Answer
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Hint: The bond angle is the angle between two atoms in a compound. The greater the bond length in the compound, the lesser the bond angle will be and the greater the electron density of the substituent atoms in the compound, the more will be the electronic repulsions and thus, the bond angle will
Complete step by step answer:
Both the \[N{H_3}\] and \[N{F_3}\] form a pyramidal shape. There is a difference in the bond angle between these two structures due to the electronegativity difference between them. The electronegativity of the nitrogen atom is more than that of H, therefore the bond cloud between the bonded nitrogen and hydrogen atom tries to attract to the nitrogen atom. Therefore, the whole electron cloud gets attracted to the central atom. When the electron cloud comes closer to the electron pair on the nitrogen atom, it suffers repulsion and the lone pair of electrons pushes away the electron cloud. This leads to an increase in the bond angle of \[N{H_3}\]. But, in the case of \[N{F_3}\] , the bond pair electrons get attracted to the fluorine atom. The distance between the electron pairs increases and hence, the repulsion decreases. So, \[N{F_3}\] has a smaller bond angle.
The bond angle of \[P{H_3}\] is less than that of \[P{F_3}\] . Both \[P{H_3}\] and \[P{F_3}\] are pyramidal in shape. Both of them consist of one lone pair of electrons on the central phosphorus atom. But \[P{F_3}\] has a partial double bond character due to back \[\pi - \] donation of electrons from p- orbitals of fluorine atom to empty d- orbitals of the phosphorus atom. This results in quite a large repulsion between \[P - F\] bonds and hence, the bond angle is large. There is no partial double bond character possible in \[P{H_3}\] (due to the lack of p-orbitals in H). The bond angle is more in \[P{F_3}\] due to the lone pair-bond pair repulsion. As the fluorine is smaller in size and more electronegative than the hydrogen atom; the bond pair of electrons shifts partially towards the fluorine atom.
Note:
There are many factors on which the bond angle depends:
1.Hybridization: The bond angle is dependent on the hybridization of the central atom of the compound.
2.Lone pair repulsion: If the number of lone pairs of electrons is more, the bonded atoms will suffer more electron repulsions which will affect the bond angle.
3.Electronegativity: The lesser the electronegativity of the central atom, the lesser will be the bond angle.
Complete step by step answer:
Both the \[N{H_3}\] and \[N{F_3}\] form a pyramidal shape. There is a difference in the bond angle between these two structures due to the electronegativity difference between them. The electronegativity of the nitrogen atom is more than that of H, therefore the bond cloud between the bonded nitrogen and hydrogen atom tries to attract to the nitrogen atom. Therefore, the whole electron cloud gets attracted to the central atom. When the electron cloud comes closer to the electron pair on the nitrogen atom, it suffers repulsion and the lone pair of electrons pushes away the electron cloud. This leads to an increase in the bond angle of \[N{H_3}\]. But, in the case of \[N{F_3}\] , the bond pair electrons get attracted to the fluorine atom. The distance between the electron pairs increases and hence, the repulsion decreases. So, \[N{F_3}\] has a smaller bond angle.
The bond angle of \[P{H_3}\] is less than that of \[P{F_3}\] . Both \[P{H_3}\] and \[P{F_3}\] are pyramidal in shape. Both of them consist of one lone pair of electrons on the central phosphorus atom. But \[P{F_3}\] has a partial double bond character due to back \[\pi - \] donation of electrons from p- orbitals of fluorine atom to empty d- orbitals of the phosphorus atom. This results in quite a large repulsion between \[P - F\] bonds and hence, the bond angle is large. There is no partial double bond character possible in \[P{H_3}\] (due to the lack of p-orbitals in H). The bond angle is more in \[P{F_3}\] due to the lone pair-bond pair repulsion. As the fluorine is smaller in size and more electronegative than the hydrogen atom; the bond pair of electrons shifts partially towards the fluorine atom.
Note:
There are many factors on which the bond angle depends:
1.Hybridization: The bond angle is dependent on the hybridization of the central atom of the compound.
2.Lone pair repulsion: If the number of lone pairs of electrons is more, the bonded atoms will suffer more electron repulsions which will affect the bond angle.
3.Electronegativity: The lesser the electronegativity of the central atom, the lesser will be the bond angle.
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