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प्रश्न
Predict the shapes of the following molecules on the basis of hybridisation.
\[\ce{BCl3, CH4 , CO2, NH3}\]
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उत्तर
In \[\ce{BCl3}\], the geometry is trigonal planar is due to sp2 hybridisation.
The shape of \[\ce{CH4}\] is tetrahedral due to sp3 hybridisation.
\[\ce{CO2}\] show linear shape because of sp hybridisation.
The geometry of \[\ce{NH3}\] is pyramidal shape and has sp3 hybridisation.
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संबंधित 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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Match the species in Column I with the type of hybrid orbitals in Column II.
| 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 |
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.
\[\begin{array}{cc}
\phantom{..........}\ce{O}\\
\phantom{..........}||\\
\ce{CH3 - CH2 - \overset{∗}{C} - H}
\end{array}\]
What is the type of hybridisation of carbon atoms marked with star.
\[\ce{CH3 - \overset{∗}{C} ≡ CH}\]
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(I) (SO3)
(II) SO2
(III) H2S
(IV) S8
The hybridisation of carbanion is:
