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Select corrcet orders for corresponding ...

Select corrcet orders for corresponding property as indicated in bracket for the following .

A

`NH_(3) gt BiH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3)` (Boiling point)

B

`H_(2)O gtH_(2) Te gt H_(2) Se gt H_S(2)` (Boiling point)

C

`NH_(3) gt PH_(3) gt AsH_(3) gt SbH_(3)` (Basic character)

D

`H_(2)O lt H_(2)S lt H_(2) Se lt H_(2)` Te (Acidic character)

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the correct orders for the properties of nitrogen and oxygen family hydrides, we will analyze each option step by step. ### Step 1: Analyze the Boiling Points of Nitrogen Family Hydrides 1. **Identify the hydrides**: The nitrogen family hydrides are NH3 (ammonia), PH3 (phosphine), AsH3 (arsine), SbH3 (stibine), and BiH3 (bismuthine). 2. **Understand the trend**: As we move down the group from nitrogen to bismuth, the van der Waals forces and molecular weights increase, leading to higher boiling points. 3. **Consider exceptions**: NH3 can form hydrogen bonds, which significantly increases its boiling point compared to other hydrides in the group. 4. **Establish the order**: The correct order based on boiling points is: - NH3 > BiH3 > SbH3 > AsH3 > PH3 ### Step 2: Analyze the Boiling Points of Oxygen Family Hydrides 1. **Identify the hydrides**: The oxygen family hydrides are H2O (water), H2S (hydrogen sulfide), H2Se (hydrogen selenide), and H2Te (hydrogen telluride). 2. **Understand the trend**: Similar to the nitrogen family, as we move down the group, the molecular weight increases, leading to higher boiling points. 3. **Consider hydrogen bonding**: H2O can form strong hydrogen bonds, which raises its boiling point significantly compared to the other hydrides. 4. **Establish the order**: The correct order based on boiling points is: - H2O > H2Te > H2Se > H2S ### Step 3: Analyze the Basic Character of Nitrogen Family Hydrides 1. **Identify the hydrides**: The basic character of the hydrides is considered for NH3, PH3, AsH3, and SbH3. 2. **Understand the trend**: As we move down the group, the size of the atoms increases, leading to a decrease in charge density and a decrease in the tendency to donate lone pairs. 3. **Establish the order**: The correct order of basicity is: - NH3 > PH3 > AsH3 > SbH3 ### Step 4: Analyze the Acidic Character of Oxygen Family Hydrides 1. **Identify the hydrides**: The acidic character of the hydrides is considered for H2O, H2S, H2Se, and H2Te. 2. **Understand the trend**: As we move down the group, the bond length increases due to the larger atomic size, leading to weaker bonds and easier release of H+ ions. 3. **Establish the order**: The correct order of acidic character is: - H2Te > H2Se > H2S > H2O ### Summary of Correct Orders 1. **Boiling Points of Nitrogen Family Hydrides**: NH3 > BiH3 > SbH3 > AsH3 > PH3 2. **Boiling Points of Oxygen Family Hydrides**: H2O > H2Te > H2Se > H2S 3. **Basic Character of Nitrogen Family Hydrides**: NH3 > PH3 > AsH3 > SbH3 4. **Acidic Character of Oxygen Family Hydrides**: H2Te > H2Se > H2S > H2O

To solve the question regarding the correct orders for the properties of nitrogen and oxygen family hydrides, we will analyze each option step by step. ### Step 1: Analyze the Boiling Points of Nitrogen Family Hydrides 1. **Identify the hydrides**: The nitrogen family hydrides are NH3 (ammonia), PH3 (phosphine), AsH3 (arsine), SbH3 (stibine), and BiH3 (bismuthine). 2. **Understand the trend**: As we move down the group from nitrogen to bismuth, the van der Waals forces and molecular weights increase, leading to higher boiling points. 3. **Consider exceptions**: NH3 can form hydrogen bonds, which significantly increases its boiling point compared to other hydrides in the group. 4. **Establish the order**: The correct order based on boiling points is: - NH3 > BiH3 > SbH3 > AsH3 > PH3 ...
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Knowledge Check

  • Select the correct order of basicity.

    A
    `CH_(3)CH_(2)^(-) gt CH_(2) =CH^(-) gt HC equiv C^(-) gt OH^(-)`
    B
    `CH_(3)CH_(2)^(-) gt HC equiv C^(-) gt CH_(2) = CH^(-) gt OH^(-)`
    C
    `CH_(3)CH_(2)^(-) gt OH^(-) gt HC equiv C^(-) gt CH_(2)=CH^(-)`
    D
    `OH^(-) gt HC equiv C^(-) gt CH_(2) =CH^(-) CH_(3)CH_(2)^(-)`
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