P-Block Elements (n & O Family) flash cards

1. 1.Write the general valence-shell electronic configuration of Group 15 (nitrogen family) elements.

   $ns^2 np^3$ (half-filled $p$ subshell). The half-filled $p^3$ configuration gives extra stability, reflected in high ionization enthalpy and low electron gain enthalpy.

   Hint: n s and n p; think half-filled stability.

2. 2.Write the general valence-shell electronic configuration of Group 16 (oxygen family, chalcogens) elements.

   $ns^2 np^4$. They are two electrons short of the noble-gas configuration, so they readily gain 2 electrons to form $M^{2-}$ ions.

   Hint: n s and n p; two short of octet.

3. 3.Name the elements of Group 15 in order and classify their metallic character.

   N, P, As, Sb, Bi. N and P are non-metals, As and Sb are metalloids, Bi is a metal. Metallic character increases down the group.

   Hint: Nitrogen to Bismuth; non-metal → metalloid → metal.

4. 4.Name the elements of Group 16 and their metallic classification.

   O, S, Se, Te, Po. O and S are non-metals, Se and Te are metalloids, Po is a radioactive metal. Metallic/covalent-to-metallic character rises down the group.

   Hint: Oxygen to Polonium (chalcogens).

5. 5.List the common oxidation states shown by Group 15 elements.

   Common: $-3, +3, +5$. Also $+1,+2,+4$ in some oxides/oxoacids. The $-3$ state stability decreases and $+3$ stability increases down the group.

   Hint: Range from $-3$ to $+5$.

6. 6.List the common oxidation states of Group 16 elements.

   $-2, +2, +4, +6$. Oxygen shows mainly $-2$ (also $-1$ in peroxides, $-1/2$ in superoxides, $+2$ in $OF_2$). $+6$ stability decreases down the group.

   Hint: $-2$ up to $+6$; O is the exception.

7. 7.Why does the stability of the $+5$ oxidation state decrease down Group 15 (N to Bi)?

   Due to the inert pair effect: the $ns^2$ electrons become increasingly reluctant to participate in bonding down the group (poor shielding by intervening $d/f$ electrons). Thus $Bi^{5+}$ is a strong oxidant; $+3$ dominates for Bi.

   Hint: Inert pair effect.

8. 8.Which is more stable, $+3$ or $+5$ state, for bismuth, and what does this imply about $Bi(V)$?

   $+3$ is far more stable for Bi. Consequently $Bi(V)$ compounds (e.g. $NaBiO_3$) are powerful oxidising agents, used to oxidise $Mn^{2+}$ to $MnO_4^-$.

   Hint: Inert pair; $NaBiO_3$ oxidant.

9. 9.How does catenation tendency vary in Group 15 and Group 16?

   Group 15: decreases N < P... (N shows little catenation; P shows more, e.g. $P_4$). Group 16: catenation is maximum for sulphur (S–S rings/chains, e.g. $S_8$), decreasing down the group; O forms only limited catenation ($O_3$, peroxides).

   Hint: S is the catenation champion; N poor.

10. 10.Why is nitrogen a gas while phosphorus (and heavier congeners) is a solid?

    Nitrogen exists as small $N_2$ molecules with a strong $N \equiv N$ triple bond and only weak van der Waals forces between molecules, so it is a gas. P and heavier elements form larger polyatomic units ($P_4$ etc.) with stronger intermolecular forces, hence solids.

    Hint: $N_2$ vs $P_4$; molecular size.

11. 11.Why is the covalency of nitrogen limited to 4 whereas phosphorus can show covalency up to 6?

    Nitrogen has no $d$ orbitals in its valence shell (n=2), limiting it to a maximum of 4 bonds. Phosphorus (n=3) has empty $3d$ orbitals available, allowing expansion of the octet and covalency up to 6 (e.g. $PF_6^-$, $PCl_5$).

    Hint: Absence vs presence of $d$ orbitals.

12. 12.State two anomalous properties of nitrogen compared to other Group 15 elements.

