Pag of 6 Fall 0: 5. PRINCIPLES OF CHEMICAL SCIENCE Problm St #4 solutions This problm st was gradd out of 99 points Qustion (out of 7 points) (i) For ach pair, dtrmin which compound has bonds with gratr ionic charactr. ( points) (a) NH or AsH : Th N-H bond is mor ionic than th As-H bond sinc th diffrnc in lctrongativity of N and H (.0 vs..) is gratr than th diffrnc btwn As and H (. vs..). (b) SiO or SO : S and O hav similar lctrongativitis (.0 and.4) whras for Si and O th diffrnc is gratr (.9 vs..4) so th Si-O bonds hav gratr ionic charactr. (c) SF 6 or IF 5 : S and I hav vry similar lctrongativitis (.6 and.7), thus th bonds ar vry similar. S-F bonds may hav a slightly highr ionic charactr, but it would b difficult to dtrmin xprimntally. (ii) In which molcul, OCO or OCCH, do you xpct th C O bond to b mor ionic? ( points) Th C O bond in OCO will hav gratr ionic charactr bcaus O is mor lctrongativ than th carbon in th CH fragmnt, so in absolut valus, th C in CO has twic as much charg rmovd as th C attachd to on O and th CH fragmnt. (iii) In which molcul do you xpct th bonds to b mor ionic, CCl 4 or SiCl 4? ( points) Th diffrnc in lctrongativity btwn C and Cl is lss than th diffrnc btwn Si and Cl, so th bonds in SiCl 4 will b hav gratr ionic charactr. Qustion (out of points) Th gasous potassium chlorid molcul has a dipol momnt of 0. D, which indicats that it is a vry polar molcul. Th sparation btwn th nucli in this molcul is.67 Å. Calculat th dipol momnt of a KCl molcul if thr wr opposit chargs of on fundamntal unit (.60 X 0-9 C) at th nucli? W can calculat th dipol momnt of th molcul if it was compltly ionic from th following formula: (.60 x 0-9 C) (.67 x 0-0 m) QR µ.6 x 0-0 C m/d.6 x 0-0 C m/d Th dipol momnt of KCl is.8 D if it wr 00% ionic in charactr..8 D Qustion (out of 6 points) Considr two diatomic molculs with 00% ionic charactr. On has a bond lngth of.75 Angstroms, th othr a bond lngth of.5 Angstroms. (a) Calculat th dipol momnt for ach molcul. ( points) Th quation for dipol momnt is: µ QR, whr Q th charg and R is th distanc sparating th two ions, if w want to hav th dipol momnt to b in dby, w nd to hav th convrsion fact of 0.08 A D -. W can now go ahad and calculat th dipol momnts for th two molculs: For th.75 A 0 bond lngth: ( ).75 A 0 QR µ 0.08 C A 0 D 0.08 A 0 8.4 D D
Problm st #4 solutions for 5., Fall 0 Pag of 6 For th.5 A 0 bond lngth: ( ).5 A 0 QR µ 0.08 C A 0 D 0.08 A 0 5.6 D D (b) Calculat th Coulombic potntial nrgy for ach molcul. ( points) W nd to rmmbr th quation for Coulombic potntial: ΔE coulomb q q, whr q is th 4πε 0 r charg on th ion multiplid by th fundamntal charg (). Now w can calculat th Coulombic potntial nrgy for ach molcul: For th.75 A 0 bond lngth: ΔE coulomb q q ( 4πε 0 r.608 x 0-9 C) (.608 x 0-9 C) 4π 8.8549 x 0 - J - C m - For th.5 A 0 bond lngth:.89 x 0-8 J (.75 x 0-0 m). x 0-8 J ΔE coulomb q q 4πε 0 r.608 x 0-9 C 4π 8.8549 x 0 - J - C m - 7.09875 x 0-9 J 7.0 x 0-9 J (.608 x 0-9 C) (.5 x 0-0 m) Qustion 4 (out of 4 points) Estimat th prcnt ionic charactr of th bond in ach of th following spcis. All th spcis ar unstabl or ractiv undr ordinary laboratory conditions, but ar obsrvd in intrstllar spac. Avrag Bond Lngth (Å) Dipol Momnt (D) (a) OH 0.980.66 (b) CH..46 (c) CN.75.45 (d) C.46 0 Th prcntag ionic charactr is calculatd by dtrmining th δ (fraction of unit charg on ach atom) and multiplying δ by 00. (a) OH 0.08A 0 D - δ [.66 D] 0.5 For OH, th prcntag ionic charactr is 5. %. 0.980 A 0 (b) CH 0.08A 0 D - δ [. A 0.46 D] 0.69 For CH, th prcntag ionic charactr is 6.9 %. (c) CN 0.08A 0 D - δ [.45 D] 0.57.75 A 0 For CN, prcntag ionic charactr is 5.7 % (d) C For C th prcntag ionic charactr is 0 %.
