
Which among the following is nonlinear?
A. $ {N_3}^{+3} $
B. $ {ClF_2}^{-} $
C. $ {Br_3}^{-} $
D. $ {BrCl_2}^{+} $
Answer
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Hint :To find the nonlinear molecule, the hybridization of each of the given molecules are determined.
To find hybridization, the number of bond pairs and number of lone pairs in a molecule should be rightly identified. Hybridization is the intermixing of atomic orbitals to form new hybrid orbitals.
Complete Step By Step Answer:
Hybridization is given by, $ \dfrac{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 positive charge and A is the negative charge.
For option A,
Hybridization is given by $ \dfrac{1}{2}(5 + 0 - 1 + 0) = 2 $ .
This implies that the hybridization is $ sp $ . Thus it is linear in shape.
For option B,
The hybridization can be given by $ \dfrac{1}{2}(7 + 2 - 0 + 1) = 5 $ .
The hybridization of $ ClF_2^ - $ molecules is $ s{p^3}d $ . But the structure is not trigonal bipyramidal.It is a linear molecule with a bond angle of $ {180^0} $ .
For option C,
The hybridization can be found out by the formula, $ \dfrac{1}{2}(7 + 2 - 0 + 1) = 5 $ . In this molecule there are a total of twenty two electrons. Even Though the molecule is $ s{p^3}d $ hybridized, according to VSEPR theory, $ Br_3^ - $ molecule is linear in shape.
For option D, the hybridization is calculated as $ \dfrac{1}{2}(7 + 2 - 1 + 0) = 4 $ .
Here the hybridization $ s{p^3} $ .But the shape will be bent. $ BrCl_2^ + $ have two bond pairs and two lone pairs in it.To avoid steric repulsion,the shape of the molecule should be bent according to VSEPR theory.
Therefore, the correct option is option D.
Note :
For linear molecules, all the non terminal atoms must be $ sp $ hybridized and Nonlinear molecules do have a cross linked chemical structure. The shape of the molecule may be tetrahedral, bent,trigonal bipyramidal and so on depending upon the number of bond pairs and number of lone pairs in it.The atoms will be arranged in a molecule ,so as to reduce the steric repulsion between bond and lone pairs according to VSEPR theory.
To find hybridization, the number of bond pairs and number of lone pairs in a molecule should be rightly identified. Hybridization is the intermixing of atomic orbitals to form new hybrid orbitals.
Complete Step By Step Answer:
Hybridization is given by, $ \dfrac{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 positive charge and A is the negative charge.
For option A,
Hybridization is given by $ \dfrac{1}{2}(5 + 0 - 1 + 0) = 2 $ .
This implies that the hybridization is $ sp $ . Thus it is linear in shape.
For option B,
The hybridization can be given by $ \dfrac{1}{2}(7 + 2 - 0 + 1) = 5 $ .
The hybridization of $ ClF_2^ - $ molecules is $ s{p^3}d $ . But the structure is not trigonal bipyramidal.It is a linear molecule with a bond angle of $ {180^0} $ .
For option C,
The hybridization can be found out by the formula, $ \dfrac{1}{2}(7 + 2 - 0 + 1) = 5 $ . In this molecule there are a total of twenty two electrons. Even Though the molecule is $ s{p^3}d $ hybridized, according to VSEPR theory, $ Br_3^ - $ molecule is linear in shape.
For option D, the hybridization is calculated as $ \dfrac{1}{2}(7 + 2 - 1 + 0) = 4 $ .
Here the hybridization $ s{p^3} $ .But the shape will be bent. $ BrCl_2^ + $ have two bond pairs and two lone pairs in it.To avoid steric repulsion,the shape of the molecule should be bent according to VSEPR theory.
Therefore, the correct option is option D.
Note :
For linear molecules, all the non terminal atoms must be $ sp $ hybridized and Nonlinear molecules do have a cross linked chemical structure. The shape of the molecule may be tetrahedral, bent,trigonal bipyramidal and so on depending upon the number of bond pairs and number of lone pairs in it.The atoms will be arranged in a molecule ,so as to reduce the steric repulsion between bond and lone pairs according to VSEPR theory.
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