Assuming pure 2s and 2p orbitals of carbon ar | vedclass

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(C) If we assume no hybridization,carbon in its excited state has one $2s$ orbital and three $2p$ orbitals $(2p_x, 2p_y, 2p_z)$.Three $C-H$ bonds are

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The shape of the molecule will be tetrahedral.

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(C) If we assume no hybridization,carbon in its excited state has one $2s$ orbital and three $2p$ orbitals $(2p_x, 2p_y, 2p_z)$.Three $C-H$ bonds are formed by the overlap of three $2p$ orbitals of carbon with the $1s$ orbitals of three hydrogen atoms. These three bonds are at $90^{\circ}$ to each other.The fourth $C-H$ bond is formed by the overlap of the $2s$ orbital of carbon with the $1s$ orbital of the fourth hydrogen atom.Since the $2s$ orbital is spherical (non-directional),this bond will have different characteristics compared to the other three bonds.Because the bond angles are $90^{\circ}$ and the bond types are different,the molecule cannot have a tetrahedral geometry.Therefore,the statement that the shape of the molecule will be tetrahedral is false.

Assuming pure $2s$ and $2p$ orbitals of carbon are used in forming $CH_4$ molecule,which of the following statements is false?

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