Consider a mixture 'X' which is made by dissolving 0.4 mol of [Co(NH3)5SO4]Br and 0.4 mol of [Co(NH3)5Br]SO4 in water to make 4 L of solution. When 2 L of mixture 'X' is allowed to react with excess of AgNO3, it forms precipitate 'Y'. The rest 2 L of mixture 'X' reacts with excess BaCl2 to form precipitate 'Z'. Which of the following statements is CORRECT.

The 4 L solution contains: 0.4 mol of [Co(NH3)5SO4]Br Gives 0.4 mol Br^- in solution. 0.4 mol of [Co(NH3)5Br]SO4 Gives 0.4 mol SO42- in solution. The solution is divided into two 2 L portions. Each 2 L portion will contain exactly half the moles of each species: Moles of Br^- in 2 L = 0.2 mol Moles of SO42- in 2 L = 0.2 mol When the first 2 L of the mixture reacts with excess AgNO3: Since there are 0.2 mol of free Br^- ions, 0.2 mol of AgBr (Y) will be formed. When the remaining 2 L of the mixture reacts with excess BaCl2: Since there are 0.2 mol of free SO42- ions, 0.2 mol of BaSO4 (Z) will be formed. Checking the options: (1) 0.2 mol of 'Z' is formed True. (2) 'Y' is BaSO4 and 'Z' is AgBr False, they are reversed. (3) 0.4 mol of 'Z' is formed False, it's 0.2 mol. (4) 0.1 mol of 'Y' is formed False, it's 0.2 mol. Only the counter-ions (outside the coordination sphere) ionize in water and participate in precipitation reactions. Always account for volume fractions when a solution is split. Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Some Basic Concepts of Chemistry

Consider a mixture 'X' which is made by dissolving 0.4 mol of [Co(NH3)5SO4]Br and 0.4 mol of [Co(NH3)5Br]SO4 in water to make 4 L of solution. When 2 L of mixture 'X' is allowed to react with excess o

The 4 L solution contains: 0.4 mol of [Co(NH3)5SO4]Br Gives 0.4 mol Br^- in solution. 0.4 mol of [Co(NH3)5Br]SO4 Gives 0.4 mol SO42- in solution. The solution is divided into two 2 L portions. Each 2 L portion will contain exactly half the moles of each species: Moles of Br^- in 2 L = 0.2 mol Moles of SO42- in 2 L = 0.2 mol When the first 2 L of the mixture reacts with excess AgNO3: Since there are 0.2 mol of free Br^- ions, 0.2 mol of AgBr (Y) will be formed. When the remaining 2 L of the mixtu

Consider a mixture 'X' which is made by dissolving 0.4 mol of [Co(NH3)5SO4]Br and 0.4 mol of [Co(NH3)5Br]SO4 in water to make 4 L of solution. When 2 L of mixture 'X' is allowed to react with excess of AgNO3, it forms precipitate 'Y'. The rest 2 L of mixture 'X' reacts with excess BaCl2 to form precipitate 'Z'. Which of the following statements is CORRECT.

The 4 L solution contains: 0.4 mol of [Co(NH3)5SO4]Br Gives 0.4 mol Br^- in solution. 0.4 mol of [Co(NH3)5Br]SO4 Gives 0.4 mol SO42- in solution. The solution is divided into two 2 L portions. Each 2 L portion will contain exactly half the moles of each species: Moles of Br^- in 2 L = 0.2 mol Moles of SO42- in 2 L = 0.2 mol When the first 2 L of the mixture reacts with excess AgNO3: Since there are 0.2 mol of free Br^- ions, 0.2 mol of AgBr (Y) will be formed. When the remaining 2 L of the mixture reacts with excess BaCl2: Since there are 0.2 mol of free SO42- ions, 0.2 mol of BaSO4 (Z) will be formed. Checking the options: (1) 0.2 mol of 'Z' is formed True. (2) 'Y' is BaSO4 and 'Z' is AgBr False, they are reversed. (3) 0.4 mol of 'Z' is formed False, it's 0.2 mol. (4) 0.1 mol of 'Y' is formed False, it's 0.2 mol. Only the counter-ions (outside the coordination sphere) ionize in water and participate in precipitation reactions. Always account for volume fractions when a solution is split. Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Some Basic Concepts of Chemistry

