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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. - Chemistry

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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?

टिप्पणी लिखिए

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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Hybridisation - Types of Hybridisation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
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

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


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Column I Column II
(i) \[\ce{SF4}\] (a) sp3d2
(ii) \[\ce{IF5}\] (b) d2sp3
(iii) \[\ce{NO^{+}2}\], (c) sp3d
(iv) \[\ce{NH^{+}4}\], (d) sp3
  (e) sp

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Column I Column II
(i) Tetrahedral (a) sp2
(ii) Trigonal (b) sp
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\[\begin{array}{cc}
\phantom{.....}\ce{O}\\
\phantom{.....}||\\
\ce{\overset{∗}{C}H2 = CH - \overset{∗}{C} - O - H}
\end{array}\]


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\[\ce{CH3 - \overset{∗}{C}H2 - OH}\]


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\[\ce{CH3 - \overset{∗}{C} ≡ CH}\]


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(I) (SO3)

(II) SO2

(III) H2S

(IV) S8


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