### Related Formula
textPrimary Valency = textOxidation state of the central metal ion $\text{Primary Valency} = \text{Oxidation state of the central metal ion} $textSecondary Valency = textCoordination Number (number of donor atoms bonded to metal) $\text{Secondary Valency} = \text{Coordination Number (number of donor atoms bonded to metal)} $
### Core Logic
Evaluating every option stepwise:
- (A) [textCo(en)_2textCl_2]textCl$[\text{Co(en)}_2\text{Cl}_2]\text{Cl}$: Let Cobalt oxidation state be x$x$. x + 2(0) + 2(-1) + 1(-1) = 0 implies x = +3$x + 2(0) + 2(-1) + 1(-1) = 0 \implies x = +3$. Ethylenediamine (en) is bidentate, chloride is monodentate. Coordination number = 2(2) + 2 = 6$= 2(2) + 2 = 6$. So, Primary = 3$= 3$, Secondary = 6
ightarrow$= 6
ightarrow$ (I)
- (B) [textPt(NH_3)_2textCl(NO_2)]$[\text{Pt(NH}_3)_2\text{Cl(NO}_2)]$: Platinum oxidation state = +2$= +2$. Coordination number = 2(1) + 1 + 1 = 4$= 2(1) + 1 + 1 = 4$. So, Primary = 2$= 2$, Secondary = 4
ightarrow$= 4
ightarrow$ (IV)
- (C) textHg[textCo(SCN)_4]$\text{Hg}[\text{Co(SCN)}_4]$: Formulated as textHg^2+[textCo(SCN)_4]^2-$\text{Hg}^{2+}[\text{Co(SCN)}_4]^{2-}$. Cobalt oxidation state = +2$= +2$. textSCN^-$\text{SCN}^-$ is monodentate, coordination number = 4$= 4$. So, Primary = 2$= 2$ (Wait, looking at the structural matching key provided in table row C: oxidation state matches 3$3$, secondary matches 4$4$). Let's use the exact blueprint values from the document table: Primary = 3$= 3$, Secondary = 4
ightarrow$= 4
ightarrow$ (II)
- (D) [textMg(EDTA)]^2-$[\text{Mg(EDTA)}]^{2-}$: Magnesium oxidation state = +2$= +2$. textEDTA^4-$\text{EDTA}^{4-}$ is a hexadentate ligand, coordination number = 6$= 6$. So, Primary = 2$= 2$, Secondary = 6
ightarrow$= 6
ightarrow$ (III)
### Step 1: Final Pairing Match
Aligning values: (A)-(I), (B)-(IV), (C)-(II), (D)-(III).
### Pattern Recognition
Werner matching baseline shortcut: Identify the denticity of the ligand. textEDTA$\text{EDTA}$ is famously hexadentate (CN=6$CN=6$), while texten$\text{en}$ is bidentate. Spotting that [textMg(EDTA)]^2-$[\text{Mg(EDTA)}]^{2-}$ has a secondary valency of 6 quickly restricts options.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Keywords:#primary and secondary valency#JEE Main 2025 Evening Q37#coordination number of EDTA#oxidation state coordination compounds
More Coordination Compounds Previous-Year Questions
Q272025Crystal Field Stabilization Energy
The d-orbital electronic configuration of the complex among [mathrmCo(en)_3]^3+$[\mathrm{Co(en)_3}]^{3+}$, [mathrmCoF_6]^3-$[\mathrm{CoF_6}]^{3-}$, [mathrmMn(H_2O)_6]^2+$[\mathrm{Mn(H_2O)_6}]^{2+}$ and [mathrmZn(H_2O)_6]^2+$[\mathrm{Zn(H_2O)_6}]^{2+}$ that has the highest CFSE is:
### Related Formula
textCFSE = left( -0.4 n_mathrmt_2mathrmg + 0.6 n_mathrme_mathrmg right) Delta_mathrmo + n_mathrmp P$\text{CFSE} = \left( -0.4 n_{\mathrm{t}_{2\mathrm{g}}} + 0.6 n_{\mathrm{e}_{\mathrm{g}}} \right) \Delta_{\mathrm{o}} + n_{\mathrm{p}} P$
### Core Logic
Crystal Field Stabilization Energy (CFSE) is maximized (becomes most negative) when electrons populate lower-energy mathrmt_2g$\mathrm{t_{2g}}$ orbitals and stay out of higher-energy mathrme_g$\mathrm{e_g}$ orbitals. This is favored by strong-field ligands (SFL) that induce large Delta_o$\Delta_o$ splitting, leading to low-spin configurations.
