Cr and Cu exceptions plus which electrons leave first in Fe³⁺ — 4s fills first but also leaves first, kaise?

6 answers & replies

• [+][–]Ishita GoyalDropperSample· 5d ago

  Here's the table you wanted 😄

  Species	Configuration	Unpaired e⁻
  Cr	[Ar] 3d⁵ 4s¹	6
  Cu	[Ar] 3d¹⁰ 4s¹	1
  Fe	[Ar] 3d⁶ 4s²	4
  Fe³⁺	[Ar] 3d⁵	5
  Cr³⁺	[Ar] 3d³	3

  For Cr: all five d orbitals singly occupied plus the 4s one. Six parallel spins, that's the maximum. For Cu³⁺... skip that, not in NCERT-level questions.

  Report
• [+][–]Farhan ShaikhDropperSample· 5d ago

  For Fe³⁺, three electrons are removed, and since 3d is the last filled you lose from 3d first, so Fe³⁺ = 3d³ 4s². Cr³⁺ would be similar. Just remove from the orbital filled last, like a stack of plates where the last one on top comes off first. That's my budget trick.

  Report

  • [+][–]Lakshmi TripathiDropperSample· 5d ago

    Farhan, that gives the wrong answer. Rule: when a transition element becomes a cation, the electrons are removed from the outermost shell (highest n) first, which is 4s, then from 3d. So Fe ($3d^64s^2$) loses the two 4s electrons to give Fe²⁺ = $3d^6$, and the third comes from 3d giving Fe³⁺ = $3d^5$. Five unpaired electrons, a half-filled d subshell, which is exactly why Fe³⁺ is more stable than Fe²⁺.

    Report

    • [+][–]Yash Tiwari11thSample· 5d agoAuthor

      Ohh, 'highest n leaves first' — bas ye line chahiye thi. Thanks Lakshmi, thanks Ishita for the table. Sab note kar liya!

      Report
• [+][–]Mehak AroraDropperSample· 5d ago

  Why half filled and full filled are stable (the part sir skipped): in a half-filled or fully-filled subshell the electrons are symmetrically distributed, which gives extra stability, and the electrons with parallel spins in the same subshell can exchange positions, giving exchange energy that lowers the energy. The more exchanges possible (like in $d^5$ with five parallel spins), the larger the stabilisation. For Cr and Cu, 4s and 3d energies are so close that moving one electron from 4s to 3d is paid for by this extra stability.

  Report
• [+][–]Aman YadavDropperSample· 5d ago

  Exam tip. Fill by $(n + l)$: lower value first. If equal, lower $n$ first. 4s: 4 + 0 = 4. 3d: 3 + 2 = 5. So 4s first.

  But for ions: lose electrons from highest $n$ first. 4s before 3d. This is a rule, memorise it. Also: Cu⁺ = $3d^{10}$, Cu²⁺ = $3d^9$ (1 unpaired, coloured). Zn²⁺ = $3d^{10}$ (colourless).

  Report