Which of the following statements are correct about Zn, Cd and mathrmHg$\mathrm{Hg}$ ?
A. They exhibit high enthalpy of atomization as the d-subshell is full.
B. Zn and Cd do not show variable oxidation state while Hg shows +mathrmI$+\mathrm{I}$ and +mathrmII$+\mathrm{II}$.
C. Compounds of Zn, Cd and Hg are paramagnetic in nature.
D. Zn, Cd and Hg are called soft metals.
Choose the most appropriate from the options given below:
A.B, D only
B.B, C only
C.A, D only
D.C, D only
Solution & Explanation
### Related Formula
textGeneral configuration of Group 12: (n-1)d^10 ns^2$$\text{General configuration of Group 12: } (n-1)d^{10} ns^2$$
### Core Logic
Analyzing each statement based on inorganic chemistry principles:
* **Statement A is false**: Because their d$d$-subshell is completely full (d^10$d^{10}$), these elements do not form strong metallic bonds. As a result, they exhibit the *lowest* enthalpy of atomization in their respective periods.
* **Statement B is true**: textZn$\text{Zn}$ and textCd$\text{Cd}$ show only a stable +2$+2$ oxidation state, whereas textHg$\text{Hg}$ exhibits variable states forming both +1$+1$ (as textHg_2^2+$\text{Hg}_2^{2+}$) and +2$+2$.
* **Statement C is false**: With a fully paired d^10$d^{10}$ subshell, their compounds lack unpaired electrons and are explicitly diamagnetic.
* **Statement D is true**: Due to weak metallic bonds, these elements have low melting points and are classified as soft metals.
### Step 1: Selection Verification
Statements B and D are true, matching choice (1).
### Pattern Recognition
Group 12 metals have a full d^10$d^{10}$ subshell, leading to exceptionally weak metallic bonding, low enthalpies of atomization, and diamagnetic characteristics.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
### Core Logic
The noble gas configuration of Xenon (Z=54$Z=54$) provides the primary core layout. For Neodymium (Z=60$Z=60$), the 6 remaining valence electrons distribute into the inner 4textf$4\text{f}$ orbital subshell rather than filling the 5textd$5\text{d}$ subshell due to shielding effects. This results in an absolute atomic ground state electronic configuration of text[Xe] 4textf^4 6texts^2$\text{[Xe]} 4\text{f}^4 6\text{s}^2$.
### Pattern Recognition
Lanthanide filling sequences generally bypass 5d$5d$ progression except for specific exceptions (La, Gd, Lu).
### Chapter Mix
Class 12 Chemistry: d-and f-Block Elements
Q63jee_main_2024_29_jan_morningPotassium Dichromate and Chromyl Chloride Test
In chromyl chloride test for confirmation of Cl^-$Cl^-$ ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10\%$10\%$H_2O_2$H_2O_2$ turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is
A.+6$+6$
B.+5$+5$
C.+10$+10$
D.+3$+3$
Solution
### Core Logic
The reaction sequence for the chromyl chloride test is:
Cl^- + K_2Cr_2O_7 + H_2SO_4 rightarrow CrO_2Cl_2$$Cl^- + K_2Cr_2O_7 + H_2SO_4 \rightarrow CrO_2Cl_2$$
The chromyl chloride gas is then passed through a basic medium (like NaOH$NaOH$) to form a yellow solution of chromate ions:
CrO_2Cl_2 xrightarrowtextBasic medium CrO_4^2- + Cl^-$$CrO_2Cl_2 \xrightarrow{\text{Basic medium}} CrO_4^{2-} + Cl^-$$
Acidification of the yellow CrO_4^2-$CrO_4^{2-}$ solution followed by the addition of H_2O_2$H_2O_2$ and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO_5$CrO_5$).
CrO_4^2- xrightarrow[textyellow solution,
1. textAcidification CrO_5 text (blue compound)$$CrO_4^{2-} \xrightarrow[\text{yellow solution},
{1. \text{Acidification}} CrO_5 \text{ (blue compound)}$$
### Step 1: Oxidation State Calculation
Potassium Dichromate and Chromyl Chloride Test diagram for Q63 - JEE Main 2024 Morning
The structure of chromium pentoxide (CrO_5$CrO_5$) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O^2-$O^{2-}$) and four peroxide oxygens (O_2^2-$O_2^{2-}$). Therefore, there are 2 peroxo linkages.
