Which of the following is the correct order o | 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{V - N}{2}$,where $V$ is the number of val

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$IF_7 < IF_5 < ClF_3 < XeF_2$

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(D) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{V - N}{2}$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of monovalent atoms bonded to it.$1$. For $IF_7$: Central atom $I$ has $7$ valence electrons. It forms $7$ bonds with $F$ atoms. $	ext{Lone pairs} = \frac{7 - 7}{2} = 0$.$2$. For $IF_5$: Central atom $I$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms. $	ext{Lone pairs} = \frac{7 - 5}{2} = 1$.$3$. For $ClF_3$: Central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms. $	ext{Lone pairs} = \frac{7 - 3}{2} = 2$.$4$. For $XeF_2$: Central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms. $	ext{Lone pairs} = \frac{8 - 2}{2} = 3$.The number of lone pairs is: $IF_7 (0) Thus,the correct order is $IF_7 < IF_5 < ClF_3 < XeF_2$.

List-$I$ (Molecule)	List-$II$ (Shape)
$A$. $NH_3$	$I$. Square pyramid
$B$. $BrF_5$	$II$. Tetrahedral
$C$. $PCl_5$	$III$. Trigonal pyramidal
$D$. $CH_4$	$IV$. Trigonal bipyramidal

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

Similar Questions

In compounds of type $ECl_3$,where $E = B, P, As$ or $Bi$,the angles $\angle Cl-E-Cl$ for different $E$ are in the order

Match List-$I$ with List-$II$:

List-$I$ (Molecule) List-$II$ (Shape)

$A$. $NH_3$ $I$. Square pyramid

$B$. $BrF_5$ $II$. Tetrahedral

$C$. $PCl_5$ $III$. Trigonal pyramidal

$D$. $CH_4$ $IV$. Trigonal bipyramidal

Choose the correct answer from the options below:

The bond angles $H-O-N$ and $O-N-O$ in the planar structure of nitric acid molecule are respectively:

The number of bond pairs of electrons and the total number of lone pairs of electrons in $XeOF_4$ are respectively:

Bond angle in $PH_{4}^{+}$ is more than that of $PH_{3}$. This is because

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