Question:
Which of the following pair of ions have equal number of unpaired electrons
Which of the following pair of ions have equal number of unpaired electrons
Show Hint
For transition metal ions, always remove electrons from the \( 4s \) orbital before the \( 3d \) orbital. Then apply Hund's rule to count the number of unpaired electrons correctly.
Updated On: Apr 4, 2026
- \( \mathrm{V^{2+}} \) and \( \mathrm{Ni^{2+}} \)
- \( \mathrm{Cr^{2+}} \) and \( \mathrm{Mn^{2+}} \)
- \( \mathrm{Fe^{2+}} \) and \( \mathrm{Sc^{2+}} \)
- \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Write the electronic configuration of each ion.
To find the number of unpaired electrons, we first write the electronic configuration of each metal ion. While forming cations, electrons are removed first from the \( 4s \) orbital and then from the \( 3d \) orbital.
\[ \mathrm{V} = [Ar]\,3d^3 4s^2 \Rightarrow \mathrm{V^{2+}} = [Ar]\,3d^3 \] \[ \mathrm{Ni} = [Ar]\,3d^8 4s^2 \Rightarrow \mathrm{Ni^{2+}} = [Ar]\,3d^8 \] \[ \mathrm{Cr} = [Ar]\,3d^5 4s^1 \Rightarrow \mathrm{Cr^{2+}} = [Ar]\,3d^4 \] \[ \mathrm{Mn} = [Ar]\,3d^5 4s^2 \Rightarrow \mathrm{Mn^{2+}} = [Ar]\,3d^5 \] \[ \mathrm{Fe} = [Ar]\,3d^6 4s^2 \Rightarrow \mathrm{Fe^{2+}} = [Ar]\,3d^6 \] \[ \mathrm{Sc} = [Ar]\,3d^1 4s^2 \Rightarrow \mathrm{Sc^{2+}} = [Ar]\,3d^1 \] \[ \mathrm{Mn^{3+}} = [Ar]\,3d^4 \] Step 2: Count the unpaired electrons in each ion.
Now we count the number of unpaired electrons in each \( d \)-configuration using Hund's rule:
\begin{itemize} \item \( \mathrm{V^{2+}} = 3d^3 \Rightarrow 3 \) unpaired electrons \item \( \mathrm{Ni^{2+}} = 3d^8 \Rightarrow 2 \) unpaired electrons \item \( \mathrm{Cr^{2+}} = 3d^4 \Rightarrow 4 \) unpaired electrons \item \( \mathrm{Mn^{2+}} = 3d^5 \Rightarrow 5 \) unpaired electrons \item \( \mathrm{Fe^{2+}} = 3d^6 \Rightarrow 4 \) unpaired electrons \item \( \mathrm{Sc^{2+}} = 3d^1 \Rightarrow 1 \) unpaired electron \item \( \mathrm{Mn^{3+}} = 3d^4 \Rightarrow 4 \) unpaired electrons \end{itemize} Step 3: Compare the pairs given in the options.
\begin{itemize} \item (A) \( \mathrm{V^{2+}} \) and \( \mathrm{Ni^{2+}} \): \( 3 \) and \( 2 \) unpaired electrons, so not equal. \item (B) \( \mathrm{Cr^{2+}} \) and \( \mathrm{Mn^{2+}} \): \( 4 \) and \( 5 \) unpaired electrons, so not equal. \item (C) \( \mathrm{Fe^{2+}} \) and \( \mathrm{Sc^{2+}} \): \( 4 \) and \( 1 \) unpaired electrons, so not equal. \item (D) \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \): both have \( 4 \) unpaired electrons, so this pair is correct. \end{itemize} Step 4: Conclusion.
Hence, the pair of ions that have equal number of unpaired electrons is \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \). Both possess the \( 3d^4 \) and \( 3d^6 \) configurations respectively, and each has \( 4 \) unpaired electrons.
Final Answer: \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \).
