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 $(3)$.

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

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

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In $PCl_{5}$ molecule,the angle $Cl-P-Cl$ present in a plane is equal to (in $^{\circ}$)

Decreasing order of bond angle is

The number of molecule$(s)$ or ion$(s)$ from the following having non-planar structure is
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Which species has the maximum number of lone pair of electrons on the central atom?

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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