P Block Elements Class 12 | Part 2 | Group 16 | Chapter 7 | Chemistry | Class 12 | CBSE|
P Block Elements Class 12
Group 16 of the periodic table consists of five elements viz, Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te) and Polonium (Po).
[He] 2s2 2p4
[Ne] 3s2 3p4
[Ar] 3d10 4s2 4p4
[Kr] 4d10 5s2 5p4
[Xe] 4f14 5d10 6s2 6p4
These elements are collectively called chalcogens.
The general electronic configuration is ns2 np4 .
The number of valence electrons are 6.
Question: Why are the group 16 elements called chalcogens?
Answer: Chalcogen means ore forming. The elements of group 16 are called chalcogens because many metals are found as oxides and sulphides and few as selenides and tellurides.
1)Atomic and Ionic Radii
Question: Atomic Radii of elements of group 16 are smaller than corresponding group 15 elements. Explain ?
Answer: The smaller radii of group 16 elements compared to group 15 elements are due to increased nuclear charge of the group 16 elements which results in greater attraction of the electrons by the nucleus.
Question: Why does, down the group, the atomic size of elements of group 16 increases ?
Answer: The increase in the atomic radii of group 16 element down the group is primarily due to increase in number of electron shells.
Question: Why is the first ionization enthalpies of elements of group 16 unexpectedly lower than group 15 elements?
Answer: This is due to the relatively symmetrical and more stable electronic configuration of the elements of group 15 as compared to the elements of group 16 (oxygen family). In other words, group 15 elements have half filled stable electronic configuration. So they require high value of ionization enthalpy for removal of first electron.
Question: Why are second ionization enthalpies of group 16 elements higher than those of corresponding elements of group15?
Answer: This is because after removal of one electron , group 16 elementys attain stable electronic configuration. So removal of second electron is difficult. Hence, II ionization enthalpies of group 16 elements is high.
Question: Why does the values of ionization enthalpy of group 16 elements decrease down the group?
Answer: This is due to increasing shielding effect and size as we move down the group.
3)Electron Gain Enthalpy
Question: Why have group16 elements large negative electron gain enthalpies?
Answer: Group 16 elements require only two electrons to attain stable noble gas configuration. Therefore, they have a high tendency to accept two additional electrons and hence have large negative electron gain enthalpies.
Question: Why is electron gain enthalpy of oxygen least negative in this group?
Answer: This is due to its small size. Its electrons which are present in outermost shell apply repulsion for incoming electron and hence the incoming electrons are not accepted with same ease as in case of other elements of this group.
Question: Why does, down the group, negative value of E.G.E decrease after sulphur?
Answer: After sulphur , size increases as well as shielding effect increases due to which effective nuclear charge is decreased. So electron accepting tendency decreases resulting in decreases electron gain enthaply.
Question: Why do group 16 elements have higher electronegativity than corresponding group 15 elements?
Answer: This is due to smaller atomic size of group 16 elements than group 15 elements. Moreover, group 16 elements need only two eletcron to attain the stable noble gas configuration.
Question: Why is electronegativity of sulphur much lower than that of oxygen?
Answer: This is probably due to an unexpected increase in size of sulphur (107 pm) as compared to that of oxygen (66 pm).
Note: Oxygen is the second most electronegative element.
Non- metallic / metallic character
Because of high ionization enthalpy values, the elements of group 16 are less metallic. However, as we move down the group, ionization enthalpy decreases and hence the metallic character increases.
Question; Explain the trend of non metallic / metallic character in group 16 elements?
Answer: Oxygen →Non metal
Sulphur → Non metal and is an insulator
Selenium → metalloid and hence semiconductor
Tellurium → metalloid and hence semiconductor
Polonium → metallic but radioactive with a short half life
Melting and Boiling points
Question: Why do melting and boiling point and densities increase regularly as we move down the group up to tellurium?
