Match the following molecules/ions in List-I | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{V - M - C + A}{2}$,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.$(A)$ For $XeF_2$: $V=8, M=2, C=0, A=0$. $	ext{Lone pairs} = \frac{8-2}{2} = 3$. Thus,$A-III$.$(B)$ For $XeO_3$: $V=8, M=0$ (Oxygen is divalent),$C=0, A=0$. $	ext{Lone pairs} = \frac{8-0}{2} = 4$ valence electrons involved in bonding,but since $O$ is divalent,$3$ double bonds use $6$ electrons,leaving $1$ lone pair. Thus,$B-IV$.$(C)$ For $XeF_4$: $V=8, M=4, C=0, A=0$. $	ext{Lone pairs} = \frac{8-4}{2} = 2$. Thus,$C-I$.$(D)$ For $PF_6^-$: $V=5, M=6, C=0, A=1$. $	ext{Lone pairs} = \frac{5-6+1}{2} = 0$. Thus,$D-II$.The correct matching is $A-III, B-IV, C-I, D-II$.

List-$I$ (molecule/ion)	List-$II$ (Number of lone pairs of electrons on the central atom)
$(A)$ $XeF_2$	$(I)$ $2$
$(B)$ $XeO_3$	$(II)$ $0$
$(C)$ $XeF_4$	$(III)$ $3$
$(D)$ $PF_6^-$	$(IV)$ $1$

List-$I$ (Species)	List-$II$ (Number of lone pairs of electrons on the central atom)
$(a)$ $XeF_{2}$	$(i)$ $0$
$(b)$ $XeO_{2}F_{2}$	$(ii)$ $1$
$(c)$ $XeO_{3}F_{2}$	$(iii)$ $2$
$(d)$ $XeF_{4}$	$(iv)$ $3$

Similar Questions

According to $VSEPR$ theory,

Which of the following molecules has the smallest bond angle?

What is the correct order of bond angles in the given molecules?

How many lone pairs of electrons are present on the chlorine atom in chlorous acid $(HClO_2)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module