Problm st #4 solutions for 5., Fall 0 Pag of 6 Qustion 5 (out of 6 points) Considr KF, which has an ionic bond. Th bond lngth is.7 x 0-0 m or 7 pm. (a) Calculat th nrgy (in kj/mol) rquird to dissociat th KF molcul into th ions K + and F -. This nrgy is th Coulomb nrgy of intraction btwn th ions of K + and F - : ΔEcoulombic Q Q 4πε 0 r ( 6.0 x0 mol - ) (.7 x 0-0 m).60 x 0-9 C 4π 8.854 x 0 C J m 640 J mol - 6.40 x 0 kj mol - Th nrgy rquird to dissociat KF into its ions is 6.40 x 0 kj mol -. Th nrgy is positiv bcaus w ar going from molcul to ions (up on th nrgy scal). (b) Th nrgy rquird to dissociat KF into nutral atoms is 498 kj/mol. Givn that th first ionization nrgy for K is 49 kj/mol, calculat th lctron affinity (in kj/mol) for F. W can calculat th lctron affinity for F by using th following quation: ΔEd Q Q 4πε 0 R [ IE (K)- EA(F) ] W ar givn th ΔE d (498 kj mol - ). W calculatd th coulomb intraction nrgy in part (a) (640 kj mol - ) and w ar givn IE for potassium (49 kj mol - ), so w can calculat th lctron affinity for F by substitution into th abov quation: ΔEd Q Q 4πε 0 r [ IE (K)-EA(F) ] 498 kj mol - 6.40 x 0 kj mol - - 49 kj mol - -EA(F) EA(F)498 kj mol - 6.40 x 0 kj mol - + 49 kj mol - 77 kj mol - Th lctron affinity for F is 77 kj mol -. (Th lctron affinity is actually 8 kj mol - ). Sinc this mthod is an approximation, th agrmnt is rasonabl. Qustion 6 (out of 6 points) Considr th SrO molcul that has an ionic bond lngth of.54 x 0-0 m. a) Calculat th nrgy in kj mol - rquird to dissociat SrO into its ions Sr + and O -. Trat th ions as point chargs. Th nrgy th qustion askd for is th ΔE coulomb, which can b calculatd from th following quation: ( ΔE coloumb 4πε 0 r.6076 x 0-9 C) ( 6.04 x 0 mol - ) 4π 8.8549 x 0 - C J - m - 6.8456 x 0-4 C mol -.748 x 0-0 C J -.6087 x 0 6 J mol - (.54 x 0-0 m).6 x 0 kj mol - Th nrgy rquird to dissociat SrO into its ions Sr + and O - is.6 x 0 kj mol -.
