(a) Arrange the following compounds in the increasing order of bond length of O-O bond O2,O2+(AsF6)-,KO2 Explain on the basis - Chemistry - States of Matter

Question

( a )   A r r a n g e   t h e   f o l l o w i n g
  c o m p o u n d s   i n   t h e   i n c r e a
sin g   o r d e r   o f   b o n d   l e n g t h
  o f   O - O   b o n d O 2 , O 2 + ( A s F 6 ) - ,
K O 2 E x p l a i n   o n   t h e   b a s i s  
o f   g r o u n d   s t a t e   e l e c t r o n i c
  c o n f i g u r a t i o n   o f   d i o x y g e n
  i n   t h e e   m o l e c u l e s ( b ) U sin g
  V S E P R   t h e o r y   d r a w   t h e
  m o l e c u l a r   s t r u c t u r e s   o f ( i
) X e F 4 ( i i ) P C l 5 ( i i i ) B r F 5

Vartika Jain · Accepted Answer

Dear Student,

(a) Greater the bond order of O in these molecules, lesser will be the bond length sinceBond length α 1Bond orderBond order=Bonding electrons - Antibonding electrons2So, O2 = σ1s2 σ2s2 σ*1s2 σ*2s2 σ2px2 π2py2π2pz2π*2py1π*2pz1Bond order=10-62=2In KO2, O2 exists as O2-O2-=σ1s2 σ2s2 σ*1s2 σ*2s2 σ2px2 π2py2π2pz2π*2py2π*2pz1Bond order=10-72=1
5O2+=σ1s2 σ2s2 σ*1s2 σ*2s2 σ2px2 π2py2π2pz2π*2py1π*2pz0Bond order=10-52=2
5So, Bond length is O2+ < O2 < O2-Kindly ask the remaining query in the next thread