CBSE 2026 · Region 2 · Set 3 · Q25 · 3 marks
(a)Explain why $\displaystyle \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ is strongly paramagnetic whereas $\displaystyle \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ is weakly paramagnetic. [Atomic number of $\displaystyle \mathrm{Fe}=26$ ] $\displaystyle \mathbf{2} \boldsymbol{+} \mathbf{1}$(b)On the basis of crystal field theory, write the electronic configuration for $\displaystyle \mathrm{d}^{5}$ ion for which $\displaystyle \Delta_{0}<\mathrm{P}$.
(a)
Explain why $\displaystyle \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ is strongly paramagnetic whereas $\displaystyle \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ is weakly paramagnetic. [Atomic number of $\displaystyle \mathrm{Fe}=26$ ] $\displaystyle \mathbf{2} \boldsymbol{+} \mathbf{1}$
(b)
On the basis of crystal field theory, write the electronic configuration for $\displaystyle \mathrm{d}^{5}$ ion for which $\displaystyle \Delta_{0}<\mathrm{P}$.
Marking-scheme solution
(a)
In [Fe(H2O)$\displaystyle 6$] $\displaystyle 3$+, H2O is a weak field ligand, 3d electrons do not pair up.
Hence it is strongly paramagnetic.
Whereas in [Fe(CN)$\displaystyle 6$] $\displaystyle 3$−, cyanide ion is a strong field ligand, 3d electrons pair up
leaving only one unpaired electron. So it is weakly paramagnetic.
(b)
t2g3eg2
Coordination CompoundsBonding in Coordination CompoundsUnderstandshort_answerhard
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CBSE Class 12 Chemistry past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.