In the formation of the CH_4 molecule,carbon | vedclass

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Question

(C) In a $CH_4$ molecule,the carbon atom is bonded to four hydrogen atoms through single covalent bonds.According to the valence bond theory,the carbo

Vedclass · Accepted Answer

$sp^3$-hybridised orbitals

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(C) In a $CH_4$ molecule,the carbon atom is bonded to four hydrogen atoms through single covalent bonds.According to the valence bond theory,the carbon atom undergoes $sp^3$ hybridisation in its excited state to form four equivalent $sp^3$ hybrid orbitals.These four $sp^3$ hybrid orbitals overlap with the $1s$ orbitals of four hydrogen atoms to form four $C-H$ sigma bonds.Therefore,carbon makes use of $sp^3$-hybridised orbitals.

List-$I$	List-$II$
$(1)$ $SF_4$	$(A)$ $sp^3d^2$
$(2)$ $IF_5$	$(B)$ $d^2sp^3$
$(3)$ $NO_2^+$	$(C)$ $sp^3d$
$(4)$ $NH_4^+$	$(D)$ $sp^3$
	$(E)$ $sp$

In the formation of the $CH_4$ molecule,carbon makes use of:

Similar Questions

The hybridization of orbitals of $N$ atom in $NO_{3}^{-}$,$NO_{2}^{+}$ and $NH_{4}^{+}$ are respectively $:$

In compound $X$,all the bond angles are exactly $109^o 28'$,$X$ is

Match the species in List-$I$ with their correct hybrid orbitals in List-$II$.

List-$I$ List-$II$

$(1)$ $SF_4$ $(A)$ $sp^3d^2$

$(2)$ $IF_5$ $(B)$ $d^2sp^3$

$(3)$ $NO_2^+$ $(C)$ $sp^3d$

$(4)$ $NH_4^+$ $(D)$ $sp^3$

$(E)$ $sp$

The molecule of $CO_2$ has a $180^{\circ}$ bond angle. It can be explained on the basis of:

The correct order of bond angle is:

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