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: \( \text{Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26, Co: 27, Ni: 28, Cu: 29, Zn: 30} \))
(Given atomic number: \( \text{Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26, Co: 27, Ni: 28, Cu: 29, Zn: 30} \))
Correct Answer: 0
Approach Solution - 1
Vanadium (V) has the highest enthalpy of atomisation (515 kJ/mol) among first-row transition elements. When it forms an oxide, it produces \( \text{V}_2\text{O}_5 \).
In \( \text{V}_2\text{O}_5 \):
Vanadium is in the \( +5 \) oxidation state.
The electron configuration of \( \text{V}^{5+} \) is \( 1s^2 2s^2 2p^6 3s^2 3p^6 \), with no unpaired electrons.
Since there are no unpaired electrons in \( \text{V}^{5+} \), the "spin-only" magnetic moment is zero.
Approach Solution -2
The set of oxides M2On with n = 3, 4, 5 corresponds to oxidation states +3, +4 and +5, respectively (since for M2On, 2×Ox(M) − 2n = 0 ⇒ Ox(M) = n). Among first-row transition metals, vanadium forms precisely V2O3 (V(+3)), V2O4 (equivalently VO2, V(+4)) and V2O5 (V(+5)). Vanadium is also noted for a very high enthalpy of atomisation in the first series, consistent with the prompt.
Step 2: Pick the amphoteric oxide from V2O3, V2O4, V2O5
Among these vanadium oxides, the one exhibiting amphoteric behaviour in many textbook treatments is V2O5. It can react with strong bases to give vanadates and participates in acid-vanadate equilibria (polyvanadates), reflecting amphoteric character.
Step 3: Determine the d-electron count in the amphoteric oxide
In V2O5, vanadium is in the +5 oxidation state. For vanadium (Z = 23), the neutral atom configuration is [Ar] 3d3 4s2. Therefore, V(+5) is 3d0 (no d electrons).
Step 4: Compute the spin-only magnetic moment
The spin-only magnetic moment is given by
μso = √(n(n + 2)) BM, where n = number of unpaired electrons.
For V(+5) with 3d0, n = 0 ⇒ μso = √(0 × 2) = 0 BM (near integer).
Final answer
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