Consider a mixture 'X' which is made by dissolving 0.4 mol of [Co(NH3)5SO4]Br and 0.4 mol of [Co(NH3)5Br]SO4 in water to make 4 L of solution. When 2 L of mixture 'X' is allowed to react with excess o

The 4 L solution contains: 0.4 mol of [Co(NH3)5SO4]Br Gives 0.4 mol Br^- in solution. 0.4 mol of [Co(NH3)5Br]SO4 Gives 0.4 mol SO42- in solution. The solution is divided into two 2 L portions. Each 2 L portion will contain exactly half the moles of each species: Moles of Br^- in 2 L = 0.2 mol Moles of SO42- in 2 L = 0.2 mol When the first 2 L of the mixture reacts with excess AgNO3: Since there are 0.2 mol of free Br^- ions, 0.2 mol of AgBr (Y) will be formed. When the remaining 2 L of the mixtu

Consider a mixture 'X' which is made by dissolving 0.4 mol of [Co(NH3)5SO4]Br and 0.4 mol of [Co(NH3)5Br]SO4 in water…

Year	2026	2025	2024	Total
Questions	19	34	15	68

Q70 jee_main_2026_28_january_morning Magnetic Properties and Hybridization

The correct statement among the following is :

A. $[Ni(CN)₄]²⁻$ $[Ni(CN)_{4}]^{2-}$ and $[NiCl₄]²⁻$ $[NiCl_{4}]^{2-}$ are diamagnetic and $Ni(CO)₄$ $Ni(CO)_{4}$ is paramagnetic.
B. $Ni(CO)₄$ $Ni(CO)_{4}$ and $[NiCl₄]²⁻$ $[NiCl_{4}]^{2-}$ are diamagnetic and $[Ni(CN)₄]²⁻$ $[Ni(CN)_{4}]^{2-}$ is paramagnetic.
C. $Ni(CO)₄$ $Ni(CO)_{4}$ and $[Ni(CN)₄]²⁻$ $[Ni(CN)_{4}]^{2-}$ are diamagnetic and $[NiCl₄]²⁻$ $[NiCl_{4}]^{2-}$ is paramagnetic.
D. $Ni(CO)₄$ $Ni(CO)_{4}$ is diamagnetic and $[NiCl₄]²⁻$ $[NiCl_{4}]^{2-}$ and $[Ni(CN)₄]²⁻$ $[Ni(CN)_{4}]^{2-}$ are paramagnetic.

Solution

Core Logic

Analyze the oxidation state, electronic configuration, and ligand field strength for each Nickel complex: $\bullet$ $[Ni(CN)₄]²⁻$ $[Ni(CN)_4]^{2-}$: $Ni²⁺$ $Ni^{2+}$ is $3d⁸$ $3d^8$. $CN^-$ $CN^-$ is a strong field ligand. It causes pairing of electrons, resulting in $dsp²$ $dsp^2$ hybridization and 0 unpaired electrons (Diamagnetic). $\bullet$ $Ni(CO)₄$ $Ni(CO)_4$: $Ni⁰$ $Ni^0$ is $3d⁸ 4s²$ $3d^8 4s^2$. $CO$ $CO$ is a strong field ligand, forcing the $4s$ $4s$ electrons into the $3d$ $3d$ subshell, resulting in a $3d¹⁰$ $3d^{10}$ configuration. It undergoes $sp³$ $sp^3$ hybridization with 0 unpaired electrons (Diamagnetic). $\bullet$ $[NiCl₄]²⁻$ $[NiCl_4]^{2-}$: $Ni²⁺$ $Ni^{2+}$ is $3d⁸$ $3d^8$. $Cl^-$ $Cl^-$ is a weak field ligand. No pairing occurs, resulting in $sp³$ $sp^3$ hybridization with 2 unpaired electrons (Paramagnetic).

Final Conclusion

$Ni(CO)₄$ $Ni(CO)_4$ and $[Ni(CN)₄]²⁻$ $[Ni(CN)_4]^{2-}$ are diamagnetic, while $[NiCl₄]²⁻$ $[NiCl_4]^{2-}$ is paramagnetic.

