Which species is planar?

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(A) To determine the geometry of the given species,we calculate the hybridization of the central atom using the formula: $H = \frac{1}{2} (V + M - C +

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$CO_3^{2-}$

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(A) To determine the geometry of the given species,we calculate the hybridization of the central atom using the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$A) \ CO_3^{2-}$: $H = \frac{1}{2} (4 + 0 - 0 + 2) = 3$. The hybridization is $sp^2$,which corresponds to a trigonal planar geometry.$B) \ SO_3^{2-}$: $H = \frac{1}{2} (6 + 0 - 0 + 2) = 4$. The hybridization is $sp^3$ with one lone pair,resulting in a trigonal pyramidal geometry.$C) \ ClO_3^{-}$: $H = \frac{1}{2} (7 + 0 - 0 + 1) = 4$. The hybridization is $sp^3$ with one lone pair,resulting in a trigonal pyramidal geometry.$D) \ BF_4^{-}$: $H = \frac{1}{2} (3 + 4 - 0 + 1) = 4$. The hybridization is $sp^3$ with no lone pairs,resulting in a tetrahedral geometry.Thus,only $CO_3^{2-}$ is planar.

List-$I$	List-$II$
$A$. $BF_3$	$II$. Trigonal planar
$B$. $ClF_3$	$III$. $T$-shape
$C$. $NH_3$	$IV$. Trigonal pyramidal
$D$. $NH_4^+$	$I$. Tetrahedral

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Match the following:
List-$I$ List-$II$
$A$. $BF_3$ $II$. Trigonal planar
$B$. $ClF_3$ $III$. $T$-shape
$C$. $NH_3$ $IV$. Trigonal pyramidal
$D$. $NH_4^+$ $I$. Tetrahedral

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