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Which type of hybridization is present in the ammonia molecule? - Chemistry

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प्रश्न

Which type of hybridization is present in the ammonia molecule?

What type of hybridisation is associated with N in NH3?

अति संक्षिप्त उत्तर
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उत्तर

The type of hybridisation present in the ammonia (NH3) molecule is sp3.

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पाठ 5: Chemical Bonding - Exercises [पृष्ठ ७९]

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

संबंधित प्रश्‍न

Select and write the most appropriate alternatives from the given choices.

The angle between the two covalent bonds is minimum in:


In Ammonia molecule the bond angle is 107° and in water molecule it is 104°35', although in both the central atoms are sp3 hybridized Explain.


F-Be-F is a liner molecule but H-O-H is angular. Explain.


In the case of bond formation in Acetylene molecule: How many covalent bonds are formed?


In the case of bond formation in Acetylene molecule: State number of sigma and pi bonds formed.


In the case of bond formation in Acetylene molecule: Name the type of Hybridization.


Define Bond Length


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:

H2S


Complete the flow chart.

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

The CORRECT order of C-X bond polarity is ____________.


Which of following bonds has maximum bond length?


What is the H-S-H bond angle in H2S?


Consider the reactions.

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

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

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

The average C-H bond enthalpy is:


Which of the following pairs have identical bond order?


The bond angle is minimum in ____________.


Identify the CORRECT option.


Why is bond order of Be2 molecule zero?

(Nb = bonding electrons, Na = antibonding electrons)


Consider the ions/molecule:

\[\ce{O^+_2, O_2, O^-_2, O^{2-}_2}\]

For increasing bond order the correct option is:


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