Answer: As we move down the group, atomic size increases. As a result, van der Waals force of attraction among their atom also increase and hence melting and boiling point regularly increase from O to Te.
Question: Why are the melting and boiling points of polonium lower than those of tellurium?
Answer: Due to maximum number of intervening d and f electrons, the inert pair effect is also maximum. Consequently, the s valence electrons in polonium are less available as compared to those in tellurium. As a result, van der Waals force of attraction will be weaker in Po than in Te and therefore, m.p and b.p of Po will be lower than that of Te.
Question: Why does oxygen exist as diatomic gas at room temperature?
Answer: Due to small size and high electronegativity , oxygen atom form pπ—pπ double bond with other oxygen atom to form O = O molecule. The intermolecular force of attraction between oxygen molecules are weak van der Waals forces and hence oxygen exists as diatomic gas at room temperature.
Question: Why do other elements (S, Se and Te) exist as octa atomic molecule?
Answer: Elements of group 16 except oxygen do not form pπ—pπ double bond due to their large size. Thus, they prefer to form single bond and have complex structures. Example S, Se and Te exists as octa-atomic molecules (S8, Se8 and Te8) having puckered 8-membered crown shaped rings.
Oxygen form pπ—pπ double bonds but other elements do not.
Sulphur and other elements of this group possess d orbitals and hence form pπ—dπ multiple bonds. To obtain effective pπ—dπ overlap, the size of d orbital must be similar to that of p orbital.
Tendency of these elements to form pπ—dπ multiple bonds decreases from S to Se.
Question: Why Sulphur has stronger tendency for catenation that oxygen?
Answer: Due to its small size, the lone pair of electron on the oxygen atom repel the bond pair of the O—O bond to a greater extent than the lone pair of electrons on the sulphur atoms in S—S bond. As a result, S—S bond is much stronger than O—O bond and hence sulphur has a much stronger tendency for catenation than oxygen.
Question: Why catenation tendency decreases from S to Po?
Answer: As the size of the atom increases down the group from S to Po, the strength of the element- element bond decreases and hence tendency for catenation decreases accordingly.
All the elements of this group show allotropy.
Question: Why Se and Te are called photosensitive elements?
Answer: The grey form of selenium and tellurium consists of parallel chains held by weak metallic bonds. In the presence of light, weak metallic bonded electrons are excited and as a result, number of free electrons increases and so does the electricity. Thus, these elements conduct electricity significantly in presence of light. That is why Se and Te are called photosensitive elements.
Negative Oxidation State:
All metal oxides are ionic and contain O2¯ ions in which oxygen shows an oxidation state of –2.
Since the electronegativities decrease as we move down the group, the tendency of these elements to show -2 oxidation state decrease from sulphur to polonium.
Oxygen also show an oxidation state of –1 in peroxides such as H2O2, –1/2 in superoxides such as KO2 , zero in O2 and O3, +1 in O2F2 and +2 in OF2.
The least electronegative element polonium does not exhibit negative oxidation state at all. It shows positive oxidation state only.
Positive Oxidation State:
Oxygen does not show positive oxidation state except in O2F2 and OF2.
Other elements of this group show positive oxidation state of +2, +4 and +6 due to promotion of electrons to vacant d orbital.
S, Se , Te and Po with oxygen are tercovalent (+4 oxidation state). These +4 compounds show both oxidising as well as reducing properties.
The compounds of S, Se, Te and Po with fluorine show an oxidation state of +6. These compounds in +6 oxidation state show only oxidising properties.
Trends in Chemical Reactivity
Reactivity toward Hydrogen (Formation of Hydrides)
General formula is H2E where E= O, S, Se, Te, Po
All the hydrides have angular shape involving sp3 hybridization of the central atom.
Question: Why the bond angles decreases from H2O to H2Te?
Answer: As we move down the group form O to Te, the size of the central atom goes on increasing and its electronegativity goes on decreasing. As a result, the position of two bond pairs shifts away and away from the central atom as we move from H2O to H2Te. Consequently, the repulsion between bond decreases as we move from H2O to H2Te . Bond angle decreases in order: H2O > H2S > H2Se > H2Te
The hydride of O, i.e., H2O is a colourless, odourless liquid while the hydrides of all the other elements are unpleasant, foul smelling, poisonous gases.
Question: Why acidic strength increases from H2O to H2Te ?
Answer: As the atomic size increases down the group, the bond length increases and hence the bond strength decreases. Consequently, the cleavage of E—H bond becomes easier. As a result, the tendency to release hydrogen as proton increases, i.e., acid strength increases down the group.
Note: H2O is least acidic while H2Po is most acidic.
Question: Why H2O is least acidic?
Answer: It is due to intermolecular hydrogen bonding in the molecule. Hydrogen atom gets trapped in hydrogen bond. Thus release tendency of H+ ion is very less.
The thermal stability of hydrides decreases from H2O to H2Te because as size of the atom E in H2E increases, the bond H—E becomes weaker and thus breaks on heating.
Hydrides of all elements of this group except that of oxygen, i.e., water is reducing agent.
Their reducing character increases from H2S to H2Te due to decrease in their thermal stability.
Melting and Boiling Points
Question: Why the m.pt and b.pt increases from H2S to H2Te?
Answer: This is due to increase in van der waals force of attraction with increasing molecular mass.
Question: Why H2O has high m.pt and b.pt than other hydrides?
Answer : This is due to presence of intermolecular hydrogen bonding in H2O.
Reactivity towards Oxygen (Formation of Oxides)
Sulphur, Selenium and tellurium when burnt in air form dioxides of formula EO2.
S8 + 8 O2 → 8SO2
Question: Why SO2 exists as gas at room temperature?
Answer: It exists as descrete molecules even in the solid state. These molecules are held together by weak van der Waals force of attraction. Therefore, SO2 is gas at room temperature.
Question: SeO2 is reducing while TeO2 is an oxidising agent. Give reasons.
Answer: Since +6 oxidation state of S is more stable than +4, therefore, it can readily donate electrons and hence acts as reducing agent. For example, it reduces ferric to ferrous salts.
Since due to inert pair effect, the stability of +6 oxidation state decreases while the stability of +4 and +2 oxidation state increases down the group. Thus, TeO2 accepts electron and thus acts as an oxidising agent.
Question: Discuss the structure and hybridization of SO2 molecule?
Answer: In SO2, S is sp2 hybridized. Two of three sp2– orbitals form two σ bondswith oxygen atom while the third contains the lone pair of electrons. SO2 has bent (angular) structure.
All the elements of this group form trioxides of formula EO3. Sulphur in SO3 is sp2 hybridised. The three sp2 orbitals of sulphur overlap with p orbitals of oxygen to form three S—O σ bonds.
The molecules of SO3 are held by weak van der waal forces of attraction. Therefore, SO3 exists as triangular planar gaseous molecule at room temperature.
Note: All trioxides are acidic in nature.
Reactivity toward Halogen (Formation of Halides)
The stability of the halides in any particular oxidation state decreases in the order: F > Cl > Br > I
Question: Why SF6 is chemically inert?
Question; Why SF6 is not hydrolysed but TeF6 is easily hydrolysed?
Answer: Its chemical inertness is due to the reason that six F atoms protect the sulphur atom from attack by the reagents to such an extent that even thermodynamically most favourable reactions like hydrolysis do not occur. However, as the size of central atom increases, steric hindrance decreases and hence the hydrolysis become easy.
The order of hydrolysis of hexafluorides is: SF6 > SeF6 > TeF6
All hexafluorides have octahedral structure.
Amongst tetrahalides, tetrafluorides are most stable. SF4 is a gas, SeF4 is a liquid while TeF4 is a solid. These fluorides have sp3d hybridization and trigonal bipyramidal structure.
Question: Why SF4 is easily hydrolysed while SF6 does not?
