Q65
jee_main_2026_21_jan_morning
Magnetic Properties of Coordination Compounds
Given below are two statements:
Statement I: Among [mathrmCu(mathrmNH_3)_4]^2+$[\mathrm{Cu}(\mathrm{NH}_{3})_{4}]^{2+}$, [mathrmNi(mathrmen)_3]^2+$[\mathrm{Ni}(\mathrm{en})_{3}]^{2+}$, [mathrmNi(mathrmNH_3)_6]^2+$[\mathrm{Ni}(\mathrm{NH}_{3})_{6}]^{2+}$ and [mathrmMn(mathrmH_2mathrmO)_6]^2+$[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$, [mathrmMn(mathrmH_2mathrmO)_6]^2+$[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$ has the maximum number of unpaired electrons.
Statement II: The number of pairs among \[NiCl_4]^2-, [Ni(CO)_4]\$\{[NiCl_{4}]^{2-}, [Ni(CO)_{4}]\}$, \[NiCl_4]^2-, [Ni(CN)_4]^2-\$\{[NiCl_{4}]^{2-}, [Ni(CN)_{4}]^{2-}\}$ and \[Ni(CO)_4], [Ni(CN)_4]^2-\$\{[Ni(CO)_{4}], [Ni(CN)_{4}]^{2-}\}$ that contain only diamagnetic species is two.
In the light of the above statements, choose the correct answer from the options given below:
- A. textStatement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$
- B. textBoth Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$
- C. textBoth Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$
- D. textStatement I is true but Statement II is false$\text{Statement I is true but Statement II is false}$
Solution
### Core Logic
Evaluating Statement I:
- [Cu(NH_3)_4]^2+$[Cu(NH_3)_4]^{2+}$: Cu^2+$Cu^{2+}$ is 3d^9$3d^9$, 1 unpaired electron.
- [Ni(en)_3]^2+$[Ni(en)_3]^{2+}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, in octahedral field, 2 unpaired electrons.
- [Ni(NH_3)_6]^2+$[Ni(NH_3)_6]^{2+}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, 2 unpaired electrons.
- [Mn(H_2O)_6]^2+$[Mn(H_2O)_6]^{2+}$: Mn^2+$Mn^{2+}$ is 3d^5$3d^5$, weak field ligand H_2O$H_2O$ leads to high spin, 5 unpaired electrons.
So [Mn(H_2O)_6]^2+$[Mn(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons. Statement I is true.
Evaluating Statement II:
- [Ni(CO)_4]$[Ni(CO)_4]$: Ni(0)$Ni(0)$ is 3d^8 4s^2$3d^8 4s^2$, strong field CO pairs electrons to 3d^10$3d^{10}$, diamagnetic (0 unpaired).
- [Ni(CN)_4]^2-$[Ni(CN)_4]^{2-}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, strong field CN^-$CN^-$ forces pairing rightarrow dsp^2$\rightarrow dsp^2$ square planar, diamagnetic (0 unpaired).
- [NiCl_4]^2-$[NiCl_4]^{2-}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, weak field Cl^-$Cl^-$ does not pair rightarrow sp^3$\rightarrow sp^3$ tetrahedral, paramagnetic (2 unpaired).
The pairs containing ONLY diamagnetic species:
- \[NiCl_4]^2-, [Ni(CO)_4]\$\{[NiCl_4]^{2-}, [Ni(CO)_4]\}$ rightarrow$\rightarrow$ 1 para, 1 dia (No)
- \[NiCl_4]^2-, [Ni(CN)_4]^2-\$\{[NiCl_4]^{2-}, [Ni(CN)_4]^{2-}\}$ rightarrow$\rightarrow$ 1 para, 1 dia (No)
- \[Ni(CO)_4], [Ni(CN)_4]^2-\$\{[Ni(CO)_4], [Ni(CN)_4]^{2-}\}$ rightarrow$\rightarrow$ Both dia (Yes)
The number of such pairs is exactly ONE. Statement II says two, so it is false.
### Step 1: Final Conclusion
Statement I is true, Statement II is false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q63
jee_main_2026_21_jan_evening
Crystal Field Stabilization Energy and Magnetic Moment
Given below are two statements:
Statement I: Crystal Field Stabilization Energy (CFSE) of [textCr(textH_2textO)_6]^2+$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ is greater than that of [textMn(textH_2textO)_6]^2+$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$.
Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide.
In the light of the above statements, choose the correct answer from the options given below:
- A. (1) \ textBoth Statement I and Statement II are true$(1) \ \text{Both Statement I and Statement II are true}$
- B. (2) \ textBoth Statement I and Statement II are false$(2) \ \text{Both Statement I and Statement II are false}$
- C. (3) textStatement I is true but Statement II is false$(3) \text{Statement I is true but Statement II is false}$
- D. (4) \ textStatement I is false but Statement II is true$(4) \ \text{Statement I is false but Statement II is true}$
Solution
### Core Logic
- Statement I: [textMn(textH_2textO)_6]^2+$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$ has d^5$d^5$ configuration with weak field ligands (CFSE = 0), whereas [textCr(textH_2textO)_6]^2+$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ has d^4$d^4$ configuration (CFSE = -0.6Delta_0$= -0.6\Delta_0$). Thus CFSE of chromium complex is greater in magnitude.
- Statement II: Potassium ferricyanide textK_3[textFe(textCN)_6]$\text{K}_3[\text{Fe}(\text{CN})_6]$ has Fe^3+$Fe^{3+}$ (d^5$d^5$, 1 unpaired electron, mu = sqrt3$\mu = \sqrt{3}$ B.M.), while sodium ferrocyanide textNa_4[textFe(textCN)_6]$\text{Na}_4[\text{Fe}(\text{CN})_6]$ has textFe^2+$\text{Fe}^{2+}$ (d^6$d^6$, 0 unpaired electrons, mu = 0$\mu = 0$). Thus statement II is true.
### Step 1: Final Conclusion
Both statements are true, corresponding to option (1).
### Pattern Recognition
Sees: CFSE calculations and magnetic moment comparisons for coordination complexes.
Trap: Miscalculating d$d$-electron count or ligand field strength.
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q74
jee_main_2026_21_jan_evening
Magnetic Moment and Unpaired Electrons
Identify the metal ions among textCo^2+$\text{Co}^{2+}$, textNi^2+$\text{Ni}^{2+}$, textFe^2+$\text{Fe}^{2+}$, textV^3+$\text{V}^{3+}$ and textTi^2+$\text{Ti}^{2+}$ having a spin-only magnetic moment value more than 3.0 text BM$3.0 \text{ BM}$. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is \_\_\_\_.
Numerical Answer. Answer: 7 to 7
Solution
### Core Logic
Let's check the d$d$-electron configurations and number of unpaired electrons (n$n$) in high spin octahedral complexes:
- textV^3+ rightarrow 3d^2$\text{V}^{3+} \rightarrow 3d^2$ (n = 2, mu = sqrt8 approx 2.83 text BM$n = 2, \mu = \sqrt{8} \approx 2.83 \text{ BM}$)
- textTi^2+ rightarrow 3d^2$\text{Ti}^{2+} \rightarrow 3d^2$ (n = 2, mu = 2.83 text BM$n = 2, \mu = 2.83 \text{ BM}$)
- textNi^2+ rightarrow 3d^8$\text{Ni}^{2+} \rightarrow 3d^8$ (n = 2, mu = 2.83 text BM$n = 2, \mu = 2.83 \text{ BM}$)
- textFe^2+ rightarrow 3d^6$\text{Fe}^{2+} \rightarrow 3d^6$ (n = 4, mu = sqrt24 approx 4.9 text BM > 3 text BM$n = 4, \mu = \sqrt{24} \approx 4.9 \text{ BM} > 3 \text{ BM}$)
- textCo^2+ rightarrow 3d^7$\text{Co}^{2+} \rightarrow 3d^7$ (n = 3, mu = sqrt15 approx 3.87 text BM > 3 text BM$n = 3, \mu = \sqrt{15} \approx 3.87 \text{ BM} > 3 \text{ BM}$)
### Step 1: Summing Unpaired Electrons
Only textFe^2+$\text{Fe}^{2+}$ (n = 4$n = 4$) and textCo^2+$\text{Co}^{2+}$ (n = 3$n = 3$) have magnetic moments > 3.0 text BM$> 3.0 \text{ BM}$.
