Question

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: \( \text{Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26, Co: 27, Ni: 28, Cu: 29, Zn: 30} \))

Answer 1

Correct Answer: 0

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.

Answer 2

Step 1: Identify the metal from its oxide formulas
The set of oxides M₂O_n with n = 3, 4, 5 corresponds to oxidation states +3, +4 and +5, respectively (since for M₂O_n, 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
μ_so = √(n(n + 2)) BM, where n = number of unpaired electrons.
For V(+5) with 3d⁰, n = 0 ⇒ μ_so = √(0 × 2) = 0 BM (near integer).

Final answer

0