Among the following complexes, the total number of diamagnetic species is ______.
[Mn(NH3)6]3+, [MnCl6]3−, [FeF6]3−, [CoF6]3−, [Fe(NH3)6]3+, and [Co(en)3]3+
[Given, atomic number: Mn = 25, Fe = 26, Co = 27;
en = H2NCH2CH2NH2]
[Mn(NH3)6]3+, [MnCl6]3−, [FeF6]3−, [CoF6]3−, [Fe(NH3)6]3+, and [Co(en)3]3+
[Given, atomic number: Mn = 25, Fe = 26, Co = 27;
en = H2NCH2CH2NH2]
Correct Answer: 1
Approach Solution - 1
Step 1: Understanding Diamagnetism
- A species is diamagnetic if all its electrons are paired.
- Transition metal complexes can be classified based on their ligand field strength and electronic configuration.
Step 2: Electronic Configurations of the Metal Ions
- Mn3+: [Ar] 3d4
- Fe3+: [Ar] 3d5
- Co3+: [Ar] 3d6
Step 3: Determining the Magnetic Properties of Each Complex
- [Mn(NH3)6]3+ (NH3 is a weak field ligand, high-spin d4 configuration) → Paramagnetic
- [MnCl6]3− (Cl- is a weak field ligand, high-spin d4 configuration) → Paramagnetic
- [FeF6]3− (F- is a weak field ligand, high-spin d5 configuration) → Paramagnetic
- [CoF6]3− (F- is a weak field ligand, high-spin d6 configuration) → Paramagnetic
- [Fe(NH3)6]3+ (NH3 is a weak field ligand, high-spin d5 configuration) → Paramagnetic
- [Co(en)3]3+ (en is a strong field ligand, low-spin d6 configuration) → Diamagnetic
Step 4: Conclusion
- Only [Co(en)3]3+ is diamagnetic.
- Total number of diamagnetic species = 1.
Approach Solution -2
To solve the problem, determine the number of diamagnetic species among the given complexes by analyzing their oxidation states, d-electron counts, ligand field strengths, and spin states.
Given complexes:
1. \([Mn(NH_3)_6]^{3+}\)
2. \([MnCl_6]^{3-}\)
3. \([FeF_6]^{3-}\)
4. \([CoF_6]^{3-}\)
5. \([Fe(NH_3)_6]^{3+}\)
6. \([Co(en)_3]^{3+}\)
Step 1: Determine metal oxidation state and d-electron count
| Complex | Metal | Oxidation State | d-electrons (dn) | Ligand Field Strength |
|---|---|---|---|---|
| [Mn(NH3)6]3+ | Mn (25) | +3 | d4 | NH3 (strong field) |
| [MnCl6]3− | Mn | +3 | d4 | Cl− (weak field) |
| [FeF6]3− | Fe (26) | +3 | d5 | F− (weak field) |
| [CoF6]3− | Co (27) | +3 | d6 | F− (weak field) |
| [Fe(NH3)6]3+ | Fe | +3 | d5 | NH3 (strong field) |
| [Co(en)3]3+ | Co | +3 | d6 | en (strong field) |
Step 2: Determine spin states and diamagnetism
- Diamagnetic means all electrons paired.
- [Mn(NH3)6]3+: d4 strong field → likely low spin → 2 unpaired electrons → paramagnetic
- [MnCl6]3−: d4 weak field → high spin → 4 unpaired electrons → paramagnetic
- [FeF6]3−: d5 weak field → high spin → 5 unpaired electrons → paramagnetic
- [CoF6]3−: d6 weak field → high spin → 4 unpaired electrons → paramagnetic
- [Fe(NH3)6]3+: d5 strong field → low spin → 1 unpaired electron → paramagnetic
- [Co(en)3]3+: d6 strong field → low spin → all electrons paired → diamagnetic
Step 3: Count diamagnetic species
Only <\(Co(en)_3]^{3+}\) is diamagnetic.
Final Answer:
\[ \boxed{1} \]
Top Questions on coordination compounds
- Calculate the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
- BITSAT - 2025
- Chemistry
- coordination compounds
- The complex $[Cr(NH_3)_4Cl_2]^+$ shows how many geometrical isomers?
- BITSAT - 2025
- Chemistry
- coordination compounds
- The number of paramagnetic complexes among $[FeF_6]^{3-}, [Fe(CN)_6]^{3-}, [Mn(CN)_6]^{3-}, [Co(C_2O_4)_3]^{3-}, [MnCl_6]^{3-}$ and $[CoF_6]^{3-}$, which involved $d^2sp^3$ hybridization is ...............
- JEE Main - 2025
- Chemistry
- coordination compounds
- An octahedral complex having molecular composition Co.5NH$_3$.Cl.SO$_4$ has two isomers A and B. The solution of A gives a white precipitate with AgNO$_3$ solution and the solution of B gives a white precipitate with BaCl$_2$ solution. The type of isomerism exhibited by the complex is,
- JEE Main - 2025
- Chemistry
- coordination compounds
- Write the IUPAC name of complex compound \( K_3[Cr(C_2O_4)_3] \).
- Bihar Board XII - 2025
- Chemistry
- coordination compounds
Questions Asked in JEE Advanced exam
A temperature difference can generate e.m.f. in some materials. Let $ S $ be the e.m.f. produced per unit temperature difference between the ends of a wire, $ \sigma $ the electrical conductivity and $ \kappa $ the thermal conductivity of the material of the wire. Taking $ M, L, T, I $ and $ K $ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $ Z = \frac{S^2 \sigma}{\kappa} $ is:
- JEE Advanced - 2025
- Dimensional Analysis
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
- JEE Advanced - 2025
- Differential equations