Justify the placement of O, S, Se, Te and Po in the same group’of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
(1)Electronic configuration:
O (At. no. = 8) = [He] 2s2 2p4
S (At. no. = 16) = [Ne] 3s2 3p4
Se (At. no. = 34) = [Ar] 3d10 4s2 4p4
Te (At. no. = 52) = [Kr] 4d10 5s2 5p4 ,
Po (At. no. = 84) = [Xe] 4f14 5d10 6s2 6p4 ,
Thus, all these elements have the same ns2 np4 (n = 2 to 6) valence shell electronic configuration, hence are justified to be placed in group 16 of the Periodic Table.
(2)Oxidation state : Two more electrons are needed to acquire the nearest noble gas configuration. Thus, the minimum oxidation state of these elements should be – 2. O and to some extent S show – 2 oxidation state. Other element being more electropositive than O and S, do not show negative oxidation state. As these contain six electrons, thus, maximum oxidation state shown by them is+ 6. Other oxidation state shown by them are + 2 and + 4. O do not show+4 and + 6 oxidation state, due to the absence of d-orbitals. Thus, on the basis of maximum and minimum oxidation states, these elements are justified to be placed in the same group 16 of the periodic table.
(3)Hydride formation: All these elements share two of their valence electrons with 1 s- orbital of hydrogen to form hydrides of the general formula EH2, i.e., H20, H2S, H2Se, H2Te and H2Po. Thus, on the basis of hydride formation, these elements are justified to be placed in the same group 16 of the Periodic Table.
Assertion (A): Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as OÏ€.
Reason (R): Oxygen forms pπ-pπ multiple bond due to small size and small bond length but pπ-pπ bonding is not possible in sulphur.
In the preparation of H2S06 by contact process, why is S03 not absorbed directly in water to form H2S04?
Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling is diamagnetic?
Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Assertion (A): NaCl reacts with concentrated H2SO4 to give colourless fumes with pungent smell. But on adding MnO2 the fumes become greenish yellow.
Reason (R): MnO2 oxidises HCl to chlorine gas which is greenish yellow.
Write balanced equations for the following:
(i) NaCl is heated witlrsulphuric acid in the presence of MnO2
(ii) Chlorine gas is passed into a solution of Nal in water.
Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?
Which of the following is correct for P4 molecule of white phosphorus?
(a) It has 6 lone pairs of electrons (b) It has six P – P single bonds
(c) It has three P – P single bonds (d) It has four lone pairs of electrons,
Which of the following statements are true?
(a) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(b) Ionisation enthalpy of.molecular oxygen is very close to that of xenon.
(c) Hydrolysis of XeF6 is a redox reaction.
(d) Xenon fluorides are not reactive.
Write a balanced chemical equation, for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen.
Phosphorus has three allotropic fonns —(i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity.
Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.
Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15 elements? (C.B.S.E. 2013)
Comment on the nature of two S-O bonds formed in S02 molecule. Are the two S-O bonds in this molecule equal ?
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising powers of F2 and Cl2.
Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Knowing the electron gain enthalpy values of O—>O– and O—>O2- as -141 and 702 kJ mol-1 respectively, how can you account for the formation of a large number of oxides having O2- species and not O–?
Give the formula and describe the structure of a noble gas species which is isostructural with: (i) ICI4– (ii) IBr2– (iii) Br03–
Comment on the nature of two S-O bonds formed in S02 molecule. Are the two S-O bonds in this molecule equal ?
Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