Pattern Recognition

Ni with strong ligands ( $CN^-, CO$ $CN^-, CO$) collapses into paired diamagnetic states. Ni with weak halogens ( $Cl^-$ $Cl^-$) stays paramagnetic and tetrahedral.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q73 jee_main_2026_28_january_morning Isomerism in Coordination Compounds

X is the number of geometrical isomers exhibited by

[Pt(NH₃)(H₂O)BrCl]

. Y is the number of optically inactive isomer(s) exhibited by

[CrCl₂(ox)₂]³⁻

Z is the number of geometrical isomers exhibited by

[Co(NH₃)₃(NO₂)₃]

The value of

X + Y + Z

is _____.

Numerical Answer. Answer: 6 to 6

Solution

Step 1: Evaluate X

$[Pt(NH₃)(H₂O)BrCl]$ $[Pt(NH_3)(H_2O)BrCl]$ is a square planar complex of the type $[Mabcd]$ $[Mabcd]$. Such complexes exhibit exactly 3 geometrical isomers (by fixing one ligand and placing the other three opposite to it, creating two cis-like and one trans-like variations relative to the fixed ligand).\nThus, $X = 3$ $X = 3$.

Step 2: Evaluate Y

$[CrCl₂(ox)₂]³⁻$ $[CrCl_2(ox)_2]^{3-}$ is an octahedral complex of the type $[M(AA)₂b₂]$ $[M(AA)_2b_2]$. It has two geometrical isomers: cis and trans.\n$\bullet$ The cis-isomer lacks a plane of symmetry and is optically active (exists as a pair of enantiomers).\n$\bullet$ The trans-isomer has a plane of symmetry and is optically inactive (meso).\nThe question asks for the number of optically inactive isomers. Thus, $Y = 1$ $Y = 1$.

Step 3: Evaluate Z

$[Co(NH₃)₃(NO₂)₃]$ $[Co(NH_3)_3(NO_2)_3]$ is an octahedral complex of the type $[Ma₃b₃]$ $[Ma_3b_3]$. It exhibits exactly 2 geometrical isomers: Facial (fac) and Meridional (mer).\nThus, $Z = 2$ $Z = 2$.

Final Conclusion

$X + Y + Z = 3 + 1 + 2 = 6$ $X + Y + Z = 3 + 1 + 2 = 6$

Pattern Recognition

$[Mabcd]$ $[Mabcd]$ planar = 3 G.I. $[Ma₃b₃]$ $[Ma_3b_3]$ octahedral = 2 G.I. (fac/mer). $[M(AA)₂b₂]$ $[M(AA)_2b_2]$ octahedral trans isomer = ALWAYS optically inactive.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q69 jee_main_2026_28_january_evening Magnetic Moments Of Complexes

The correct increasing order of spin-only magnetic moment values of the complex ions

[MnBr₄]²⁻

(A),

[Cu(H₂O)₆]²⁺

(B),

[Ni(CN₄)]²⁻

(C) and

[Ni(H₂O)₆]²⁺

(D) is:

A. $(1) A = B < C < D$ $(1)\text{ } A = B < C < D$
B. $(2) A = B < D < C$ $(2)\text{ } A = B < D < C$
C. $(3) C = D < B < A$ $(3)\text{ } C = D < B < A$
D. $(4) C < B < D < A$ $(4)\text{ } C < B < D < A$

Solution

Related Formula

μ = \sqrt(n(n+2)) B.M.

where $n$ $n$ is the number of unpaired electrons.

Core Logic

(A) $[MnBr₄]²⁻$ $[MnBr_{4}]^{2-}$: $Mn²⁺$ $Mn^{2+}$ has $3d⁵$ $3d^5$ configuration. $Br^-$ $Br^-$ is a weak field ligand (WFL). It forms a tetrahedral complex with 5 unpaired electrons ( $n=5$ $n=5$).

(B) $[Cu(H₂O)₆]²⁺$ $[Cu(H_{2}O)_{6}]^{2+}$: $Cu²⁺$ $Cu^{2+}$ has $3d⁹$ $3d^9$ configuration. It has exactly 1 unpaired electron ( $n=1$ $n=1$).

(C) $[Ni(CN₄)]²⁻$ $[Ni(CN_{4})]^{2-}$: $Ni²⁺$ $Ni^{2+}$ has $3d⁸$ $3d^8$ configuration. $CN^-$ $CN^-$ is a strong field ligand (SFL). It forms a square planar complex ( $dsp²$ $dsp^2$) forcing pairing. Unpaired electrons $n=0$ $n=0$.