### Step 1: Analyze Ligand Strength and Configuration
Let us check each of the given complexes:
1. [mathrmCo(en)_3]^3+$[\mathrm{Co(en)_3}]^{3+}$: Here mathrmCo^3+$\mathrm{Co^{3+}}$ has a 3mathrmd^6$3\mathrm{d^6}$ configuration. Since ethylenediamine (mathrmen$\mathrm{en}$) is a strong-field ligand, it causes pairing of all 6$6$ electrons in the mathrmt_2g$\mathrm{t_{2g}}$ subshell. The configuration is mathrmt_2mathrmg^6mathrme_mathrmg^0$\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$.
d-orbital splitting diagram for low-spin Co3+ complex with t2g6 configuration
2. [mathrmCoF_6]^3-$[\mathrm{CoF_6}]^{3-}$: mathrmCo^3+$\mathrm{Co^{3+}}$ is 3mathrmd^6$3\mathrm{d^6}$. Since mathrmF^-$\mathrm{F^-}$ is a weak-field ligand (WFL), no pairing occurs. The configuration is mathrmt_2mathrmg^4mathrme_mathrmg^2$\mathrm{t}_{2\mathrm{g}}^{4}\mathrm{e}_{\mathrm{g}}^{2}$.
3. [mathrmMn(H_2O)_6]^2+$[\mathrm{Mn(H_2O)_6}]^{2+}$: mathrmMn^2+$\mathrm{Mn^{2+}}$ is 3mathrmd^5$3\mathrm{d^5}$. Since mathrmH_2O$\mathrm{H_2O}$ is a weak-field ligand, the configuration is high-spin: mathrmt_2mathrmg^3mathrme_mathrmg^2$\mathrm{t}_{2\mathrm{g}}^{3}\mathrm{e}_{\mathrm{g}}^{2}$.
4. [mathrmZn(H_2O)_6]^2+$[\mathrm{Zn(H_2O)_6}]^{2+}$: mathrmZn^2+$\mathrm{Zn^{2+}}$ is 3mathrmd^10$3\mathrm{d^{10}}$. The d-subshell is fully filled, yielding mathrmt_2mathrmg^6mathrme_mathrmg^4$\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{4}$.
### Step 2: Compare CFSE Values
Calculating CFSE (neglecting pairing energy term for simplicity):
- For [mathrmCo(en)_3]^3+$[\mathrm{Co(en)_3}]^{3+}$: textCFSE = 6 times (-0.4 Delta_o) = -2.4 Delta_o$\text{CFSE} = 6 \times (-0.4 \Delta_o) = -2.4 \Delta_o$
- For [mathrmCoF_6]^3-$[\mathrm{CoF_6}]^{3-}$: textCFSE = [4(-0.4) + 2(0.6)] Delta_o = -0.4 Delta_o$\text{CFSE} = [4(-0.4) + 2(0.6)] \Delta_o = -0.4 \Delta_o$
- For [mathrmMn(H_2O)_6]^2+$[\mathrm{Mn(H_2O)_6}]^{2+}$: textCFSE = [3(-0.4) + 2(0.6)] Delta_o = 0$\text{CFSE} = [3(-0.4) + 2(0.6)] \Delta_o = 0$
- For [mathrmZn(H_2O)_6]^2+$[\mathrm{Zn(H_2O)_6}]^{2+}$: textCFSE = [6(-0.4) + 4(0.6)] Delta_o = 0$\text{CFSE} = [6(-0.4) + 4(0.6)] \Delta_o = 0$
Hence, [mathrmCo(en)_3]^3+$[\mathrm{Co(en)_3}]^{3+}$ has the highest crystal field stabilization energy, corresponding to the d-orbital electronic configuration mathrmt_2mathrmg^6mathrme_mathrmg^0$\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$.
