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Chemistry · 2024 · 4 marks
The d-and f-Block ElementsGeneral Properties of the Transition Elements (d-Block)4 marksAnalysecase study
CBSE 2024 · Region 2 · Set 1 · Q30
Transition metals have incomplete d-subshell either in neutral atom or in their ions. The presence of partly filled d-orbitals in their atoms makes…
Transition metals have incomplete d-subshell either in neutral atom or in their ions. The presence of partly filled d-orbitals in their atoms makes transition elements different from that of the non-transition elements. With partly filled d-orbitals, these elements exhibit certain characteristic properties such as display of a variety of oxidation states, formation of coloured ions and entering into complex formation with a variety of ligands. The transition metals and their compounds also exhibit catalytic properties and paramagnetic behaviour. The transition metals are very hard and have low volatility. An examination of the $\displaystyle \mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{\mathrm{o}}$ values shows the varying trends :
Answer the following questions :
| $\displaystyle \mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{\mathrm{o}}$ | |
| V | -$\displaystyle 1.18$ |
| Cr | -$\displaystyle 0.91$ |
| Mn | -$\displaystyle 1.18$ |
| Fe | -$\displaystyle 0.44$ |
| Co | -$\displaystyle 0.28$ |
| Ni | -$\displaystyle 0.25$ |
| Cu | +$\displaystyle 0.34$ |
| Zn | -$\displaystyle 0.76$ |
(a)
On what basis can we say that Cu is a transition element but Zn is not ? (Atomic number : $\displaystyle \mathrm{Cu}=29, \mathrm{Zn}=30$ ) $\displaystyle 1$
(b)
Why do transition elements show variety of oxidation states ?
(c)
Why do $\displaystyle \mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{\mathrm{o}}$ values show irregular trend from Vanadium to Zinc ?
(ii)
How is the variability in oxidation states of transition metals different from that of the non-transition elements ?
Of the $\displaystyle \mathrm{d}^{4}$ species, $\displaystyle \mathrm{Cr}^{2+}$ is strongly reducing while $\displaystyle \mathrm{Mn}^{3+}$ is strongly oxidizing. Why ? (Atomic number : $\displaystyle \mathrm{Cr}=24, \mathrm{Mn}=25$ )
(ii)
Complete the following ionic equation :\[2 \mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O}+\mathrm{I}^{-} \longrightarrow \]
Marking-scheme solution
(a)
Cu has incomplete d-orbital in +$\displaystyle 2$ oxidation state whereas Zn has fully filled d- orbital in ground state as well as in +$\displaystyle 2$ oxidation state.
(b)
Because both (n-$\displaystyle 1$)d and ns subshell electrons take part in the bond formation due to their comparable energies/ due to the presence of unpaired electrons in d-orbitals.
(c)
Because of irregular values of \(\displaystyle \left(\Delta_{\mathrm{i}} \mathrm{H}_{1}+\Delta_{\mathrm{i}} \mathrm{H}_{2}\right)\) and sublimation enthalpies.
(ii)
In transition metals, oxidation states differ by +$\displaystyle 1$ whereas in non-transition metals differ by +2.
Because \(\displaystyle \mathrm{Cr}^{2+}\) will be converted to \(\displaystyle \mathrm{Cr}^{3+}\) which has more stable half filled \(\displaystyle \mathrm{t}_{2 \mathrm{~g}}\) configuration while \(\displaystyle \mathrm{Mn}^{3+}\) changes to \(\displaystyle \mathrm{Mn}^{2+}\) which has more stable half-filled \(\displaystyle \mathrm{d}^{5}\) configuration.
(ii)
\(\displaystyle 2 \mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O}+\mathrm{I}^{-} \longrightarrow 2 \mathrm{MnO}_{2}+2 \mathrm{OH}^{-}+\mathrm{IO}_{3}{ }^{-}\)
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CBSE Class 12 Chemistry past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.