A molecule XY_2 contains two sigma,two pi-bon | vedclass

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(C) The central atom $X$ has two $\sigma$-bonds and two $\pi$-bonds. This implies that $X$ is involved in two double bonds (or one single and one trip

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trigonal planar

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(C) The central atom $X$ has two $\sigma$-bonds and two $\pi$-bonds. This implies that $X$ is involved in two double bonds (or one single and one triple bond,but given $XY_2$ and the presence of one lone pair,the structure is $Y=X=Y$). The total number of electron domains around $X$ is determined by the number of $\sigma$-bonds and lone pairs. Here,there are $2$ $\sigma$-bonds and $1$ lone pair,totaling $3$ electron domains. According to $VSEPR$ theory,$3$ electron domains correspond to $sp^2$ hybridization with a trigonal planar geometry. Since one of these domains is a lone pair,the molecular shape is bent,but the arrangement of the electron pairs (lone pair and bond pairs) is trigonal planar.

List-$I$	List-$II$
$(1)$ ${H_3O^+}$	$(A)$ Linear
$(2)$ ${HC \equiv CH}$	$(B)$ Angular
$(3)$ ${ClO_2^-}$	$(C)$ Tetrahedral
$(4)$ ${NH_4^+}$	$(D)$ Trigonal bipyramidal
	$(E)$ Pyramidal

$A$ molecule $XY_2$ contains two $\sigma$,two $\pi$-bonds and one lone pair of electron in the valence shell of $X$. The arrangement of lone pair as bond pair is

Similar Questions

The correct order of the bond angle for the given bond is:

Match the species given in List-$I$ with their correct geometric shapes given in List-$II$.

List-$I$ List-$II$

$(1)$ ${H_3O^+}$ $(A)$ Linear

$(2)$ ${HC \equiv CH}$ $(B)$ Angular

$(3)$ ${ClO_2^-}$ $(C)$ Tetrahedral

$(4)$ ${NH_4^+}$ $(D)$ Trigonal bipyramidal

$(E)$ Pyramidal

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