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Give a reason for HF is a polar molecule

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प्रश्न

Give a reason for HF is a polar molecule

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उत्तर

  1. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
  2. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial −ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and the H-F bond becomes polar. Hence, H-F is a polar molecule.
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अध्याय 5: Chemical Bonding - Exercises [पृष्ठ ७९]

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बालभारती Chemistry [English] Standard 11 Maharashtra State Board
अध्याय 5 Chemical Bonding
Exercises | Q 3. (H)(b) | पृष्ठ ७९

संबंधित प्रश्न

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Predict the shape and bond angles in the following molecule:

CF4


Predict the shape and bond angles in the following molecule:

NF3


Predict the shape and bond angles in the following molecule:

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Using data from the Table, answer the following:

Examples C2H6 Ethane C2H4 Ethene C2H2 Ethyne
Structure

\[\begin{array}{cc} \backslash \phantom{......}/\phantom{.}\\ \ce{—C – C —}\\  /\phantom{......}\backslash\phantom{.}\end{array}\]

\[\begin{array}{cc} \backslash \phantom{......}/\\ \ce{C \text{=} C}\\  /\phantom{......}\backslash\end{array}\]

\[\ce{- C ≡ C -}\]
Type of bond between carbons single double triple
Bond length (nm) 0.154 0.134 0.120
Bond Enthalpy kJ mol-1 348 612 837
  1. What happens to the bond length when unsaturation increases?
  2. Which is the most stable compound?
  3. Indicate the relationship between bond strength and Bond enthalpy.
  4. Comment on the overall relation between Bond length, Bond Enthalpy, and Bond strength and stability.

Complete the flow chart.

Molecular Formula Structural Formula Shape/ Geometry Bond angle
BeCl2     180°
  O=C=O Linear  
C2H2      

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\[\ce{C_{(s)} + 2H2_{(g)} -> CH4_{(g)}}\], ΔH = −x kcal

\[\ce{C_{(g)} + 4H_{(g)} -> CH4_{(g)}}\], ΔH = −x1 kcal

\[\ce{CH4_{(g)} -> CH3_{(g)} + H_{(g)}}\], ΔH = +y kcal

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What is the bona order of O2 molecules?


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