Identify the diamagnetic octahedral complex ions from below ;
A. [Mn(CN)\(_6\)]\(^{3-}\)
B. [Co(NH\(_3\))\(_6\)]\(^{3+}\)
C. [Fe(CN)\(_6\)]\(^{4-}\)
D. [Co(H\(_2\)O)\(_3\)F\(_3\)]
Choose the correct answer from the options given below :
Identify the diamagnetic octahedral complex ions from below ;
A. [Mn(CN)\(_6\)]\(^{3-}\)
B. [Co(NH\(_3\))\(_6\)]\(^{3+}\)
C. [Fe(CN)\(_6\)]\(^{4-}\)
D. [Co(H\(_2\)O)\(_3\)F\(_3\)]
Choose the correct answer from the options given below :
Show Hint
- B and D Only
- A and D Only
- A and C Only
- B and C Only
The Correct Option is D
Approach Solution - 1
To determine which of the given complexes are diamagnetic, we must first understand the electronic configuration and the nature of the ligands.
- [Mn(CN)6]3-:
- Manganese (Mn) in [Mn(CN)6]3- is in the +3 oxidation state, giving it a configuration of \(d^4\) (Mn: [Ar] 3d5 4s2, so Mn3+: [Ar] 3d4).
- CN- is a strong field ligand and can cause pairing of electrons in the 3d orbitals.
- However, since the d-configuration is \(d^4\), complete pairing leading to diamagnetism will not occur.
- [Co(NH3)6]3+:
- Cobalt (Co) in [Co(NH3)6]3+ is in the +3 oxidation state, resulting in a \(d^6\) configuration (Co: [Ar] 3d7 4s2, so Co3+: [Ar] 3d6).
- NH3 is also a strong field ligand, causing the pairing of d electrons.
- This results in a completely paired electronic configuration, making the complex diamagnetic.
- [Fe(CN)6]4-:
- Iron (Fe) in [Fe(CN)6]4- is in the +2 oxidation state, giving it a configuration of \(d^6\) (Fe: [Ar] 3d6 4s2, so Fe2+: [Ar] 3d6).
- CN-, being a strong field ligand, similarly causes the pairing of the 3d electrons in this case.
- This again leads to a completely paired electronic configuration, making the complex diamagnetic.
- [Co(H2O)3F3]:
- This complex exhibits weak field ligands (such as H2O and F-), so it will not lead to the pairing of electrons.
- For Co3+, it would have a \(d^6\) configuration, and due to weak fields, electrons remain unpaired, resulting in a paramagnetic complex.
Hence, the diamagnetic complexes are [Co(NH3)6]3+ (B) and [Fe(CN)6]4- (C).
Therefore, the correct answer is: B and C Only.
Approach Solution -2
Let's analyze each complex ion to determine its magnetic properties. A diamagnetic complex has no unpaired electrons. diamagnetic complex ion has all its electrons paired. Let's analyze each complex:
A. [Mn(CN)₆]³⁻:
- Mn is in the +3 oxidation state: Mn³⁺ has a d⁴ configuration.
- CN⁻ is a strong field ligand, so it will cause pairing in the +3 oxidation state. Mn³⁺ has an electronic configuration of d⁴.
- CN⁻ is a strong field ligand, so the complex is low spin.
- In an octahedral low-spin complex, the d⁴ configuration will have the electrons fill the t₂g orbitals first: ( of electrons (low spin).
- The d⁴ configuration in a strong field octahedral complex will have the configuration t₂g⁴ eg⁰, meaning all four electrons aret₂g)⁴ (eg)⁰. Thus, there will be 4 unpaired electrons in the t2g orbitals. The electronic configuration is \(t_{2g}^4e_g^0\).
- Since there are unpaired in the t₂g orbitals.
- Therefore, there are 2 unpaired electrons after filling in, hence paramagnetic
- t2g⁴ eg⁰ (↑↓ ↑↓ __) --> two unpaired
- electrons, this complex is paramagnetic.
B. [Co(NH₃)₆]³⁺:
- Co is in the +3 oxidation state. Co³⁺ has an electronic configuration of d⁶.
- Co is in the +3 oxidation state: Co³⁺ has a d⁶ configuration.
- NH₃ is a strong field ligand, so it will cause pairing of electrons (low spin).
- NH₃ is a strong field ligand, so the complex is low spin.