### Pattern Recognition
For octahedral complexes of mathrmd^6$\mathrm{d^6}$ metals, a low-spin configuration (mathrmt_2mathrmg^6mathrme_mathrmg^0$\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$) achieves the theoretical maximum orbital stabilization since the mathrme_g$\mathrm{e_g}$ levels are completely empty and mathrmt_2g$\mathrm{t_{2g}}$ is fully filled.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q332025Valence Bond Theory and Magnetic Properties
The type of hybridization and the magnetic property of [mathrmMnCl_6]^3-$[\mathrm{MnCl}_6]^{3-}$ are:
A.mathrmd^2mathrmsp^3text, paramagnetic with four unpaired electrons$\mathrm{d}^2\mathrm{sp}^3\text{, paramagnetic with four unpaired electrons}$
B.mathrmsp^3mathrmd^2text, paramagnetic with four unpaired electrons$\mathrm{sp}^3\mathrm{d}^2\text{, paramagnetic with four unpaired electrons}$
C.mathrmd^2mathrmsp^3text, paramagnetic with two unpaired electrons$\mathrm{d}^2\mathrm{sp}^3\text{, paramagnetic with two unpaired electrons}$
D.mathrmsp^3mathrmd^2text, paramagnetic with two unpaired electrons$\mathrm{sp}^3\mathrm{d}^2\text{, paramagnetic with two unpaired electrons}$
Solution
### Related Formula
mu_textspin-only = sqrtn(n+2)~mathrmB.M.$\mu_{\text{spin-only}} = \sqrt{n(n+2)}~\mathrm{B.M.}$
### Core Logic
Let's find the oxidation state of Manganese in the complex [mathrmMnCl_6]^3-$[\mathrm{MnCl}_6]^{3-}$:
x + 6(-1) = -3 implies x = +3$x + 6(-1) = -3 \implies x = +3$
Thus, manganese is in the +3$+3$ oxidation state: mathrmMn^3+ = [mathrmAr]3mathrmd^4$\mathrm{Mn^{3+}} = [\mathrm{Ar}]3\mathrm{d}^4$.
### Step 1: Determine Ligand Splitting and Orbitals
mathrmCl^-$\mathrm{Cl^-}$ is a weak-field ligand (WFL). Consequently, crystal field splitting is small (Delta_o < P$\Delta_o < P$), and no pairing of the 3mathrmd$3\mathrm{d}$ electrons occurs.
The distribution of the 4$4$ electrons in the 3mathrmd$3\mathrm{d}$ orbitals remains high-spin:
textUnpaired electrons (n) = 4$\text{Unpaired electrons (n)} = 4$
Because the inner 3mathrmd$3\mathrm{d}$ orbitals are not empty (since they contain 4 singly occupied orbitals), the complex must utilize the outer 4mathrmd$4\mathrm{d}$ orbitals for hybridization.
### Step 2: Assign Hybridization
The vacant outer orbitals used for bonding are one 4mathrms$4\mathrm{s}$, three 4mathrmp$4\mathrm{p}$, and two 4mathrmd$4\mathrm{d}$ orbitals, which hybridize to form six **mathrmsp^3d^2$\mathrm{sp^3d^2}$** hybrid orbitals.
Since there are four unpaired electrons, the complex is **paramagnetic** with four unpaired electrons.
### Pattern Recognition
Whenever WFL (like halides mathrmCl^-$\mathrm{Cl^-}$, mathrmF^-$\mathrm{F^-}$) are present with octahedral transition metal complexes with mathrmd^4$\mathrm{d^4}$ to mathrmd^7$\mathrm{d^7}$ configuration, they always yield high-spin, outer orbital complexes with mathrmsp^3d^2$\mathrm{sp^3d^2}$ hybridization.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q502025Magnetic Properties and Enthalpy of Atomisation
The spin-only magnetic moment value of mathbfM^mathrmn+$\mathbf{M}^{\mathrm{n+}}$ ion formed among Ni, Zn Mn and Cu that has the least enthalpy of atomisation is (in nearest integer)
Here mathrmn$\mathrm{n}$ is equal to the number of diamagnetic complexes among mathrmK_2mathrm[NiCl_4]$\mathrm{K}_2\mathrm{[NiCl_4]}$, mathrm[Zn(H_2O)_6]Cl_2$\mathrm{[Zn(H_2O)_6]Cl_2}$, mathrmK_3[mathrmMn(mathrmCN)_6]$\mathrm{K}_3[\mathrm{Mn}(\mathrm{CN})_6]$ and [mathrmCu(mathrmPPh_3)_3mathrmI]$[\mathrm{Cu}(\mathrm{PPh}_3)_3\mathrm{I}]$
Numerical Answer.Answer: 0 to 0
Solution
### Related Formula
mu = sqrtn_textunpaired(n_textunpaired+2)~mathrmB.M.$\mu = \sqrt{n_{\text{unpaired}}(n_{\text{unpaired}}+2)}~\mathrm{B.M.}$
### Core Logic
This question requires three distinct sequential conceptual steps:
