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Statement-1: K(2)CO(3) cannot be prepare...

Statement-1: `K_(2)CO_(3)` cannot be prepared by Le-blanc method.
Statement-2: Sodium nitrate when heated gives `NO_(2)` gas.
Statement-3: `NaO_(2)` is paramagnetic while `Na_(2)O_(2)` is diamagnetic.

A

FFT

B

FTT

C

TFT

D

TTF

Text Solution

AI Generated Solution

The correct Answer is:
To analyze the statements provided in the question, we will evaluate each statement one by one and determine their validity based on the principles of chemistry. ### Step 1: Evaluate Statement 1 **Statement 1:** `K2CO3` cannot be prepared by Le-Blank method. - The Le-Blank method is primarily used to prepare sodium carbonate (`Na2CO3`) from sodium chloride (`NaCl`). The process involves reacting `NaCl` with sulfuric acid (`H2SO4`) to form sodium sulfate (`Na2SO4`), which is then reacted with carbon to produce sodium carbonate. - However, since potassium (`K`) and sodium (`Na`) are both alkali metals and share similar chemical properties, `K2CO3` can indeed be prepared using a similar method by substituting `KCl` for `NaCl`. - Therefore, **Statement 1 is false**. ### Step 2: Evaluate Statement 2 **Statement 2:** Sodium nitrate when heated gives `NO2` gas. - The thermal decomposition of sodium nitrate (`NaNO3`) typically produces sodium nitrite (`NaNO2`) and oxygen (`O2`), not nitrogen dioxide (`NO2`). - The reaction can be represented as: \[ 2 \text{NaNO}_3 \rightarrow 2 \text{NaNO}_2 + \text{O}_2 \] - Since `NO2` is not produced in this reaction, **Statement 2 is false**. ### Step 3: Evaluate Statement 3 **Statement 3:** `NaO2` is paramagnetic while `Na2O2` is diamagnetic. - Sodium superoxide (`NaO2`) contains unpaired electrons, making it paramagnetic. Paramagnetic substances are attracted to magnetic fields due to the presence of unpaired electrons. - On the other hand, sodium peroxide (`Na2O2`) has all its electrons paired, resulting in it being diamagnetic. Diamagnetic substances are not attracted to magnetic fields. - Therefore, **Statement 3 is true**. ### Conclusion Based on the evaluations: - Statement 1: False - Statement 2: False - Statement 3: True Thus, the overall answer is **FFT** (False, False, True).
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k_(2)CO_(3) cannot be prepared by solvay process because

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Knowledge Check

  • Give reason for the statement [Ni(CN)_(4)]^(2-) is diamagnetic while [NiCl_(4)]^(2-) is paramagnetic in nature .

    A
    In `[NiCl_(4)]^(2-)` no unpaired electrons are present while in `[Ni(CN)_(4)]^(2-)` two unpaired electrons are present
    B
    In `[Ni(CN)_(4)]^(2-)` , no unpaired electrons are present while in `[NiCl_(4)]^(2-)` two unpaired electrons are present .
    C
    `[NiCl_(4)]^(2-)` shows `dsp^(2)` hybridisation hence it is paramagnetic .
    D
    `[Ni(CN)_(4)]^(2-)` shows `sp^(3)` hybridisation hence it is diamagnetic .
  • Which nitrate will decompose to give NO_(2) on heating?

    A
    `NaNO_(3)`
    B
    `KNO_(3)`
    C
    `RbNO_(3)`
    D
    `LiNO_(3)`
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