    (1) N forms $p\pi$–$p\pi$ multiple bonds ($N \equiv N$, $N=O$) whereas heavier ones prefer single bonds. (2) N has no $d$ orbitals, so it cannot expand its octet (max covalency 4). Also small size, high electronegativity, no catenation.

    Hint: $p\pi$–$p\pi$ bonding and no $d$ orbitals.

13. 13.Why does nitrogen form $p\pi$–$p\pi$ multiple bonds readily but phosphorus prefers single bonds?

    Nitrogen's small size and short bond length allow effective sideways ($p\pi$–$p\pi$) overlap. Phosphorus is larger, so its $3p$ orbitals overlap poorly; it prefers to form more single bonds ($P_4$) rather than $P \equiv P$.

    Hint: Small size = good $p$-orbital overlap.

14. 14.State two anomalous properties of oxygen relative to other Group 16 elements.

    (1) $O_2$ is diatomic with $O=O$ double bond ($p\pi$–$p\pi$); S,Se,Te form larger $S_8$-type rings. (2) O has no $d$ orbitals, so maximum covalency 2 (rarely 4); S can show 4 and 6. Also strong H-bonding by O.

    Hint: $O_2$ vs $S_8$; no $d$ orbitals; H-bonding.

15. 15.Why is water a liquid but $H_2S$ a gas at room temperature?

    Oxygen's small size and high electronegativity let water molecules form extensive intermolecular hydrogen bonding, giving a high boiling point (liquid). Sulphur is larger/less electronegative, so $H_2S$ has only weak van der Waals forces and is a gas.

    Hint: Hydrogen bonding in $H_2O$.

16. 16.Explain the order of thermal stability of Group 15 hydrides $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$.

    As the central atom size increases down the group, the M–H bond length increases and bond strength (bond enthalpy) decreases, so thermal stability decreases. $BiH_3$ is the least stable.

    Hint: Bond strength falls with size.

17. 17.Give the trend of reducing character of Group 15 hydrides and explain.

    Reducing character increases: $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$. Weaker M–H bonds down the group release hydrogen more easily, increasing reducing power.

    Hint: Opposite to thermal stability.

18. 18.Explain why $NH_3$ has a higher boiling point than $PH_3$.

    $NH_3$ molecules are hydrogen-bonded (N is small and highly electronegative), requiring extra energy to separate; $PH_3$ has only weak van der Waals forces. Hence $NH_3$ boils higher despite lower molar mass.

    Hint: Intermolecular H-bonding.

19. 19.Compare and explain the bond angles of $NH_3$ ($107°$) and $PH_3$ ($\approx 93.5°$).

    In $NH_3$, high electronegativity of N and $sp^3$ hybridisation give $107°$. In $PH_3$, P uses nearly pure $p$ orbitals for bonding (little hybridisation) so H–P–H is close to $90°$ ($93.5°$). Bond angle decreases down the group.

    Hint: $sp^3$ vs almost pure $p$ orbitals.

20. 20.Why does basicity of Group 15 hydrides decrease down the group ($NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$)?

    Basicity depends on availability/density of the lone pair for donation. As atomic size increases, the lone pair occupies a larger, more diffuse orbital, decreasing its donor ability, so basicity falls.

    Hint: Lone-pair availability.

21. 21.How is dinitrogen ($N_2$) prepared in the laboratory from ammonium nitrite?

    By heating an aqueous solution of ammonium nitrite (formed from $NH_4Cl + NaNO_2$): $NH_4NO_2 \rightarrow N_2 + 2H_2O$. The unstable ammonium nitrite decomposes to give nitrogen gas.

    Hint: $NH_4NO_2$ decomposition.

22. 22.How is very pure $N_2$ obtained (two thermal-decomposition methods)?

    (1) $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$. (2) $2NaN_3 \rightarrow 2Na + 3N_2$ (thermal decomposition of sodium azide) — gives very pure nitrogen.

    Hint: Ammonium dichromate; sodium azide.

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