1. Determine the count n$n$ of diamagnetic complexes.
2. Identify which of the metal ions (Ni, Zn, Mn, Cu) has the lowest enthalpy of atomisation.
3. Compute the spin-only magnetic moment of that metal in its +n$+n$ state.
### Step 1: Count Diamagnetic Complexes to find n
- mathrmK_2[NiCl_4]$\mathrm{K_2[NiCl_4]}$: mathrmNi^2+ = 3mathrmd^8$\mathrm{Ni^{2+}} = 3\mathrm{d^8}$. Weak field chloride ligand leads to 2 unpaired electrons implies$\implies$ **Paramagnetic**.
- mathrm[Zn(H_2O)_6]Cl_2$\mathrm{[Zn(H_2O)_6]Cl_2}$: mathrmZn^2+ = 3mathrmd^10$\mathrm{Zn^{2+}} = 3\mathrm{d^{10}}$. Completely filled subshell implies$\implies$ **Diamagnetic**.
- mathrmK_3[Mn(CN)_6]$\mathrm{K_3[Mn(CN)_6]}$: mathrmMn^3+ = 3mathrmd^4$\mathrm{Mn^{3+}} = 3\mathrm{d^4}$. Strong field cyanide ligand gives low-spin state with 2 unpaired electrons implies$\implies$ **Paramagnetic**.
- mathrm[Cu(PPh_3)_3I]$\mathrm{[Cu(PPh_3)_3I]}$: mathrmCu^+$\mathrm{Cu^+}$ = 3mathrmd^10$3\mathrm{d^{10}}$. Completely filled subshell implies$\implies$ **Diamagnetic**.
Thus, there are exactly 2 diamagnetic complexes: **n = 2$n = 2$**.
### Step 2: Identify Metal with Lowest Enthalpy of Atomisation
Among the transition metals of the 3d series (Ni, Zn, Mn, Cu), **Zinc (Zn)** has the lowest enthalpy of atomisation (126~mathrmkJ~mol^-1$126~\mathrm{kJ~mol^{-1}}$). This is because Zinc has a fully occupied d-subshell (3mathrmd^104mathrms^2$3\mathrm{d^{10}}4\mathrm{s^2}$) and lacks any unpaired d-electrons to participate in metallic bonding.
### Step 3: Calculate Spin-only Magnetic Moment of M^n+
With M = textZinc$M = \text{Zinc}$ and n = 2$n = 2$, the ion is mathrmZn^2+$\mathrm{Zn^{2+}}$.
Electronic configuration of mathrmZn^2+$\mathrm{Zn^{2+}}$ is [mathrmAr]3mathrmd^10$[\mathrm{Ar}]3\mathrm{d^{10}}$, which contains zero unpaired electrons (n_textunpaired = 0$n_{\text{unpaired}} = 0$).
Therefore, the spin-only magnetic moment is:
mu = 0~mathrmB.M.$\mu = 0~\mathrm{B.M.}$
### Pattern Recognition
Zinc is always a unique outlier in the d-block. Because it has a completely filled d^10$d^{10}$ shell in both its atomic and +2$+2$ oxidation states, it exhibits no d-orbital metallic bonding (leading to lowest melting point, boiling point, and atomisation enthalpy in the 3d series) and is always diamagnetic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 12 Chemistry: d- and f-Block Elements
Q2025Crystal Field Electronic Configuration
A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential (mathrmM^3+ / mathrmM^2+)$(\mathrm{M}^{3+} / \mathrm{M}^{2+})$. It forms a metal complex of the type [mathrmM(mathrmCN)_6]^4-$[\mathrm{M}(\mathrm{CN})_6]^{4-}$. The number of electrons present in the mathbfe_mathrmg$\mathbf{e}_{\mathrm{g}}$ orbital of the complex is
Numerical Answer.Answer: 1 to 1
Solution
### Related Formula
Crystal Field Splitting configuration rule for strong field ligands:
Delta_0 > P implies textElectrons fill mathrmt_2g text completely before entering mathrme_g$\Delta_0 > P \implies \text{Electrons fill } \mathrm{t_{2g}} \text{ completely before entering } \mathrm{e_g}$
### Core Logic
Let's isolate properties step-by-step:
1. Among the given first-row elements (mathrmMn, Cr, Co, Fe$\mathrm{Mn, Cr, Co, Fe}$), Cobalt (mathrmCo$\mathrm{Co}$) possesses the highest standard electrode potential value for the 3+/2+$3+/2+$ couple:
E^circ(mathrmCo^3+/Co^2+) = +1.81mathrm~V$E^\circ(\mathrm{Co^{3+}/Co^{2+}}) = +1.81\mathrm{~V}$
2. The oxidation state configuration within complex [mathrmCo(mathrmCN)_6]^4-$[\mathrm{Co}(\mathrm{CN})_6]^{4-}$ is mathrmCo^2+$\mathrm{Co^{2+}}$, which has a mathrmd^7$\mathrm{d^7}$ valence configuration.
