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Arrangethe elementsN, P , O and S in theorder of - (i) increasingfirstionisationenthalpy. (ii)increasingnonmetalliccharacter .Givereason for thearrangementassigned . |
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Answer» Solution :Arrangetheelements N, P, Oand S intodifferentgroupsand periodsin order of theirincreasingatomicnumbers we have `{:("Group " to ,,15,,16),("Period " 2,,N,,O),("Period " 3 ,,P ,,S):}` (i) theelectronicconfigurationof `N (1 s^(2) 2s^(2) 2p_(y)^(1) 2p_(y)^(1) 2p_(z)^(1))`in which2p-orbitals areexactly half - filledismorestablethan theelectronicconfigurationof `O (1s^(2)2s^(2)2p_(x)^(2) 2p_(y)^(1) 2p_(z)^(1))` in which2p- ORBITALSARE neither half- fillednorcompletelyfilled.Thereforeit is difficult toremove anelectron from N andthan from Oin spites ofthe fact theO (+ 8)has highest nuclearcharge than `N ( +7)` . As aresult `Delta_(i) H_(1)` ofN is higherthant thatofO. Similarlythe electronicconfigurationof `P (1s^(2)2s^(2)2p^(6)3S^(2)3p_(x)^(1) 3p_(y)^(1) 3p_(z)^(1))` in which3p- orbirtals are exactlyhalf- filledis morestablethan theelectronicconfigurationof `S (1s^(2) 2s^(2)2p^(6) 3s^(2)3p_(x)^(2)3p_(y)^(1)2p_(z)^(1))` in which 3p-orbitals are neitherhalf - fillednorcompletelyfilled.Thereforeis is difficultto removeon electronfromPthan fromS inspiteof the factthat S (+ 16)has highernuclearchargethan P (+ 15). Thus `Delta_(i)H_(1)` ofP ishigherthan thatS. Furthersince `Delta_(i)H_(1)` decreasesdown agrouptherefore`Delta_(i)H_(i)` ofN isgreaterthan thatof Pand thatof O isgreaterthan thatof S. Combiningthe tworesultstogetherthe firstionizationenthalpies of N, P, Oandincreases in the order `: S ltP lt O lt N` (ii) Sincenon metalliccharacterincreasesalonga periodand decreasesdown AGROUP, thereforenonmetalliccharacterincreases in THEORDER`: Plt Slt N lt O` |
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