JEE Main 2006 — D And F Block Elements Question with Solution
From: AIEEE 2006
Question
The “spin-only” magnetic moment [in units of Bohr magneton, (µB)] of Ni2+ in aqueous solution would
be : (Atomic number of Ni = 28)
Choose an option
Show full solutionCorrect option: A
Correct answer
A2.84
Step-by-step explanation
The number of unpaired electrons in
Water is weak ligand hence no pairing will take place spin magnetic moment
\eqalign{ & = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( {2 + 2} \right)} \cr & = \sqrt 8 = 2.82 \cr}
Water is weak ligand hence no pairing will take place spin magnetic moment
\eqalign{ & = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( {2 + 2} \right)} \cr & = \sqrt 8 = 2.82 \cr}
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This is a previous-year question from JEE Main 2006, covering the D And F Block Elements chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.