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.
Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15 elements? (C.B.S.E. 2013)
Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling is diamagnetic?
Why does nitrogen show catenation properties less than phosphorus ? (C.B.S.E. Foreign 2009)
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).
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.
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.
In the ring test of NO3 ion.Fe2+ion reduces nitrate ion to nitric oxide, which combines with Fe2+ (aq.) ion to form brown complex. Write the reactions involved in the formation of brown ring.
Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.
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.
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.
What are the oxidation states of phosphorus in the following: –
(i) H3PO3 (ii)PCl3
(iii) Ca3P2(iv)Na3PO4
(v) POF3
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.
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 [he formation of a large number of oxides having O2- species and not O–?
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.
On addition of cone. H2SO4 to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because
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
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.
