Match the following:A. SnCl_2I. Trigonal plan | vedclass

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(A) $Sn$ belongs to group $14$ and has four valence electrons,out of which two are used for forming two $Sn-Cl$ bonds and the remaining two form a lon

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$A-III, B-IV, C-II, D-I$

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(A) $Sn$ belongs to group $14$ and has four valence electrons,out of which two are used for forming two $Sn-Cl$ bonds and the remaining two form a lone pair of electrons.Thus,$SnCl_2$ will be angular or bent in shape $(A \rightarrow III)$.$NH_3$ has three bond pairs and one lone pair,so it acquires a trigonal pyramidal shape $(B \rightarrow IV)$.$I_3^{-}$ has a linear geometry as the two bond pairs lie along a straight line while the three lone pairs occupy the equatorial positions $(C \rightarrow II)$.$SO_3$ has three bond pairs only,so it forms a trigonal planar shape $(D \rightarrow I)$.Therefore,the correct matching is $A-III, B-IV, C-II, D-I$.

$A$. $SnCl_2$	$I$. Trigonal planar
$B$. $NH_3$	$II$. Linear
$C$. $I_3^{-}$	$III$. Angular
$D$. $SO_3$	$IV$. Trigonal pyramidal

Match the following:
$A$. $SnCl_2$ $I$. Trigonal planar
$B$. $NH_3$ $II$. Linear
$C$. $I_3^{-}$ $III$. Angular
$D$. $SO_3$ $IV$. Trigonal pyramidal

The correct answer is:

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Match the following:$A$. $SnCl_2$$I$. Trigonal planar$B$. $NH_3$$II$. Linear$C$. $I_3^{-}$$III$. Angular$D$. $SO_3$$IV$. Trigonal pyramidalThe correct answer is: