid HNO3). Sb2O5 is only weakly acidic/amphoteric.

Basicity of group-15 hydrides decreases down the group, so NH3 is the strongest base.

In the hydrides of group 15 (NH3, PH3, AsH3, SbH3), basicity depends on the availability of the lone pair on the central atom for donation. As we go down the group the atomic size increases, the lone pair occupies a larger, more diffuse orbital, and its electron density (and hence donor ability) falls. Therefore basic strength decreases:

NH3 > PH3 > AsH3 > SbH3

Nitrogen is the smallest, so its lone pair is the most concentrated and most readily donated — NH3 is the strongest base.

Nitrogen lacks d orbitals in the second shell, so its maximum covalency is 4.

Nitrogen is a second-period element; its valence shell (n = 2) has only 2s and 2p orbitals and no d orbitals. It can therefore form a maximum of four covalent bonds by using its four available orbitals, as in the ammonium ion NH4+ (three normal N-H bonds plus one coordinate bond). It cannot expand its octet beyond four bonds, unlike phosphorus, which can reach a covalency of 5 or 6 using d orbitals. Hence the maximum covalency of nitrogen is 4.

Count the electrons: 2 + 2 + 1 + 1 + 1 = 7. The element with atomic number 7 is nitrogen.

Add up the electrons in 1s2 2s2 2px1 2py1 2pz1 = 2 + 2 + 1 + 1 + 1 = 7 electrons.

The atom with 7 electrons (and Z = 7) is nitrogen. Note the three 2p electrons occupy the three 2p orbitals singly (Hund's rule), giving nitrogen its half-filled, stable 2p3 configuration.

• Hydrogen (Z = 1), oxygen (Z = 8) and fluorine (Z = 9) do not match 7 electrons.

Reducing character of group 15 hydrides increases down the group, so NH3 is the weakest reducing agent.

The hydrides of group 15 elements are NH3, PH3, AsH3, SbH3 and BiH3. Reducing character of a hydride depends on the strength of the E-H bond (E = group 15 element) - the weaker the E-H bond, the more easily it is broken to release H, and the stronger the reducing power.

Down the group, atomic size increases and the E-H bond becomes progressively longer and weaker, so bond dissociation enthalpy decreases from NH3 to BiH3. Hence reducing character increases in the order:

NH3 < PH3 < AsH3 < SbH3 < BiH3

Since NH3 has the strongest (shortest) N-H bond of the series, it is the hardest to oxidise and therefore the weakest reducing agent.

Basicity of group 15 hydrides decreases down the group.

Among the group 15 hydrides, basic strength decreases down the group: $NH_3 > PH_3 > AsH_3 > BiH_3$. This is because as the central atom gets larger and less electronegative down the group, its lone pair becomes more diffuse and less available for donation to a proton, weakening its Lewis basicity. $NH_3$ is thus the strongest base.

Nitrogen and oxygen are non-metals.

Nitrogen (group 15) and oxygen (group 16) lie in the upper-right p-block of the periodic table. They are electronegative, tend to gain electrons, exist as gaseous diatomic molecules (N2, O2), and do not show metallic properties like lustre or electrical conductivity — hence they are non-metals.

Inert-pair effect → Bi is most stable as +3.

In group 15 the stability of the +5 state falls and that of +3 rises down the group (inert-pair effect on the 6s² electrons). Bismuth, the heaviest, is therefore most stable in the +3 state, and Bi(V) compounds (e.g. NaBiO₃) are powerful oxidisers.