9.e Magnetism and Molecular Orbitals Key PDF

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Magnetic Properties of Ions Substances that are considered paramagnetic are weakly attracted to a magnetic field. To be paramagnetic, a substance must have unpaired electrons. On the other hand, diamagnetic substances have all of their electrons paired and are weakly repelled by a magnetic field. Example 1: Determine if each of the following is paramagnetic or diamagnetic: Y, Y2+, Y3+ Plan: To determine if the atom/ion is para- or diamagnetic, we need to consider the electron configuration and determine if tall the electrons are paired or if there are any unpaired. If all sublevels are filled, all electrons will be paired and it will be diamagnetic. If any sublevel is partially filled, it will contain unpaired electrons and the substance will be paramagnetic. Solution: Y = [Kr] 5s2 4d1

paramagnetic

Y2+ = [Kr]4d1

paramagnetic

Y3+ = [Kr]

all core electrons are paired so diamagnetic

1) For each of the following write the condensed electron configuration and determine if it is para- or dia-magnetic. a. Pb4+ [Xe]4f145d10 All orbitals with electrons are filled, so all paired up and diamagnetic b. Zr [Kr]5s24d2 The electrons in 4d are unpaired so paramagnetic c. Zr4+ [Kr] The electrons core electrons are all paired so diamagnetic d. Co [Ar]4s23d7 - three unpaired in 3d so paramagnetic (actually cobalt is ferromagnetic) e. Co2+ [Ar] 3d7 - still three unpaired in 3d so paramagentic f. Co3+ [Ar] 3d6 - still has unpaired in 3d so paramagnetic

MODi a gr a ms Mol e c u l a ro r bi t a l sa r ec r e a t e df r o mt hec o mb i na t i o nofa t omi co r b i t a l so ns e pa r a t ea t o ms .El e c t r on s fil l i n gt heo r bi t a l sf o l l o wt h es a mer ul e sa swh e nfil l i n ga t omi co r b i t a l s . Ex a mp l e2:Po pul a t et hemo l e c ul a ror b i t a ld i a gr a mf o rF2-,d e t e r mi nei fi ti sp a r a -o rdi a ma gne t i c ,a n d d e t e r mi n et h ebo ndo r d e r Pl a n–Co un tupt hev a l e n c ee l e c t r o nsf o rt h es ub s t a n c e ,a ddt h e mt ot hec e n t e rs t a c ko ft hed i a gr a ma s t h i sr e pr e s e n t st heb on di n gor b i t a l s ,i ft he r ea r eu n pa i r e de l e c t r on si ti sp a r a ma g ne t i c ,d e t e r mi n eb on d o r d e rb ys u bt r a c t i n ga n t i bo nd i n gf r om b on di n ge l e c t r onsa n ddi v i d eb y2 . So l u t i o n: Tot a lv a l e n c ee l e c t r ons=15

MOd i a gr a m wi t h1 5e l e c t r ons :

Pa r a ma g ne t i cs i nc et h e r ei sa nu n p a i r e de l e c t

Bo ndo r d e r=( 8–7 ) / 2=0 . 5

F

F–F

F

2 ) Fo re a c hoft hef o l l o wi n g ,fil li nt h eMOd i a gr a m wi t ht hee l e c t r onsa n dd e t e r mi net h ebo ndo r de r a n dma gne t i cpr o pe r t i e so ft h es ub s t a n c e . a . Co mp l e t et heMOd i a gr a mf o rO2 i .Bo ndo r d e r=_ __2 ___ ___ i i .Par a ma g ne t i co rd i a ma g ne t i c ?

b . Co mp l e t et heMOd i a gr a mf o rO22i .Bo ndo r d e r=_ __1 ___ ___ i i .Pa r a ma g ne t i cordi a ma g ne t i c ?

3 ) Wh e nt woa t omst ha ta r en oti de nt i c a lb ond ,t he i ra t o mi co r bi t a l sa r eno ta tt hes a mee ne r g ya nda MOd i a gr a ms u c ha st heo neb e l o wf o rNOr e s u l t .Ev e nt ho u ght h ea t omi co r bi t a l sa r en ota tt h e s a mee n e r gy ,t hes a mep r i n c i p l e sa pp l ywhe nc o mbi n i n ga t o mi co r b i t a l st oma k emol e c ul a r o r bi t a l sa n di nfil l i n gt hemo l e c ul a ro r bi t a l swi t he l e c t r o ns .

c . Fi l lt h eMOd i a g r a m wi t ht hee l e c t r onsi nNO+. i .Bo ndo r d e r=_ ___ 3__ ___ i i .Pa r a ma g ne t i cordi a ma g ne t i c ?

d . Fi l lt h eMOd i a g r a m wi t ht hee l e c t r onsi nNO-. i .Bo ndo r d e r=_ ___ 2__ ___ i i .Par a ma g ne t i co rd i a ma g ne t i c ?

e . Fi l lt h eMOd i a g r a m wi t ht hee l e c t r onsi nNO. i .Bo ndo r d e r=_ __2 . 5__ ___ _ i i .Par a ma g ne t i co rd i a ma g ne t i c ?...