To find the number of unpaired electrons, we first write the electronic configuration of each metal ion. While forming cations, electrons are removed first from the \( 4s \) orbital and then from the \( 3d \) orbital.
\[ \mathrm{V} = [Ar]\,3d^3 4s^2 \Rightarrow \mathrm{V^{2+}} = [Ar]\,3d^3 \] \[ \mathrm{Ni} = [Ar]\,3d^8 4s^2 \Rightarrow \mathrm{Ni^{2+}} = [Ar]\,3d^8 \] \[ \mathrm{Cr} = [Ar]\,3d^5 4s^1 \Rightarrow \mathrm{Cr^{2+}} = [Ar]\,3d^4 \] \[ \mathrm{Mn} = [Ar]\,3d^5 4s^2 \Rightarrow \mathrm{Mn^{2+}} = [Ar]\,3d^5 \] \[ \mathrm{Fe} = [Ar]\,3d^6 4s^2 \Rightarrow \mathrm{Fe^{2+}} = [Ar]\,3d^6 \] \[ \mathrm{Sc} = [Ar]\,3d^1 4s^2 \Rightarrow \mathrm{Sc^{2+}} = [Ar]\,3d^1 \] \[ \mathrm{Mn^{3+}} = [Ar]\,3d^4 \] Step 2: Count the unpaired electrons in each ion.
Now we count the number of unpaired electrons in each \( d \)-configuration using Hund's rule:
\begin{itemize} \item \( \mathrm{V^{2+}} = 3d^3 \Rightarrow 3 \) unpaired electrons \item \( \mathrm{Ni^{2+}} = 3d^8 \Rightarrow 2 \) unpaired electrons \item \( \mathrm{Cr^{2+}} = 3d^4 \Rightarrow 4 \) unpaired electrons \item \( \mathrm{Mn^{2+}} = 3d^5 \Rightarrow 5 \) unpaired electrons \item \( \mathrm{Fe^{2+}} = 3d^6 \Rightarrow 4 \) unpaired electrons \item \( \mathrm{Sc^{2+}} = 3d^1 \Rightarrow 1 \) unpaired electron \item \( \mathrm{Mn^{3+}} = 3d^4 \Rightarrow 4 \) unpaired electrons \end{itemize} Step 3: Compare the pairs given in the options.
\begin{itemize} \item (A) \( \mathrm{V^{2+}} \) and \( \mathrm{Ni^{2+}} \): \( 3 \) and \( 2 \) unpaired electrons, so not equal. \item (B) \( \mathrm{Cr^{2+}} \) and \( \mathrm{Mn^{2+}} \): \( 4 \) and \( 5 \) unpaired electrons, so not equal. \item (C) \( \mathrm{Fe^{2+}} \) and \( \mathrm{Sc^{2+}} \): \( 4 \) and \( 1 \) unpaired electrons, so not equal. \item (D) \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \): both have \( 4 \) unpaired electrons, so this pair is correct. \end{itemize} Step 4: Conclusion.
Hence, the pair of ions that have equal number of unpaired electrons is \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \). Both possess the \( 3d^4 \) and \( 3d^6 \) configurations respectively, and each has \( 4 \) unpaired electrons.
Final Answer: \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \).
Was this answer helpful?
0
0
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
View More Questions
Top JEE Main Atomic Structure Questions
- Choose the correct option having all the elements with \( d^{10} \) electronic configuration from the following:
- The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is:
- $\textbf{Match List I with List II}$
$\begin{array}{|c|c|c|} \hline \text{List I (Spectral Series for Hydrogen)} & \text{List II (Spectral Region/Higher Energy State)} \\ \hline \text{A. Lyman} & \text{I. Infrared region} \\ \text{B. Balmer} & \text{II. UV region} \\ \text{C. Paschen} & \text{III. Infrared region} \\ \text{D. Pfund} & \text{IV. Visible region} \\ \hline \end{array}$
$\text{Choose the correct answer from the options given below :-}$ - Number of spectral lines obtained in He+ spectra, when an electron makes transition from fifth excited state to first excited state will be __________.
- Which of the following electronic configurations would be associated with the highest magnetic moment?
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions