Class 12 The p-Block Elements - Group 16 Elements

Q: The formation of the oxide ion, \$O^{2-}(g)\$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:\$O(g)+e^-\rightarrow O^-(g); \Delta_fH^{\ominus}=-141kJ mol^{-1}\$\$O^-(g)+e^-\rightarrow O^{2-}(g); \Delta_fH^{\ominus}=+780 kJ mol^{-1}\$Thus, process of formation of \$O^{2-}\$ in gas phase is unfavourable even though \$O^{2-}\$ is isoelectronic with neon. It is due to the fact that:

The formation of the oxide ion, \$O^{2-}(g)\$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:\$O(g)+e^-\rightarrow O^-(g); \Delta_fH^{\ominus}=-141\ kJ mol^{-1}\$\$O^-(g)+e^-\rightarrow O^{2-}(g); \Delta_fH^{\ominus}= +780\ kJ mol^{-1}\$Thus, the process of formation of \$O^{2-}\$ in the gas phase is unfavorable even though \$O^{2-}\$ is isoelectronic with neon. It is due to the fact that electron repulsion outweighs the stability gained by achieving noble gas configuration. When an electron is added to \$O^-\$ anion, there is strong electrostatic repulsion between the two negative charges. Due to this, the second electron gain enthalpy of oxygen is positive.

Q: An amorphous solid (X) burns in air to form a gas (Y) which turns lime water milky. This gas decolourises aqueous solution of acidified \$ KMnO_4\$. Gas (Y) reacts with oxygen to give another gas (Z) which is responsible for acid rain. X, Y, and Z are:

Hence as shown in above image: X=\$S\$ , Y=\$SO_2\$ , Z=\$SO_3\$Therefore, option B is correct.

Q: The increasing order of electron gain enthalpy of the following substances is: \$O < S < Se < Te\$If the order of bond dissociation enthalpy is correct, enter 1 else 0.

Electron gain enthalpy is the amount of energy released when an element accepts an electron. The process is endothermic.\$\text {Lesser the electronic repulsion faced by an electron to get occupied in the outermost shell higher the amount of }\$\$\text {energy released.}\$On moving down the group, the electron gain enthalpy decreases from S to Po. Oxygen has an unexpectedly low value of electron gain enthalpy.Since oxygen is smaller in size than S, so incoming \$e^-\$ on oxygen faces more electronic repulsion on oxygen outermost surface ( due to more \$e^-\$ density) as a result electron affinity decreases.The increasing order of electron gain enthalpy with negative sign is \$O (-141.4 kJ/mol) < Po (-174 kJ/mol) < Te (-190.0 kJ/mol) < Se (-195.5 kJ/mol) < S (-208.8 kJ/mol)\$ Hence the correct answer is 0.

Q: Explain -\$H_2 O\$ is Liquid but \$H_2S\$ is a gas.

\$H_2S\$ is a gas and water is liquid at room temperature. Because, in water there is strong hydrogen bonding, and the molecules get extensively associated by weak electrostatic forces. So, water exists as a liquid at room temperature with a high boiling point. Whereas, Sulphur is less electronegative than oxygen, and the S-H bond is much less polar than the O-H bond. Hence, there is no hydrogen bonding in hydrogen sulphide, and it exists as a gas.

Q: The oxidation state of sulphur in the anions \$S_{2}O_{4}^{2-}\$, \$S_{2}O_{4}^{2-}\$ and \$S_{2}O_{6}^{2-}\$ follows the order:

Oxidation state of \$S_{2}O_{4}^{2-}\$\$2(x)+4(-2)=-2\$ \$2x=8-2\$ \$2x=6\$\$x=3\$Oxidation state of \$SO_{3}^{2-}\$\$x+3(-2)=-2\$ \$x=6-2\$ \$x=4\$Oxidation state of \$S_{2}O_{6}^{2-}\$\$2(x)+6(-2)=-2\$ \$2x=12-2\$ \$2x=10\$\$x=5\$ So the oxidation state of sulphur in the anions \$S_{2}O_{4}^{2-}\$, \$S_{2}O_{4}^{2-}\$ and \$S_{2}O_{6}^{2-}\$ follows the order.\$S_{2}O_{4}^{2-}\$<\$SO_{3}^{2-}\$<\$S_{2}O_{6}^{2-}\$.Hence option B is correct.

Q: Among \$PbS, \ CuS, \ HgS, \ MnS, \ Ag_2S, \ NiS, \ CoS, \ Bi_2S_3\$ and SnS\$_2\$, the total number of black coloured sulphides are:

The total number of black colored sulfides is 6. These include \$PbS, CuS, HgS, NiS, CoS, Bi_2S_3\$. \$Bi_2S_3\$ in its crystalline form is dark brown but \$Bi_2S_3\$ precipitate obtained is black in color.Hence, the correct option is A

Q: The correct boiling point order is:

True (1)Higher BP`s mean higher forces (because more thermal energy needed to separate the molecules in liquid state) Even though Sulfur is more electronegative than Se or Te (so you would expect a higher dipole moment and so higher attractive forces between molecules) the size of the molecules is a bigger factor in their interactions. More energy is needed simply because the molecules of the hydrides get bigger as you go down the Group. Oxygen is so much more electronegative that the molecules have H-bonding in water. This give rise to almost a different (and higher) order of intermolecular force, which gives rise to such a high BP in water.

Q: In which of the following arrangements, the given sequence is not strictly according to the property indicated against it?

The following sequence is not strictly according to the property indicated against it.\${H}_{2}O {H}_{2}S>{H}_{2}Se>{H}_{2}Te\$; decreasing \$p{K}_{a}\$ values.In Hydrides, as we move top to bottom, acidic nature increases so with the increase in the acidity from top to bottom, the \${K}_{a}\$ value increases and \$p{K}_{a}\$ value decreases.

Q: Give reason oxygen is a gas but sulphur is a solid ?

The intermolecular forces in oxygen are weak van der Waals forces, which causes it to exist as gas while sulphur form strong \$S=S\$ double bonds hence exist as a puckered structure held together by strong covalent bonds and exist as a polyatomic molecule, hence exists as a solid.

Q: Among the hydrides of group \$16\$ elements, which has the lowest boiling point?

The boiling points of the hydrides of group 16 elements is as follows:\$H_2O\$ \$\rightarrow\$ \$100^{\circ}C\$\$H_2S\$ \$\rightarrow\$ \$- (60^{\circ }C)\$\$H_2Se\$ \$\rightarrow\$ \$-(41.25^{\circ}C)\$\$H_2Te\$ \$\rightarrow\$ \$-(2.2^{\circ}C)\$Gradual increase in boiling points down the rest of the group happens as the relative molecular mass of the molecules increases.Also, boiling point of water is \$\textbf{anomalously}\$ high.Hence \$H_2S\$ has lowest boiling point here.