Answer: SF6 does not undergo hydrolysis because six fluorine atoms protect the sulphur atom from attack of water due to steric hindrance, SF4 readily undergo hydrolysis because four F atoms cannot protect the S atom from attack by water.
Question: How tetrahalides of group 16 elements can act both as lewis acid as well as lewis base?
Ans: The tetrahalides can act both as Lewis bases due to the presence of a lone pair of electrons and lewis acids because the central atom can extend its co-ordination number to six.
All elements except selenium form dihalides. These dihalides are formed by sp3 hybridisation. However, due to presence of two lone pair of electrons they have bent structure.
Question: Why SCl2 and H2O show different bond angles while both have bent (angular) shape?
Answer: In SCl2, the bond angle is little smaller (103º) as compared to that in H2O (104.5º) . This is due to reason that S is less electronegative than O. As a result, bond pair of two S—Cl bonds lie away from S atom in SCl2 as compared to those of O—H bonds in H2O. Consequently, bond pair – bond pair repulsion decreases and hence the bond angle decreases to 103º in SCl2 from 104.5º in H2O.
Dioxygen is a colourless, odourless and tasteless gas.
It is paramagnetic inspite of having even number of electrons.
It has three isotopes: 168O , 178O and 188O
Reaction with Metals:
4Na + O2 → 2Na2O
2Ca +O2 → 2CaO
2Mg + O2 → 2MgO
4Al +3O2 → 2Al2O3
2Fe +3O2 → 2Fe2O3
Reaction with non-metals:
P4 + 5O2 → P4O10
S + O2 → SO2
N2 +O2 → 2NO
Reaction with sulphur dioxide: 2SO2 + O2 → 2SO3
Reaction with Hydrogen chloride: 4HCl +O2 → 2H2O + Cl2
Question: Why dioxygen requires initiation by external heating?
Answer: The bond dissociation energy of dioxygen is high and hence the reaction requires initiation by external heating. However, when the reaction starts, it continues on its own. This is due to the reason that the chemical reactions of dioxygen are exothermic and the heat liberated during the reaction is sufficient ot carry out the reactions.
It is an allotropic form of oxygen and also called trioxygen.
Ozone is obtained by passing silent electric discharge through pure, cold and dry oxygen in a specially designed appartus called ozoniser. During this reaction, conversion of dioxygen to ozone is 10% and the product is called ozonised oxygen.
Since the formation of ozone from dioxygen is an exothermic process, therefore, it is necessary to use silent electric discharge in its preparation. A silent electric discharge produces less heat and thus prevent the decomposition of ozone back to oxygen.
Structure of Ozone
The central oxygen atom is sp2 hybridized containing a lone pair of electrons. As a result, ozone has angular structure with bond angle of 117º.
Question: Account for following : The two O—O bond lengths in the ozone molecule are equal.
Answer: Ozone may be regarded as resonance hybrid of the two resonating structures. Because of resonance , both the O—O bond lengths have partial double bond character. In other words, both the O—O bond lengths are equal.
Ozone is a pale blue gas having a strong characteristic smell.
In small concentration, it is harmless. However, if concentration rises above 100 ppm breathing becomes uncomfortable.
Ozone is diamagnetic.
It is about 1.67 times heavier than air.
Decomposition: Ozone is thermodynamically unstable as compared to oxygen since its decomposition into oxygen results in liberation of heat (ΔH is -ve) and an increase in entropy (ΔS is +ve). These two effect reinforce each other esulting in large negative gibb’s free energy for decomposition of ozone to oxygen.
2O3 (Ozone) → 3O2
Question: How is O3 estimated quantitatively?
Answer: When O3 is treated with excess of KI buffered with borate buffer, I2 is liberated quantitatively.
2I¯ +H2O +O3 → 2OH‾ + I2 + O2
The I2 thus liberated is treated against a standard solution of sodium thiosulphate using starch as an indicator and the amount of ozone can thus be calculated.