JEE Main 2022 · ChemistryNumerical answerNumerical valueMediumConceptual

JEE Main 26 June 2022, Shift 2, Chemistry Q89

Question 89 of 90 in this shift, Chemistry question 29 of 30, Section B.

Reaction of

[\mathrm{Co(H_2O)_6}]^{2+}

with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in

t_{2g}

-orbitals of the product is ________.

Official answer

NTA final key.

Chapter: Coordination Compounds
Topic: Crystal Field Theory
Idea tested: Octahedral Field Splitting

Same idea in other shifts

Asked 4× in all

5 Apr 2026, Shift 1 · Q62Given below are two statements: Statement I: Each electron in $e_g$ orbitals destabilizes the orbitals by $+0.6\,\Delta_o$ and each…EasyTwo statements
15 Apr 2023, Shift 1 · Q89The homoleptic and octahedral complex of $\mathrm{Co^{2+}}$ and $\mathrm{H_2O}$ has __________ unpaired electron(s) in the…MediumNumerical value
25 Jun 2022, Shift 1 · Q89If $[\mathrm{Cu(H_2O)_4}]^{2+}$ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.