Sum of unpaired electrons = 4 + 3 = 7$= 4 + 3 = 7$.
### Pattern Recognition
Sees: high spin octahedral complexes and spin-only magnetic moment threshold checks.
Trap: Misidentifying high-spin versus low-spin electron pairing in d^6$d^6$ or d^7$d^7$ configurations.
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q51
jee_main_2026_22_january_morning
Crystal Field Theory
Consider the transition metal ions Mn^3+$Mn^{3+}$, Cr^3+$Cr^{3+}$, Fe^3+$Fe^{3+}$ and Co^3+$Co^{3+}$ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is
- A. Cr^3+ > Fe^3+ > Co^3+ > Mn^3+$Cr^{3+} > Fe^{3+} > Co^{3+} > Mn^{3+}$
- B. Mn^3+ > Fe^3+ > Co^3+ > Cr^3+$Mn^{3+} > Fe^{3+} > Co^{3+} > Cr^{3+}$
- C. Fe^3+ > Co^3+ > Mn^3+ > Cr^3+$Fe^{3+} > Co^{3+} > Mn^{3+} > Cr^{3+}$
- D. Cr^3+ > Mn^3+ > Fe^3+ > Co^3+$Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$
Solution
### Related Formula
For low spin octahedral complexes, electrons pair up in the t_2g$t_{2g}$ orbitals before occupying e_g$e_{g}$ orbitals. Configuration follows the splitting t_2g$t_{2g}$ (lower energy) and e_g$e_{g}$ (higher energy).
### Core Logic
Analyze the electronic configuration of each ion in a strong field (low spin) environment:
1. Co^3+ rightarrow 3d^6 Rightarrow t_2g^2,2,2 e_g^0,0$Co^{3+} \rightarrow 3d^{6} \Rightarrow t_{2g}^{2,2,2} e_{g}^{0,0}$
Unpaired electrons = 0
2. Fe^3+ rightarrow 3d^5 Rightarrow t_2g^2,2,1 e_g^0,0$Fe^{3+} \rightarrow 3d^{5} \Rightarrow t_{2g}^{2,2,1} e_{g}^{0,0}$
Unpaired electrons = 1
3. Cr^3+ rightarrow 3d^3 Rightarrow t_2g^1,1,1 e_g^0,0$Cr^{3+} \rightarrow 3d^{3} \Rightarrow t_{2g}^{1,1,1} e_{g}^{0,0}$
Unpaired electrons = 3
4. Mn^3+ rightarrow 3d^4 Rightarrow t_2g^2,1,1 e_g^0,0$Mn^{3+} \rightarrow 3d^{4} \Rightarrow t_{2g}^{2,1,1} e_{g}^{0,0}$
Unpaired electrons = 2
### Step 1: Final Conclusion
Arranging in decreasing order of unpaired electrons:
Cr^3+ (3) > Mn^3+ (2) > Fe^3+ (1) > Co^3+ (0)$Cr^{3+} (3) > Mn^{3+} (2) > Fe^{3+} (1) > Co^{3+} (0)$
### Pattern Recognition
For low spin d^4$d^4$ to d^7$d^7$ configurations, pairing occurs early. d^3$d^3$ (Cr^3+$Cr^{3+}$) always has 3 unpaired electrons regardless of field strength, making it the highest among these.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 12 Chemistry: d and f Block Elements