Let the oxidation state of Chromium be x$x$.
x + 1(-2) + 4(-1) = 0$$x + 1(-2) + 4(-1) = 0$$x - 2 - 4 = 0$x - 2 - 4 = 0$x = +6$x = +6$
Thus, the oxidation state of Cr in CrO_5$CrO_5$ is +6$+6$.
### Pattern Recognition
A classic oxidation state trap. Calculating simply via formula CrO_5$CrO_5$ yields x - 10 = 0 implies x = +10$x - 10 = 0 \implies x = +10$, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Class 11 Chemistry: Redox Reactions
### Core Logic
Potassium permanganate (KMnO_4$KMnO_4$) is a strong oxidizing agent. When heated to 513mathrmK$513\mathrm{K}$, it undergoes thermal decomposition to give potassium manganate (K_2MnO_4$K_2MnO_4$), manganese dioxide (MnO_2$MnO_2$), and oxygen gas (O_2$O_2$).
The balanced chemical equation is:
2KMnO_4 xrightarrowDelta K_2MnO_4 + MnO_2 + O_2$$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$$
### Step 1: Final Identification
The products formed along with O_2$O_2$ are K_2MnO_4$K_2MnO_4$ (green) and MnO_2$MnO_2$ (black).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
In alkaline medium. MnO_4^-$MnO_4^-$ oxidises I^-$I^-$ to
A.IO_4^-$IO_4^-$
B.IO^-$IO^-$
C.I_2$I_2$
D.IO_3^-$IO_3^-$
Solution
### Core Logic
The behavior of the permanganate ion (MnO_4^-$MnO_4^-$) varies with the pH of the medium.
In a faintly alkaline or neutral medium, MnO_4^-$MnO_4^-$ oxidizes iodide (I^-$I^-$) completely to iodate (IO_3^-$IO_3^-$) while getting reduced to manganese dioxide (MnO_2$MnO_2$).
The balanced ionic equation is:
2MnO_4^- + H_2O + I^- rightarrow 2MnO_2 + 2OH^- + IO_3^-$$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-$$
### Pattern Recognition
Rule of thumb for I^-$I^-$ oxidation by KMnO_4$KMnO_4$:
In acidic medium: I^- rightarrow I_2$I^- \rightarrow I_2$
In alkaline/neutral medium: I^- rightarrow IO_3^-$I^- \rightarrow IO_3^-$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Q70jee_main_2024_30_january_eveningProperties of Transition Metal Compounds
The orange colour of K_2Cr_2O_7$K_2Cr_2O_7$ and purple colour of KMnO_4$KMnO_4$ is due to
B.textd rightarrow textd transition in KMnO_4 text and charge transfer transitions in K_2textCr_2textO_7$\text{d} \rightarrow \text{d transition in KMnO}_4 \text{ and charge transfer transitions in K}_2\text{Cr}_2\text{O}_7$
C.textd rightarrow textd transition in K_2textCr_2textO_7 text and charge transfer transitions in KMnO_4.$\text{d} \rightarrow \text{d transition in K}_2\text{Cr}_2\text{O}_7 \text{ and charge transfer transitions in KMnO}_4.$
D.textd rightarrow textd transition in both.$\text{d} \rightarrow \text{d transition in both.}$
Solution
### Core Logic
In K_2Cr_2O_7$K_2Cr_2O_7$, Chromium is in the +6$+6$ oxidation state, which means its electronic configuration is d^0$d^0$. Since there are no d-electrons, d-d$d-d$ transitions cannot occur. The orange color is due to ligand-to-metal charge transfer (LMCT) from oxygen to chromium.
Similarly, in KMnO_4$KMnO_4$, Manganese is in the +7$+7$ oxidation state, which also corresponds to a d^0$d^0$ configuration. Again, no d-d$d-d$ transitions are possible. The intense purple color is due to ligand-to-metal charge transfer (LMCT) from oxygen to manganese.
### Step 1: Final Conclusion
Both compounds owe their colors to charge transfer transitions.
### Pattern Recognition
Compounds of transition metals in their highest oxidation states (where they have d^0$d^0$ configurations, like Cr^+6$Cr^{+6}$, Mn^+7$Mn^{+7}$, V^+5$V^{+5}$) are deeply colored primarily due to Charge Transfer spectra, NOT d-d transitions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: The d and f Block Elements
More d and f Block Elements Questions — jee_main_2024_29_january_evening
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