Problm st #4 solutions for 5., Fall 0 Pag 4 of 6 b) Writ 5 ractions which whn summd will rsult in th raction: SrO Sr + O. Four of ths ractions ar rlatd to th nrgis for th procsss givn in th tabl blow. IE of Sr 549 kj/mol IE of Sr 064 kj/mol EA of O 4 kj/mol EA of O - -844 kj/mol Thn, calculat th bond dissociation nrgy of CaO into its nutral atoms, Ca and O. Fiv possibl ractions ar: Raction ΔE Sr + + - Sr ΔE A -(IE of Sr) -549 kj mol - Sr + + - Sr + ΔE B -(IE of Sr) -064 kj mol - O - O + - ΔE c +4 kj mol - O - O - + - ΔE D -844 kj mol - SrO Sr + + O - ΔE coulomb 60 kj mol - CaO Ca + O W can now calculat th nrgy rquird to dissociat into nutral atoms (which is calld th dissociation nrgy); ΔEΔEA + ΔEB + ΔEC + ΔED + ΔECoulomb + ( 064 kj mol - ) + ( 4 kj mol - ) + ( 844 kj mol - ) + ( 60 kj mol - ) 549 kj mol - 94 kj mol - Th dissociation nrgy is 94 kj mol -. Qustion 7 (out of points) Bcaus of its particl natur, light can xrt prssur on a surfac. Suppos a.5-w lasr is focusd on a circular ara 0.5 mm in radius. Th wavlngth of th lasr light is 569 nm. Assum that th photons travl prpndicular to th surfac and ar prfctly rflctd, and thir momnta chang sign aftr rflction from th wall. Using rsults from th kintic thory of gass, calculat th prssur xrtd by th light on th surfac. Th momntum p of a photon is rlatd to its nrgy through E pc. Th fall of photons on a surfac xrts a prssur. Us th dbrogli rlation to comput th momntum of th photons: p h λ 6.66 x 0-4 J s 569 x 0-9 m.645 x 0-7 kg m s - Each photon bouncs away from th wall with momntum of th sam magnitud but opposit sign. Th chang in momntum pr collision is: Δp p p.645 x 0-7 kg m s - (-.645 x 0-7 kg m s - ). x 0 7 kg m s - Th numbr of photons colliding pr scond quals th powr of th lasr (.5 watt.5 J s - ) dividd by th nrgy dlivrd pr photon. Th nrgy transportd by on 550 nm photon is: E hc -4 ( 6.66 x 0 J s ) (.9979 x 0 8 m s - ) λ 569 x 0-9.49 x 0-9 J m Clarly, it rquirs lots of such photons to transport.5 J s - :
Problm st #4 solutions for 5., Fall 0 Pag 5 of 6.5 J photon s x.49 x 0-9 J 4.0 x 08 photons scond As dvlopd in Sction 4-, th total forc quals th chang in momntum pr collision multiplid by th numbr of collisions pr scond: F. x 0-7 kg m s - x ( 4.0 x 0 8 s - ).00 x 0-8 kg m s - Prssur is dfind as forc dividd by ara. Th ara of th circular wall is πr whr r is th radius. Hnc: P F A F -8.00 x 0 kg m s - πr π 0.5 x 0 - m 0.45 Pa 0.4 Pa (corrct sig figs) So th prssur xrtd by th light on th surfac is 0. Pa. Qustion 8 (out of 6 points) In an optical atomic trap, bams of tightly focusd lasr light rplac th physical walls of convntional containrs. Thy brifly (for 0.500 s) xrt nough prssur to confin 695 sodium atoms in a volum of.0 X 0 5 m. Th tmpratur of this gas is 0.0004 K, th lowst tmpratur vr rachd for a gas. a) Comput th root-man-squar spd of th atoms in this confinmnt. Th root-man-squar spd of th molculs of a gas is RT M whr M is th molar mass of th gas. W substitut M 0.099 kg mol -, R 8.5 J mol - K -, and T 0.0004 