Which compound has the maximum number of lone | vedclass

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(D) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - M - C + A)$,where $V$ is the n

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$[I_3]^-$

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(D) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$A) [ClO_3]^-$: Central atom $Cl$ $(V=7)$. $O$ is divalent $(M=0)$. Anionic charge $A=1$. $	ext{Lone pairs} = \frac{1}{2} (7 - 0 - 0 + 1) = 4$ electrons,which corresponds to $1$ lone pair.$B) XeF_4$: Central atom $Xe$ $(V=8)$. $F$ is monovalent $(M=4)$. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 2$ lone pairs.$C) SF_4$: Central atom $S$ $(V=6)$. $F$ is monovalent $(M=4)$. $	ext{Lone pairs} = \frac{1}{2} (6 - 4) = 1$ lone pair.$D) [I_3]^-$: Central atom $I$ $(V=7)$. Two $I$ atoms are attached $(M=2)$. Anionic charge $A=1$. $	ext{Lone pairs} = \frac{1}{2} (7 - 2 + 1) = 3$ lone pairs.Comparing the values,$[I_3]^-$ has the maximum number of lone pairs $(3)$.

$A$. Trigonal planar	$I$. $PCl_5$
$B$. Tetrahedral	$II$. $SF_6$
$C$. Trigonal bipyramidal	$III$. $BF_3$
$D$. Octahedral	$IV$. $CCl_4$

Which compound has the maximum number of lone pairs of electrons on the central atom?

Similar Questions

Which of the following is the correct order of increasing number of lone pair of electrons on the central atom?

Match the molecular geometries in List-$I$ with their corresponding molecules in List-$II$:
$A$. Trigonal planar $I$. $PCl_5$
$B$. Tetrahedral $II$. $SF_6$
$C$. Trigonal bipyramidal $III$. $BF_3$
$D$. Octahedral $IV$. $CCl_4$

The correct match for $A, B, C, D$ is:

The number of molecules having a lone pair of electrons on the central atom in the following is: $BF_3, SF_4, SiCl_4, XeF_4, NCl_3, XeF_6, PCl_5, HgCl_2, SnCl_2$

Provide the shapes of $H_2O$,$NH_3$,and $SO_2$ including their lone pairs of electrons.

Match the following:
$($Molecule$/$Ion$)$ $($Shape$)$
$A. I_3^- \rightarrow 4.$ Linear
$B. ClF_3 \rightarrow 1.$ $T-$shaped
$C. H_2O \rightarrow 3.$ Angular
$D. SF_4 \rightarrow 2.$ See$-$Saw

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