(D) $[Ni(H₂O)₆]²⁺$ $[Ni(H_{2}O)_{6}]^{2+}$: $Ni²⁺$ $Ni^{2+}$ has $3d⁸$ $3d^8$ configuration. $H₂O$ $H_2O$ is a weak field ligand (WFL). In an octahedral field, it leaves 2 unpaired electrons ( $n=2$ $n=2$).

Step 1: Final Conclusion

Unpaired electrons (n): (C) = 0 (B) = 1 (D) = 2 (A) = 5 Increasing order of magnetic moment: C < B < D < A.

Pattern Recognition

Nickel + Strong Ligand ( $CN^-$ $CN^-$) = Square Planar, Diamagnetic ( $n=0$ $n=0$). Nickel + Weak Ligand ( $H₂O$ $H_2O$) = Octahedral, Paramagnetic ( $n=2$ $n=2$). Copper (II) is always $d⁹$ $d^9$ ( $n=1$ $n=1$). Manganese (II) with WFL is high spin $d⁵$ $d^5$ ( $n=5$ $n=5$).

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q75 jee_main_2026_28_january_evening Crystal Field Theory

The number of isoelectronic species among

Sc³⁺

,

Cr²⁺

,

Mn³⁺

,

Co³⁺

and

Fe³⁺

is 'n'. If 'n' moles of AgCl is formed during the reaction of complex with formula

CoCl₃(en)₂NH₃

with excess of

AgNO₃

solution, then the number of electrons present in the

t2g

orbital of the complex is ____.

Numerical Answer. Answer: 6 to 6

Solution

Core Logic

Step 1: Find 'n' (number of isoelectronic species) Electrons count: $Sc³⁺ = 21 - 3 = 18$ $Sc^{3+} = 21 - 3 = 18$ $Cr²⁺ = 24 - 2 = 22$ $Cr^{2+} = 24 - 2 = 22$ $Mn³⁺ = 25 - 3 = 22$ $Mn^{3+} = 25 - 3 = 22$ $Co³⁺ = 27 - 3 = 24$ $Co^{3+} = 27 - 3 = 24$ $Fe³⁺ = 26 - 3 = 23$ $Fe^{3+} = 26 - 3 = 23$ The isoelectronic species are $Cr²⁺$ $Cr^{2+}$ and $Mn³⁺$ $Mn^{3+}$ (both have 22 electrons). So, $n = 2$ $n = 2$.

Step 2: Complex formulation. Reaction produces 'n' moles of AgCl, so 2 moles of $Cl^-$ $Cl^-$ are ionizable outside the coordination sphere. Complex formula is $[Co(en)₂(NH₃)Cl]Cl₂$ $[Co(en)_2(NH_3)Cl]Cl_2$.

Step 3: Crystal field splitting. The central metal ion is $Co³⁺$ $Co^{3+}$ ( $3d⁶$ $3d^6$ configuration). The ligands are $en$ $en$ (strong field) and $NH₃$ $NH_3$ (strong field), which will cause complete pairing of electrons. $Cl^-$ $Cl^-$ inside the sphere is weak, but the overwhelming presence of $en$ $en$ and $NH₃$ $NH_3$ combined with $+3$ $+3$ oxidation state makes it a strong field (low spin) complex. $Co³⁺$ $Co^{3+}$ ( $3d⁶$ $3d^6$) low spin splitting: $t2g⁶ eg⁰$ $t_{2g}^{6} \; e_g^{0}$.

Step 1: Final Conclusion

The number of electrons present in the $t2g$ $t_{2g}$ orbital of the complex is 6.

Pattern Recognition

$Co³⁺$ $Co^{3+}$ in predominantly N-donor environment forms low spin octahedral complexes.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Structure of Atom

Q27 jee_main_2025_02_april_evening Crystal Field Stabilization Energy

The d-orbital electronic configuration of the complex among

[Co(en)₃]³⁺

,

[CoF₆]³⁻

,

[Mn(H₂O)₆]²⁺

and

[Zn(H₂O)₆]²⁺

that has the highest CFSE is:

A. $t2g⁶eg⁰$
B. $t2g⁶eg⁴$
C. $t2g³eg²$
D. $t2g⁴eg²$

Solution

Related Formula

CFSE = ( -0.4 n_t2g + 0.6 n_eg ) Δₒ + nₚ P

Core Logic

Crystal Field Stabilization Energy (CFSE) is maximized (becomes most negative) when electrons populate lower-energy $t2g$ $\mathrm{t_{2g}}$ orbitals and stay out of higher-energy $eg$ $\mathrm{e_g}$ orbitals. This is favored by strong-field ligands (SFL) that induce large $Δₒ$ $\Delta_o$ splitting, leading to low-spin configurations.

Step 1: Analyze Ligand Strength and Configuration

Let us check each of the given complexes:

$[Co(en)₃]³⁺$ $[\mathrm{Co(en)_3}]^{3+}$: Here $Co³⁺$ $\mathrm{Co^{3+}}$ has a $3d⁶$ $3\mathrm{d^6}$ configuration. Since ethylenediamine ( $en$ $\mathrm{en}$) is a strong-field ligand, it causes pairing of all $6$ $6$ electrons in the $t2g$ $\mathrm{t_{2g}}$ subshell. The configuration is $t2g⁶eg⁰$ .

d-orbital splitting diagram for low-spin Co3+ complex with t2g6 configuration

$[CoF₆]³⁻$ $[\mathrm{CoF_6}]^{3-}$: $Co³⁺$ $\mathrm{Co^{3+}}$ is $3d⁶$ $3\mathrm{d^6}$. Since $F^-$ $\mathrm{F^-}$ is a weak-field ligand (WFL), no pairing occurs. The configuration is $t2g⁴eg²$ .
$[Mn(H₂O)₆]²⁺$ $[\mathrm{Mn(H_2O)_6}]^{2+}$: $Mn²⁺$ $\mathrm{Mn^{2+}}$ is $3d⁵$ $3\mathrm{d^5}$. Since $H₂O$ $\mathrm{H_2O}$ is a weak-field ligand, the configuration is high-spin: $t2g³eg²$ .
$[Zn(H₂O)₆]²⁺$ $[\mathrm{Zn(H_2O)_6}]^{2+}$: $Zn²⁺$ $\mathrm{Zn^{2+}}$ is $3 d¹⁰$ $3\mathrm{d^{10}}$. The d-subshell is fully filled, yielding $t2g⁶eg⁴$ .

Step 2: Compare CFSE Values

Calculating CFSE (neglecting pairing energy term for simplicity):

For $[Co(en)₃]³⁺$ $[\mathrm{Co(en)_3}]^{3+}$: $CFSE = 6 \times (-0.4 Δₒ) = -2.4 Δₒ$
For $[CoF₆]³⁻$ $[\mathrm{CoF_6}]^{3-}$: $CFSE = [4(-0.4) + 2(0.6)] Δₒ = -0.4 Δₒ$
For $[Mn(H₂O)₆]²⁺$ $[\mathrm{Mn(H_2O)_6}]^{2+}$: $CFSE = [3(-0.4) + 2(0.6)] Δₒ = 0$
For $[Zn(H₂O)₆]²⁺$ $[\mathrm{Zn(H_2O)_6}]^{2+}$: $CFSE = [6(-0.4) + 4(0.6)] Δₒ = 0$

Hence, $[Co(en)₃]³⁺$ $[\mathrm{Co(en)_3}]^{3+}$ has the highest crystal field stabilization energy, corresponding to the d-orbital electronic configuration $t2g⁶eg⁰$ .

Pattern Recognition

For octahedral complexes of $d⁶$ $\mathrm{d^6}$ metals, a low-spin configuration ( $t2g⁶eg⁰$ ) achieves the theoretical maximum orbital stabilization since the $eg$ $\mathrm{e_g}$ levels are completely empty and $t2g$ $\mathrm{t_{2g}}$ is fully filled.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Solution & Explanation

Core Logic

Step 1: Reaction with Silver Nitrate

Step 2: Reaction with Barium Chloride

Step 3: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Coordination Compounds Previous-Year Questions — Page 4

Solution

Core Logic

Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Step 1: Evaluate X

Step 2: Evaluate Y

Step 3: Evaluate Z

Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analyze Ligand Strength and Configuration

Step 2: Compare CFSE Values

Pattern Recognition

Chapter Mix

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