K to comput RT M 0.50 m s - (.4 x 0 4 K) 8.5 J mol - K - 0.099 kg mol - 0.5 m s - 0.5 m s -. Common rrors hr ar to us M in g mol - instad of kg mol - or to us R in th wrong units. b) Assum idal gas bhavior to comput th prssur xrtd on th walls of th optical bottl in this xprimnt. Sinc w ar making th assumption that th Na atoms ar acting as an idal gas, w can us th quation PV nrt. W hav to first dtrmin V. Th volum of th trap is xprssd in litrs ( L 0 - m ) and w can convrt th volum of.0 x 0-5 m to L; Th amount of its contnts is xprssd in mols 695 atoms n 6.0 x 0 atoms mol.54 x - 0- mol Thn, th problm bcoms a substitution into PV nrt;
Problm st #4 solutions for 5., Fall 0 Pag 6 of 6 0.0806 L atm mol - K - (.4 0 4 K ) P.54 0 mol.0 0 L.7 0 4 atm Th atoms xrt a prssur of. x 0-4 atm. Qustion 9 (out of points) Equal numbrs of atoms of Hlium-4 and Xnon-9 xrt th sam prssur at constant volum and tmpratur, dspit th larg diffrnc in th masss. Basd on kintic thory, xplain what this tlls you about th vlocitis of th atoms. Sinc prssur is dfind as th forc xrtd on a unit ara of th walls of a containr by atoms of th gas, an avrag hlium atom must xrt th sam forc as an avrag xnon atom. According to th dfinition of an idal gas, ths collisions must b lastic. Sinc forc xrtd in an lastic collision by an impact is proportional to th momntum of th atom, hlium atoms must mov much fastr than xnon atoms to xrt th sam forc. Qustion 0 (out of 6 points) Th scap vlocity ncssary for objcts to lav th gravitational fild of th arth is. km/s. Calculat th ratio of th scap vlocity to th root-man-squar spd of hlium, argon, and xnon atoms at 00 K. Dos your rsult hlp xplain th low abundanc of th light gas hlium in th atmosphr? Explain. To calculat th rms spds of hlium, argon, and xnon atoms at 000 K ( significant figur!), w substitut th molar masss in kg mol - of th thr gass succssivly into th xprssion, taking T 000 K and R 8.5 J mol - K -. For H Ratio of For Ar Ratio of For X 00 K 8.5 J mol- K - 4.006 x 0 kg mol -.7 x 0 m s - (corrct sig figs) scap vlocity root-man squar vlocity.70 x 0 m s -. x 0 m s - 0.06 0. 00 K 8.5 J mol- K - 9.948 x 0 kg mol -. x 0 m s - (corrct sig figs) scap vlocity root-man squar vlocity. x 0 m s - 00. x 0 m s - 0.0465 0.0 (corrct sig figs).70 x 0 m s -.7 x 0 m s -
Problm st #4 solutions for 5., Fall 0 Pag 7 of 6 00 K 8.5 J mol- K -.9 x 0 kg mol - 6.5 x 0 m s - (corrct sig figs) Ratio of 646.5 m s - scap vlocity root-man squar vlocity 6.46 x 0 m s - 00. x 0 m s - 0.0577 0.058 (corrct sig figs) Th ratio of th scap vlocity to th root man squar spds ar 0. (for H), 0.0 (for Ar) and 0.058 (for X). Hlium is much mor likly to scap than th havir gass so this mans that it will not b found at high abundanc in our atmosphr. Qustion (out of points) A confind gas of Ar is hatd so that its prssur is tripld. If th volum rmains th sam, how much kintic nrgy dos th gas gain or los? Think about th idal gas law, PVnRT. If th prssur is doubld, volum stays th sam, w can dtrmin how th tmpratur