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.
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.
Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
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).
Comment on the nature of two S-O bonds formed in S02 molecule. Are the two S-O bonds in this molecule equal ?
Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Arrange the following in the order of property indicated for each set: –
(i) F2 , Cl2 , Br2 , I2 – increasing bond dissociation enthalpy.
(ii) HF, HCI, HBr, HI – increasing acid . strength.
(iii) NH3, PH3, AsH3, SbH3, BiH3 – increasing Sol. base strength.
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).
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.
Write a balanced chemical equation, for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen.
White phosphorus reacts with chlorine and the product hydrolysis in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.
What happens when sulp’hur dioxide is passed through an aqueous solution of Fe(III) salt?
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions . involved.
With which neutral molecule is ClO– isoelectronic? Is this molecule Lewis acid or base ? (Pb. Board 2009)
What are the oxidation states of phosphorus in the following: –
(i) H3PO3 (ii)PCl3
(iii) Ca3P2(iv)Na3PO4
(v) POF3
Which of the following orders are correct as per the properties mentioned against each?
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.
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.