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: \( \text{Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26, Co: 27, Ni: 28, Cu: 29, Zn: 30} \))
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: \( \text{Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26, Co: 27, Ni: 28, Cu: 29, Zn: 30} \))
Updated On: Jan 13, 2026
Hide Solution
Verified By Collegedunia
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.
Was this answer helpful?
1
0
Hide Solution
Verified By Collegedunia
Approach Solution -2
Step 1: Identify the metal from its oxide formulas
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
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
0
Was this answer helpful?
0
0
Top Questions on d -and f -Block Elements
- At temperature \(T\) K, \(2\) moles of liquid \(A\) and \(3\) moles of liquid \(B\) are mixed. The vapour pressure of the ideal solution formed is \(320\) mm Hg. At this stage, one mole of \(A\) and one mole of \(B\) are added to the solution. The vapour pressure is now measured as \(328.6\) mm Hg. The vapour pressures (in mm Hg) of pure \(A\) and pure \(B\) respectively are:
- JEE Main - 2026
- Chemistry
- d -and f -Block Elements
- Consider the general reaction given below at 400 K: xA(g) ⇌ yB(g). The values of K___p and K___c are studied under the same condition of temperature but variation in x and y.
(i) K___p = 85.87 and K___c = 2.586
(ii) K___p = 0.862 and K___c = 28.62.
The values of x and y in (i) and (ii) respectively are :- JEE Main - 2026
- Chemistry
- d -and f -Block Elements
- Among the following oxides of 3d elements, the number of mixed oxides are ____________.
\[ \mathrm{Ti_2O_3,\ V_2O_4,\ Cr_2O_3,\ Mn_3O_4,\ Fe_3O_4,\ Fe_2O_3,\ Co_3O_4} \]- JEE Main - 2026
- Chemistry
- d -and f -Block Elements
- 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.- JEE Main - 2026
- Chemistry
- d -and f -Block Elements
- Why is the ability of oxygen greater than fluorine to stabilise higher oxidation states of transition metals?
- CBSE CLASS XII - 2025
- Chemistry
- d -and f -Block Elements
View More Questions
Questions Asked in JEE Main exam
- The sum of all possible values of \( n \in \mathbb{N} \), so that the coefficients of \(x, x^2\) and \(x^3\) in the expansion of \((1+x^2)^2(1+x)^n\) are in arithmetic progression is :
- JEE Main - 2026
- Integration
- In a microscope of tube length $10\,\text{cm}$ two convex lenses are arranged with focal lengths $2\,\text{cm}$ and $5\,\text{cm}$. Total magnification obtained with this system for normal adjustment is $(5)^k$. The value of $k$ is ___.
- JEE Main - 2026
- Optical Instruments
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?
- JEE Main - 2026
- Organic Chemistry
- Let \( ABC \) be an equilateral triangle with orthocenter at the origin and the side \( BC \) lying on the line \( x+2\sqrt{2}\,y=4 \). If the coordinates of the vertex \( A \) are \( (\alpha,\beta) \), then the greatest integer less than or equal to \( |\alpha+\sqrt{2}\beta| \) is:
- JEE Main - 2026
- Coordinate Geometry
- Three charges $+2q$, $+3q$ and $-4q$ are situated at $(0,-3a)$, $(2a,0)$ and $(-2a,0)$ respectively in the $x$-$y$ plane. The resultant dipole moment about origin is ___.
- JEE Main - 2026
- Electromagnetic waves
View More Questions