CBSE 2022 · Region 4 · Set 2 · Q5 · 3 marks
(i)Write the electronic configuration of $\displaystyle \mathrm{d}^{4}$ on the basis of crystal field splitting theory, if $\displaystyle \Delta_{0}<\mathrm{P}$.(ii)$\displaystyle \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ with square-planar structure is diamagnetic and $\displaystyle \left[\mathrm{NiCl}_{4}\right]^{2-}$ with tetrahedral geometry is paramagnetic. Give reason to support the statement. [Atomic number : $\displaystyle \mathrm{Ni}=28$ ](iii)Write the number of ions produced in the solution from the following complex :$\displaystyle \left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}$(i)Calculate the spin only magnetic moment of the complex $\displaystyle \left[\mathrm{FeF}_{6}\right]^{3-}$. (Atomic number of $\displaystyle \mathrm{Fe}=26$ )(ii)Write the IUPAC name of the given complex : $\displaystyle \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$(iii)Why is the complex $\displaystyle \left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$ more stable than $\displaystyle \left[\mathrm{CoF}_{6}\right]^{3-}$ ?
(i)
Write the electronic configuration of $\displaystyle \mathrm{d}^{4}$ on the basis of crystal field splitting theory, if $\displaystyle \Delta_{0}<\mathrm{P}$.
(ii)
$\displaystyle \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ with square-planar structure is diamagnetic and $\displaystyle \left[\mathrm{NiCl}_{4}\right]^{2-}$ with tetrahedral geometry is paramagnetic. Give reason to support the statement. [Atomic number : $\displaystyle \mathrm{Ni}=28$ ]
(iii)
Write the number of ions produced in the solution from the following complex :$\displaystyle \left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}$
(i)
Calculate the spin only magnetic moment of the complex $\displaystyle \left[\mathrm{FeF}_{6}\right]^{3-}$. (Atomic number of $\displaystyle \mathrm{Fe}=26$ )
(ii)
Write the IUPAC name of the given complex : $\displaystyle \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$
(iii)
Why is the complex $\displaystyle \left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$ more stable than $\displaystyle \left[\mathrm{CoF}_{6}\right]^{3-}$ ?
Marking-scheme solution
(i)
$\displaystyle t_{2\mathrm{g}}^{3} e_{\mathrm{g}}^{1}$
(ii)
$\displaystyle \mathrm{Ni}^{2+}$ is $\displaystyle 3d^{8}$ and has $\displaystyle dsp^{2}$ hybridisation. Cyanide is a strong field ligand; electrons pair up so it is diamagnetic. In $\displaystyle \left[\mathrm{NiCl}_{4}\right]^{2-}$, $\displaystyle \mathrm{Cl}^{-}$ is a weak field ligand, electrons do not pair up, hence hybridisation is $\displaystyle sp^{3}$. Thus, it is paramagnetic.
(iii)
$\displaystyle 3$
(i)
$\displaystyle \mathrm{Fe}^{3+}$ — No. of unpaired electrons = $\displaystyle 5$
\[\begin{aligned}
\mu &=\sqrt{n(n+2)}=\sqrt{5(5+2)} \\
&=5.916 \mathrm{~BM}
\end{aligned}
\]
(ii)
Pentaamminechloridocobalt(III) chloride
(iii)
$\displaystyle \left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$, chelate effect / formation of cyclic structure.
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CBSE Class 12 Chemistry past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.