Identify the correct statement regarding geom | vedclass

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Question

(C) In $CH_4$,the central carbon atom is $sp^3$ hybridized and forms $4$ sigma bonds with hydrogen atoms. It has $0$ lone pairs and a tetrahedral geom

Vedclass · Accepted Answer

Both have same geometry with no lone pair of electrons each.

Vedclass · Answer

(C) In $CH_4$,the central carbon atom is $sp^3$ hybridized and forms $4$ sigma bonds with hydrogen atoms. It has $0$ lone pairs and a tetrahedral geometry.In $SiCl_4$,the central silicon atom is $sp^3$ hybridized and forms $4$ sigma bonds with chlorine atoms. It has $0$ lone pairs and a tetrahedral geometry.Therefore,both molecules have the same tetrahedral geometry and contain no lone pairs of electrons on the central atom.

List-$I$	List-$II$
$A. BrF_5$	$I. AB_4E$,see-saw
$B. SF_4$	$II. AB_4E_2$,square planar
$C. XeF_4$	$III. AB_5E$,square pyramidal
$D. ClF_3$	$IV. AB_3E_2$,$T$-shape

Identify the correct statement regarding geometry and lone pair of electrons present in $CH_4$ and $SiCl_4$.

Similar Questions

Match the following:

List-$I$ List-$II$

$A. BrF_5$ $I. AB_4E$,see-saw

$B. SF_4$ $II. AB_4E_2$,square planar

$C. XeF_4$ $III. AB_5E$,square pyramidal

$D. ClF_3$ $IV. AB_3E_2$,$T$-shape

The correct answer is:

The bond angles in methane $(I)$,ammonia $(II)$,and trimethylamine $(III)$ increase in the order:

The structure of $Cl_2O_7$ is:

$XH_3 + H^{+} \longrightarrow XH_4^{+} (X = N, P, As, Sb)$. In this process,the maximum increase in bond angle will be in the case of $:-$

Which of the following molecules has trigonal planar geometry?

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