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.
With which neutral molecule is ClO– isoelectronic? Is this molecule Lewis acid or base ? (Pb. Board 2009)
Which of the following acid forms three series of salts?
(a) H2PO2 (b) H3BO3 (C)H3PO4(d)H3PO3
Why does nitrogen show catenation properties less than phosphorus ? (C.B.S.E. Foreign 2009)
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) Among halogens, radius ratio between iodine and fluorine is maximum.
(b) Leaving F – F bond, all halogens have weaker X – X bond than X – X’ bond in interhalogens.
(c) Among interhalogen compounds maximum number of atoms ate present in iodine fluoride.
(d) Interhalogen compounds are more reactive than halogen compounds.
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.
Nitric acid forms an oxide of nitrogen on reaction with P4O10. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed.
Assertion (A): HI cannot be prepared by the reaction of KI with concentrated H2SO4.
Reason (R): HI has lowest H – X bond strength among halogen acids.
Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15 elements? (C.B.S.E. 2013)
What happens when sulp’hur dioxide is passed through an aqueous solution of Fe(III) salt?
Comment on the nature of two S-O bonds formed in S02 molecule. Are the two S-O bonds in this molecule equal ?
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.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions . involved.
Which of the following statements are correct?
(a) All three N – O bond lengths in HNO3 are equal.
(b) All P – Cl bond lengths in PCl5 molecule in gaseous state are equal.
(c) P4 molecule in white phosphorus have angular strain therefore white phosphorus is very reactive.
(d) PCl5 is ionic in solid state in which cation is tetrahedral and anion is octahedral.
Write a balanced chemical equation, for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen.
Explain why does the stability of oxoacids of chlorine increase in the order given below:
HClO < HClO2 < HClO3 < HClO4
P4O6 reacts with water according to equation P4O6 + 6H2O Calculate the volume of 0.1 M NaOH solution required to neutralize the acid formed by dissolving 1.1 g of P4O6 in H2O.
Assertion (A): HNO3 makes from passive.
Reason (R): HNO3 forms a protective layer of ferric nitrate on the surface of iron.
Assertion (A): SF6 cannot be hydrolysed but SF4 can be.
Reason (R): Six F atoms in SF6 prevent the attack of H2O on sulphur atom of SF6.
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–?
Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
