One of the following conversions results in t | vedclass

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(C) For the conversion $CH_4 \rightarrow C_2H_6$,there is no change in the hybridization of the carbon atom. Both $CH_4$ and $C_2H_6$ have $sp^3$ hybr

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$BF_3$ to $BF_4^-$

Vedclass · Answer

(C) For the conversion $CH_4 \rightarrow C_2H_6$,there is no change in the hybridization of the carbon atom. Both $CH_4$ and $C_2H_6$ have $sp^3$ hybridization and tetrahedral geometry.For the conversion $NH_3 \rightarrow NH_4^+$,nitrogen in $NH_3$ is $sp^3$ hybridized with a pyramidal shape. In $NH_4^+$,it remains $sp^3$ hybridized but the geometry becomes tetrahedral.For the conversion $H_2O \rightarrow H_3O^+$,oxygen in $H_2O$ is $sp^3$ hybridized with a bent shape. In $H_3O^+$,it remains $sp^3$ hybridized but the shape becomes pyramidal.For the conversion $BF_3 \rightarrow BF_4^-$,boron in $BF_3$ is $sp^2$ hybridized with a trigonal planar geometry. In $BF_4^-$,the hybridization changes to $sp^3$ and the geometry changes to tetrahedral.

One of the following conversions results in the change of hybridization and geometry:

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