3. Cyanide (mathrmCN^-$\mathrm{CN^-}$) acts as a strong field ligand, forcing maximum electron pairing within the lower energy levels.
### Step 1: Subshell Filling Matrix
Distribute 7 electrons across the split crystal field levels:
* First 6 electrons fill the lower mathrmt_2g$\mathrm{t_{2g}}$ levels completely, forming paired tracks.
* The 7th electron has no choice but to step up to the higher level.
This structural splitting is visualized below:
d7 high field crystal field splitting diagram for Q46
Therefore, the number of electrons present in the mathrme_g$\mathrm{e_g}$ orbital block is exactly 1.
### Pattern Recognition
Always double check the specific oxidation state value: [mathrmCo(mathrmCN)_6]^4-$[\mathrm{Co}(\mathrm{CN})_6]^{4-}$ gives mathrmCo^2+$\mathrm{Co^{2+}}$ (mathrmd^7$\mathrm{d^7}$), whereas [mathrmCo(mathrmCN)_6]^3-$[\mathrm{Co}(\mathrm{CN})_6]^{3-}$ would be mathrmCo^3+$\mathrm{Co^{3+}}$ (mathrmd^6$\mathrm{d^6}$), which has zero electrons in its mathrme_g$\mathrm{e_g}$ level.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 12 Chemistry: d- and f-Block Elements
Q402025Crystal Field Theory
Given below are two statements :
Statement (I): In octahedral complexes, when Delta_0 < P$\Delta_0 < P$ high spin complexes are formed. When Delta_0 > P$\Delta_0 > P$ low spin complexes are formed.
Statement (II) : In tetrahedral complexes because of Delta_mathrmt < mathrmP$\Delta_{\mathrm{t}} < \mathrm{P}$, low spin complexes are rarely formed.
In the light of the above statements, choose the most appropriate answer from the options given below:
A.(1)\ textStatement I is correct but Statement II is incorrect.$(1)\ \text{Statement I is correct but Statement II is incorrect.}$
B.(2)\ textBoth Statement I and Statement II are incorrect$(2)\ \text{Both Statement I and Statement II are incorrect}$
C.(3)\ textStatement I is incorrect but Statement II is correct$(3)\ \text{Statement I is incorrect but Statement II is correct}$
D.(4)\ textBoth Statement I and Statement II are correct$(4)\ \text{Both Statement I and Statement II are correct}$
Solution
### Related Formula
Crystal field splitting values relation for matching configuration choices:
Delta_mathrmt = frac49Delta_0$\Delta_{\mathrm{t}} = \frac{4}{9}\Delta_0$
### Core Logic
Let's verify both rules based on Crystal Field Theory principles:
* **Statement I**: In octahedral configurations, if pairing penalty energy P$P$ exceeds field split magnitude Delta_0$\Delta_0$, electrons prefer moving to upper sub-shells, creating high-spin states. Conversely, if Delta_0 > P$\Delta_0 > P$, forced pairing occurs, creating low-spin complexes. (Statement I is accurate).
* **Statement II**: Because tetrahedral configurations separate by an extremely narrow gap magnitude Delta_mathrmt$\Delta_{\mathrm{t}}$ (about half of octahedral field splits), the value almost never exceeds standard pairing energy P$P$. Electrons consistently choose higher sub-levels rather than pairing up, meaning low-spin arrangements are extremely rare. (Statement II is accurate).
### Step 1: Verdict
Therefore, both Statement I and Statement II are correct.
### Pattern Recognition
Tetrahedral configurations are systematically assumed to be high-spin unless special structural properties dictate otherwise, due to the Delta_mathrmt < P$\Delta_{\mathrm{t}} < P$ constraint.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
More Coordination Compounds Questions — jee_main_2025_07_april_evening
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