has changd: PV P V T T know that P P and V V simplifying th rlationship: P T P T thrfor : T T So, th tmpratur has tripld. Bcaus E RT, th gas tripls its kintic nrgy Qustion (out of 4 points: for part (a), for part (b)) Six Rb atoms, with spds of.,.4,.6,.6,.0 and.5 m/s, ar trappd in a lasr trap. (a) Dtrmin th most probabl spd for this distribution of six atoms. Th most probabl spd is.6 m s -. Thr ar two atoms with that spd. (b) Calculat th root man squar spd for this distribution of six atoms. Th root man squar spd is: ( v ) (. m s - ) + (.4 m s - ) + (.6 m s - ) + (.0 m s - ) + (.5 m s - ).44 m s - +.074 m s - + 5. m s - + 4.0 m s - + 6.5 m s - 6 8.77 m s -.7687 m s -.8 m s - 6 Th root man squar spd for th six Rb atoms is.8 m s -. 6
Problm st #4 solutions for 5., Fall 0 Pag 8 of 6 (c) Calculat th tmpratur that corrsponds to your answr for (b). From our lcturs # nots, w know that: RT M For th quation: V rms that w calculatd in part (b).8 m s - R which is th univrsal gas constant 8.447 J K - mol - (Us this valu bcaus a Joul is a Nwton.mtr, and a Nwton m kg s - which onc you cancl units givs: RT M K m kg s - K mol - kg mol - m s M which is th molar mass in kg mol - 85.47 x 0 - kg mol - T is th unknown: Now w can solv for T: RT M.8 m s - (8.447 J K- mol - T 85.47 x 0 - kg.4 m s - (8.447 J K- mol - T 85.47 x 0 - kg T0.0 K Qustion (out of points: for ach spd) Considr N gas at T98 K. For vry million molculs travling at th root man squar spd, approximatly how many travl at 0. ; at. ; at ; at. This is a problm looking at th Maxwll-Boltzmann spd distribution function; ΔN N f ( v )Δv, whr f ( v ) 4π M ( v) xp Mv πrt RT So w can compar th functions for th diffrnt spds rmmbring that m is 0.80 x 0 - kg mol -, R is 8.5 J mol - K -, T is 98 K and v is going to b. W hav to first dtrmin vrms RT M (8.45 J mol- K - )(98 K) 0.80 x 0 - kg Substituting th first ratios; 606.9 m s -
Problm st #4 solutions for 5., Fall 0 Pag 9 of 6 (, 000, 000) rms vrms 0.80 x 0 - kg mol - ΔN0. vrms 0.vrms xp ( 8.5 J mol - K - 606.9 ) ( 98 K ) (, 000, 000) rms ΔN0. vrms, 000, 000 rms ΔN0. vrms (, 000, 000) rms.65 ΔN0. vrms 00 xp ( 4.0705 x 0-6 ).64656 x 0 5 ( 00) xp(.48490) ( 00) (.65) ( 60.69) ΔN0. vrms 44, 44 For vry,000,000 molculs travling at, 40 thousand molculs travl at 0.. W can now do th sam calculation for. : (, 000, 000) rms vrms ( 0.80 x 0 - kg mol - ) ΔN. vrms.vrms xp 8.5 J mol - K - (, 000, 000) rms ΔN. vrms (, 000, 000) rms ΔN. vrms 98 K 0.6944 xp ( 4.0705 x 0-6 ).607 x 0 5 606.9 0.6944 xp ( 0.659955) ( 0.6944 ) (.9470).446 ΔN. vrms 7.4445 x 0 5 ( 78.9) 7.4 x 0 5 For vry,000,000 molculs travling at, 7.4 x 0 5 molculs travl at.. W can now do th sam calculation for : (, 000, 000) rms vrms ( 0.80 x 0 - kg mol - ) ΔNvrms vrms xp 8.5 J mol - K - (, 000, 000) rms ΔNvrms (, 000, 000) rms ΔNvrms ΔNvrms 44, 450 98 K 0.5 xp ( 4.0705 x 0-6 ).050 x 0 6 ( 0.5) xp ( 4.49969) ( 0.5) (89.9896).49 606.9 (.8) For vry million molculs travling at, thr ar 40 thousand molculs travling at twic th.
