The molecular shapes of $S{F_4},Si{F_4}$ and $IC{l_4}^ - $ are
A.Same with $1,1$ and $1$ lone pairs of electrons on the central atoms respectively.
B.Same with $2,0$ and $1$ lone pairs of electrons on the central atoms respectively.
C.Different with $1,0$ and $2$ lone pairs of electrons on the central atoms respectively.
D.Different with $0,1$ and $2$ lone pairs of electrons on the central atoms respectively.
Answer
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Hint: Lone pairs of electrons: It is defined as the pair of valence electrons which is not shared with another atom in a covalent bond. They are found in the outermost electron shell of atoms. For example: an oxygen atom has two lone pairs and can form two bonds during chemical bonding.
Complete step by step solution:
First of all let us talk about lone pairs and bond pairs.
Lone pairs of electrons: It is defined as the pair of valence electrons which is not shared with another atom in a covalent bond. They are found in the outermost electron shell of atoms. For example: an oxygen atom has two lone pairs and can form two bonds during chemical bonding.
Bond pairs of electrons: When covalent bonds are formed between two atoms then each atom shares one electron to each other. And the electrons which are being shared by two atoms are known as bond pairs. For example: an oxygen atom has two lone pairs and can form two bonds (i.e. two bond pairs) during chemical bonding.
Now let us talk about the compounds given in the question.
$Si{F_4}$: In this molecule total number of bond pairs is $4$ and number of lone pairs is $0$. And the shape of this molecule is tetrahedral.
$S{F_4}$: In this molecule the number of lone pairs is $1$ and the total number of bond pairs is $4$. And the shape of the molecule is trigonal bipyramidal.
$IC{l_4}^ - $: In this molecule total number of bond pairs is $4$ and number of lone pairs is $2$. And the shape of this molecule is square planar.
So the molecular shapes of $S{F_4}, Si{F_4}$ and $IC{l_4}^ - $ are different with $1,0$ and $2$ lone pairs of electrons on the central atoms respectively.
Hence, option (C) is correct.
Note: If a lone pair of electrons are present in a molecule then the geometry of the molecule changes from the expected geometry. For example: the expected geometry of $S{F_4}$ is see-saw but due to the presence of one lone pair the actual geometry is trigonal bipyramidal.
Complete step by step solution:
First of all let us talk about lone pairs and bond pairs.
Lone pairs of electrons: It is defined as the pair of valence electrons which is not shared with another atom in a covalent bond. They are found in the outermost electron shell of atoms. For example: an oxygen atom has two lone pairs and can form two bonds during chemical bonding.
Bond pairs of electrons: When covalent bonds are formed between two atoms then each atom shares one electron to each other. And the electrons which are being shared by two atoms are known as bond pairs. For example: an oxygen atom has two lone pairs and can form two bonds (i.e. two bond pairs) during chemical bonding.
Now let us talk about the compounds given in the question.
$Si{F_4}$: In this molecule total number of bond pairs is $4$ and number of lone pairs is $0$. And the shape of this molecule is tetrahedral.
$S{F_4}$: In this molecule the number of lone pairs is $1$ and the total number of bond pairs is $4$. And the shape of the molecule is trigonal bipyramidal.
$IC{l_4}^ - $: In this molecule total number of bond pairs is $4$ and number of lone pairs is $2$. And the shape of this molecule is square planar.
So the molecular shapes of $S{F_4}, Si{F_4}$ and $IC{l_4}^ - $ are different with $1,0$ and $2$ lone pairs of electrons on the central atoms respectively.
Hence, option (C) is correct.
Note: If a lone pair of electrons are present in a molecule then the geometry of the molecule changes from the expected geometry. For example: the expected geometry of $S{F_4}$ is see-saw but due to the presence of one lone pair the actual geometry is trigonal bipyramidal.
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