The three-dimensional structure of CH_4 is gi | vedclass

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(A) In the $CH_4$ molecule,the carbon atom is $sp^3$ hybridized with a tetrahedral geometry.In the provided structure:$1$. The bonds $X$ and $Y$ are r

Vedclass · Accepted Answer

$A$ (In-plane)$B$ (Away)$C$ (Towards)$X, Y$$Z$$W$

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(A) In the $CH_4$ molecule,the carbon atom is $sp^3$ hybridized with a tetrahedral geometry.In the provided structure:$1$. The bonds $X$ and $Y$ are represented by simple lines,indicating they lie in the plane of the paper.$2$. The bond $Z$ is represented by a dashed wedge,indicating it projects away from the observer (behind the plane).$3$. The bond $W$ is represented by a solid wedge,indicating it projects towards the observer (in front of the plane).Therefore,the correct mapping is: In-plane $(A)$ = $X, Y$; Away from observer $(B)$ = $Z$; Towards observer $(C)$ = $W$.

The three-dimensional structure of $CH_4$ is given below. The bonds are labelled as $W, X, Y$ and $Z$. What are the bonds lying in the plane $(A)$,projecting out of the plane away from the observer $(B)$ and projecting out of the plane towards the observer $(C)$?

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$A$ (In-plane)	$B$ (Away)	$C$ (Towards)
$X, Y$	$Z$	$W$

$A$ (In-plane)	$B$ (Away)	$C$ (Towards)
$Z$	$W$	$X, Y$

$A$ (In-plane)	$B$ (Away)	$C$ (Towards)
$X, Y$	$W$	$Z$

$A$ (In-plane)	$B$ (Away)	$C$ (Towards)
$W$	$X, Y$	$Z$