A first row transition metal with the highest enthalpy of atomisation, upon reaction with oxygen at high temperature, forms oxides of formula $ \text{M}_2\text{O}_n $ (where $ n = 3, 4, 5 $). The 'spin-only' magnetic moment value of the amphoteric oxide from the above oxides is ___ BM (near integer).
(Given atomic number: $ \text{Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26, Co: 27, Ni: 28, Cu: 29, Zn: 30} $)

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

Among first-row transition elements, Vanadium (V) exhibits the highest enthalpy of atomisation at 515 kJ/mol. Its oxide is $ \text{V}_2\text{O}_5 $.
In $ \text{V}_2\text{O}_5 $:
Vanadium possesses a $ +5 $ oxidation state.
The electron configuration of $ \text{V}^{5+} $ is $ 1s^2 2s^2 2p^6 3s^2 3p^6 $, featuring no unpaired electrons.
Consequently, the "spin-only" magnetic moment of $ \text{V}^{5+} $ is zero due to the absence of unpaired electrons.

Correct Answer: 0

Solution and Explanation

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