Problm st #4 solutions for 5., Fall 0 Pag 0 of 6 W can now do th sam calculation for : (, 000, 000) rms ΔNvrms (, 000, 000) rms ΔNvrms (, 000, 000) rms ΔNvrms vrms vrms xp ( 0.80 x 0 - kg mol - ) 8.5 J mol - K - 98 K 0.xp ( 4.0705 x 0-6 ).9467 x 0 6 0. xp (.999) ( 0.) (66) 8067. 606.9 ( 80.7) ΔNvrms 55. For vry million molculs travling at, thr ar 55 molculs travling at thr tims th. Hr is anothr way to st up th problm using symbols:
Problm st #4 solutions for 5., Fall 0 Pag of 6 Qustion 4 out of 6 points) (a) Th following graph shows th rang of molcular spds for thr gass, H, N and Ar, as givn by th Maxwll distribution. Each of th curvs (, and ) corrspond to 5 K. Match th curvs with th gass, and provid a brif xplanation. Curv, and corrspond to Ar, N and H, rspctivly, bcaus according to th Maxwll distribution of spds, th gratr th molar mass, th lowr th avrag spd and th narrowr th sprad of spds. (b) Th following graph shows th rang of molcular spds for Gas in th abov qustion, at 5 K, 5 K and 45 K as givn by th Maxwll distribution. Match th curvs with th tmpraturs, and provid a brif xplanation. Curv 4, 5 and 6 corrspond to 5 K, 5 K and 45 K, rspctivly, bcaus according to th Maxwll distribution of spds, th highr th tmpratur, th highr th avrag spd and th broadr th sprad of spds. Qustion 5 (out of points) Th numbr of molculs in a gas sampl that hav th most probably spd (v mp ) at a tmpratur T is on-quartr th numbr of th sam typ of molculs that hav th most probably spd at 55 K. What is th tmpratur? Using th Maxwll distribution of spds
Problm st #4 solutions for 5., Fall 0 Pag of 6 M f (v) 4π π RT v Mv Mv RT df (v) dv M Mv RT π RT 8πv + 4πv RT 0 Solving for ν mp in trms of T: v mp RT M Substitution into f (v) givs: M RT f (v mp ) 4π π RT M M 4π π RT Th problm asks for th tmpratur that satisfis this rlationship: f (v mp ) T 4 f (v mp ) 55 M 4π π RT 4 4π M πr(55 K) T 4 55 K T 6 55 K T 6(55 K) 5680 K Th scond tmpratur is 5680 K Qustion 6 (out of 6 points) Considr mthan (CH 4 ) gas at a tmpratur T. (a) Calculat th ratio, to significant figurs, of th numbr of mthan molculs that hav th avrag spd v to th numbr of molculs that hav twic th avrag spd, v. Lt s first considr th Maxwll-Boltzmann spd distribution; ΔN N f (u)δu, whr f (u) 4π m πk B T u xp mu k B T, W can compar th fraction of total numbr of molculs travling at th avrag spd v and compar that to th fraction of total numbr of molculs travling at v xp M ( v ) f v RT ( v ) ( v) f v RT M v xp
Problm st #4 solutions for 5., Fall 0 Pag of 6 v RT( ( v ) 4( v) ) ( v) xp M M v xp 4 RT 4 xp M RT xp 4 π.5000 8RT πm ( 45.59).978.4 Th ratio of th numbr of molculs with spd v to that of v is.4. (b) Calculat th ratio, to significant figurs, of th numbr of mthan molculs that hav th avrag nrgy E to th numbr of molculs that hav twic th avrag nrgy, E. W nd to look at th nrgy distribution quation: E f (E) π E RT πrt W can substitut into th quation and look at th ratios: f (E) f (E) π πrt π πrt E E E E RT ( E) E RT + E RT E RT RT 0.707 RT E RT ( 4.4869).690.7
Problm st #4 solutions for 5., Fall 0 Pag 4 of 6 Th ratio of mthan molculs that hav th avrag nrgy E to th numbr of molculs that hav twic th avrag nrgy, E is.7. Qustion 7 (out of 4 points) Th following figur is a plot of th potntial nrgy vrsus sparation distanc for two nutral nobl-gas atoms: Lnnard-Jons potntial for th H-H (black) and Ar-Ar dimrs. Commonly rfrrd to as th Lnnard-Jons potntial, th quation that dscribs ths functions is: whr ε and σ ar constants that dpnd on th systm of intrst. Th L-J potntial is a usful modl bcaus it capturs th ssntial physics of intraction at long and short sparation: attraction btwn th atoms at long r and rpulsion at short. a) Dtrmin which trms in V(r) corrspond to attraction and