- In an octahedral low-spin complex, the d⁶ configuration will have the electrons fill the t₂g orbitals first: (t₂g)⁶ (eg)⁰. Thus, there will be no unpaired electrons. The electronic The d⁶ configuration in a strong field octahedral complex will have the configuration t₂g⁶ eg⁰, meaning all six electrons are in the t₂g orbitals.
- Therefore, there are no unpaired electrons, hence diamagnetic.
- t2g⁶ eg⁰ (↑↓ ↑↓ ↑
- The d⁶ configuration in a strong field octahedral complex will have the configuration t₂g⁶ eg⁰, meaning all six electrons are in the t₂g orbitals.
- Therefore, there are no unpaired electrons, hence diamagnetic.
- t2g⁶ eg⁰ (↑↓ ↑↓ ↑↓ )--> zero unpaired
- Since there are no unpaired electrons, this complex is diamagnetic.
C. [Fe(CN)₆]⁴⁻:
- Fe is in the +2 oxidation state. Fe²⁺ has an electronic configuration of d⁶.
- Fe is in the +2 oxidation state: Fe²⁺ has a d⁶ configuration.
- CN⁻ is a strong field ligand, so it will cause pairing of electrons (low spin).
- The d⁶ configuration of d⁶.
- CN⁻ is a strong field ligand, so the complex is low spin.
- In an octahedral low-spin complex, the d⁶ configuration will have the electrons fill the t₂g orbitals first: (t₂g)⁶ (eg)⁰. Thus, there configuration in a strong field octahedral complex will have the configuration t₂g⁶ eg⁰, meaning all six electrons are in the t₂g orbitals.
- Therefore, there are no unpaired electrons, hence diamagnetic.
- t2g⁶ eg⁰ (↑↓ ↑↓ ↑↓ )--> zero
- strong field ligand, so it will cause pairing of electrons (low spin).
- In an octahedral low-spin complex, the d⁶ configuration will have the electrons fill the t₂g orbitals first: (t₂g)⁶ (eg)⁰. Thus, there will be no unpaired electrons. The electronic configuration is \(t_{2g}^6e_g^0\).
- Since there are no unpaired electrons, this complex is diamagnetic.
D. [Co(H₂O)₃F₃]:
- The complex is neutral. unpaired
- Co is in the +3 oxidation state: Co³⁺ has a d⁶ configuration.
- H₂O is a weak field ligand and F is also a weak field ligand, so it will NOT cause pairing of electrons Co is in +3 oxidation state.
- H₂O is a weak field ligand, and F is also a weak field ligand. So, this is likely a high-spin complex.
- Co³⁺ is d⁶, so in a high-spin complex, the configuration would be (t₂g) (high spin).
- Therefore, there are 4 unpaired electrons, hence paramagnetic
- The d⁶ configuration in a weak field octahedral complex will have the configuration t₂g⁴ eg²
- t2g⁴ eg² (↑↓ ↑ ↑ ↑) --> Four unpaired electrons
Thus, the diamagnetic complexes are B and C.
Final Answer: The final answer is B and C.
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:
Top JEE Main Coordination chemistry Questions
- Match List I with List II:
List I (Substances) List II (Element Present) (A) Ziegler catalyst (I) Rhodium (B) Blood Pigment (II) Cobalt (C) Wilkinson catalyst (III) Iron (D) Vitamin B12 (IV) Titanium
Choose the correct answer from the options given below : - Given below are two statements:
Statement (I): A solution of \([Ni(H_2O)_6]^{2+}\) is green in color.
Statement (II): A solution of \([Ni(CN)_4]^{2-}\) is colorless.
In the light of the above statements, choose the most appropriate answer from the options given below. - Which of the following complex is homoleptic?
- Match List-I with List-II.
List-I
(Complex ion)List-II
(Spin only magnetic moment in B.M.)(A) [Cr(NH$_3$)$_6$]$^{3+}$ (I) 4.90 (B) [NiCl$_4$]$^{2-}$ (II) 3.87 (C) [CoF$_6$]$^{3-}$ (III) 0.0 (D) [Ni(CN)$_4$]$^{2-}$ (IV) 2.83
Choose the correct answer from the options given below: - Total number of unpaired electrons in the complex ion [Co(NH$_3$)$_6$]$^{3+}$ and [NiCl$_4$]$^{2-}$ is
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,