Skip to content
Question of 132

Q.Fluorine exhibits only -1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Explain.

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2022Subjective· 2mImportance★★★★★
0% · 0/132 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Fluorine lacks accessible d-orbitals and is the most electronegative element of all, so it is restricted to the -1 state; the other halogens have valence-shell d-orbitals available and can expand their octet to show positive oxidation states.

Fluorine is the smallest and most electronegative element in the periodic table. Its valence shell is the second shell (n = 2), which has only 2s and 2p orbitals available - there are no 2d orbitals. Because it has no low-lying, energetically accessible d-orbitals, fluorine cannot expand its octet beyond eight electrons, so it can never use more than one unpaired electron for bonding, and being the most electronegative element it never loses electrons to (or shares electrons unequally in favour of) any other element. Consequently fluorine can only gain (or share, pulling density towards itself) one electron, showing exclusively the -1 oxidation state in all its compounds.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.