rpulsion. ( points) σ Lt us look at th quation from th qustion: V (r) 4ε r σ 6 r Th attractiv intraction is going to stabiliz th formation of th molcul, so it has th σ ngativ sign in front of it. Th attractiv intraction trm is thrfor: r Th rpulsiv intraction is going to dstabiliz th dimr, so it will hav th positiv sign σ in front of it. Th rpulsiv intraction is thrfor: r. b) What dos th dpth of th potntial nrgy wll corrspond to (indicatd by th black arrow)? ( point) Th dpth of th potntial nrgy wll,( ε ) is th binding nrgy btwn two nutral nobl atoms (or th dissociation nrgy of th dimr) 6
Problm st #4 solutions for 5., Fall 0 Pag 5 of 6 Th potntial nrgy minimum for th hlium dimr is at an intrmolcular distanc of 0.87 nm, whil th minimum for argon is 0.80 nm. c) Why dos argon hav a gratr sparation distanc at th minimum than hlium? ( point) Argon has a largr atomic radium (lowr in th group on th priodic tabl), thrfor (in hard sphr modl) unabl to com closr to ach othr. d) (Bonus) Was Lnnard-Jons on prson or two? Answr ONE! Qustion 8 (out of 4 points) Nam th typs of attractiv forcs that will contribut to th intractions among atoms, molculs, or ions in th following substancs. Indicat th on(s) you xpct to prdominat. (a) KF Potassium and fluorin diffrnt considrably in lctrongativity. Th bonding in th compound potassium fluorid is xpctd to b ionic. Ths ion-ion attractions prdominat in this compound; disprsion forcs ar also prsnt. (b) Cl (c) HBr (d) Kr Disprsion forcs ar th only intrmolcular forcs possibl btwn molculs of Cl. Dipol-Dipol attractions prdominat in th intractions btwn molculs in hydrogn bromid. Th positiv (H) nd of on molcul is attractd by th ngativ (Br) nd of anothr, but rplld by its positiv (H) nd. Disprsion forcs ar also prsnt. Disprsion forcs ar th only forcs oprating among th atoms in a sampl of Krypton. Singl Kr atoms hav compltly symmtrical (sphrically symmtrical) distributions of charg. Two nighboring Kr atoms induc tmporary dipols in ach othr that caus thm to attract ach othr. Qustion 9 (out of points) Prdict whthr a bromid ion will b most strongly attractd to a sodium ion, a molcul of hydrogn chlorid, or an atom of krypton. A bromid ion should b most strongly attractd to a sodium ion. Th attraction btwn ions of unlik charg such as ths is strongr than th ion-dipol attraction btwn Br - and HCl and th ion-inducd dipol attraction btwn Br - and Kr. Qustion 0 (out of points) Undr room conditions, fluorin and chlorin ar gass, bromin is a liquid, and iodin is a solid. Explain th origin of this trnd in th physical stat of th halogns. Th boiling point and mlting point of a substanc dpnd on th strngth of th attractiv forcs opration amount th particls that compris th liquid or solid substanc. Ths forcs tnd to incras with incrasing molar mass in a group of rlatd substancs. Th halogns ar crtainly closly rlatd chmically. Thrfor thir boiling and mlting points tnd to ris with incrasing molar mass. Qustion (out of 4 points) List th following substancs in ordr of incrasing normal boiling points T b, and xplain your rasoning: NO, NH, N, RbCl. Substancs with th strongst intrmolcular forcs rquir th highst tmpratur to mak thm boil. Liquid RbCl has strong Coulomb (lctrostatic) forcs holding its ions togthr. It has th highst boiling point. Liquid NH has dipol-dipol attractions, as dos liquid NO. In liquid NH, ths ar particularly strong. Thy ar hydrogn bonds. In liquid NO, th dipol-dipol
Problm st #4 solutions for 5., Fall 0 Pag 6 of 6 attractions ar wakr. Liquid NH boils at a highr tmpratur than liquid NO. Inducd dipolinducd dipol forcs ar th only intrmolcular attractions in liquid non. Consquntly, it has th lowst boiling point of all. Th ordr of th molculs in incrasing boiling points ar: N < NO < NH < RbCl.