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Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre.

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प्रश्न

Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar?

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उत्तर १

According to VSEPR theory, if CH4 were square planar, the bond angle would be 90°. For tetrahedral structure, the bond angle is 109°28′. Therefore, in square planar structure, repulsion between bond pairs would be more and thus the stability will be less.

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उत्तर २

Electronic configuration of carbon atom:

6C: 1s2 2s2 2p2

In the excited state, the orbital picture of carbon can be represented as:

Hence, carbon atom undergoes sp3 hybridization in CH4 molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be dsp2. However, an atom of carbon does not have d-orbitalsto undergo dsp2 hybridization. Hence, the structure of CH4 cannot be square planar.

Moreover, with a bond angle of 90° in square planar, the stability of CH4 will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for CH4.

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अध्याय 4: Chemical Bonding and Molecular Structure - EXERCISES [पृष्ठ १३४]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 11
अध्याय 4 Chemical Bonding and Molecular Structure
EXERCISES | Q 4.21 | पृष्ठ १३४

संबंधित प्रश्न

Draw a diagram showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.


What is the total number of sigma and pi bonds in the following molecules?

C2H2


Distinguish between a sigma and a pi bond.


Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs.


The types of hybrid orbitals of nitrogen in \[\ce{NO^{+}2}\] , \[\ce{NO^{-}3}\] and \[\ce{NH^{+}4}\] respectively are expected to be ______.


Predict the shapes of the following molecules on the basis of hybridisation.

\[\ce{BCl3, CH4 , CO2, NH3}\]


Match the shape of molecules in Column I with the type of hybridisation in Column II.

Column I Column II
(i) Tetrahedral (a) sp2
(ii) Trigonal (b) sp
(iii) Linear (c) sp3

Discuss the concept of hybridisation. What are its different types in a carbon atom.


What is the type of hybridisation of carbon atoms marked with star.

\[\begin{array}{cc}
\phantom{.....}\ce{O}\\
\phantom{.....}||\\
\ce{\overset{∗}{C}H2 = CH - \overset{∗}{C} - O - H}
\end{array}\]


What is the type of hybridisation of carbon atoms marked with star.

\[\ce{CH3 - \overset{∗}{C}H2 - OH}\]


What is the type of hybridisation of carbon atoms marked with star.

\[\begin{array}{cc}
\phantom{..........}\ce{O}\\
\phantom{..........}||\\
\ce{CH3 - CH2 - \overset{∗}{C} - H}
\end{array}\]


In the given reaction, 

the number of sp2 hybridised carbon (s) in compound 'X' is ______.


In which of the following species S atom assumes sp3 hybrid state?

(I) (SO3)

(II) SO2

(III) H2S

(IV) S8


The hybridisation of carbanion is:


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