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.
Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
In the preparation of H2S06 by contact process, why is S03 not absorbed directly in water to form H2S04?
Explain why does the stability of oxoacids of chlorine increase in the order given below:
HClO < HClO2 < HClO3 < HClO4
PCl5 reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH3 solution. Write the reactions involved to explain what happens.
On heating ammonium dichromate and barium azide separately we get
(a) N2 in both cases
(b) N2 with ammonium dichromate and NO with barium azide
(c) N2O with ammonium dichromate and N2 with barium azide
(d) N2O with ammonium dichromate and N2O with barium azide
Which of the following statements are correct?
(a) S – S bond is present in H2S2O6.
(b) In peroxosulphuric acid (H2SO5) sulphur is in +6 oxidation state.
(c) Iron powder along with Al2O3 and K2O is used as a catalyst in the preparation of NH3 by Haber's process
(d) Change in enthalpy is positive for the preparation of SO3 by catalytic oxidation of SO2.
Write a balanced chemical equation, for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen.
Phosphorus forms a number of oxoacids. Out of these oxoacids phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
Assertion (A): HNO3 makes from passive.
Reason (R): HNO3 forms a protective layer of ferric nitrate on the surface of iron.
Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15 elements? (C.B.S.E. 2013)
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions . involved.
Why does nitrogen show catenation properties less than phosphorus ? (C.B.S.E. Foreign 2009)
Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions . involved.
The HNH angle value is higher than HPH, H AsH and HSbH angles. Why?
(Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only s-p bonding , between hydrogen and other elements of the group).
