Question

The incorrect order of adjacent bond angle is :
(A)- $C{{O}_{2}} > B{{F}_{3}} > C{{H}_{4}}$
(B)- $N{{O}_{2}}^{+} > N{{O}_{3}}^{-} > N{{O}_{2}}^{-}$
(C)- $Xe{{F}_{2}} > Xe{{F}_{4}} > Xe{{O}_{4}}$
(D)- $P{{H}_{3}} > As{{H}_{3}} > Sb{{H}_{3}}$

Answer 1

Hint: The bond angle of the molecules can be predicted by determining the geometry of the molecules. The molecule will attach to the central atom such that it experiences minimum repulsion and more attraction. To minimize the repulsion, the atom/molecule will try to move farther and thereby increasing bond length and changing the bond angle.

Complete answer:
-For determining the bond angle of the molecule, we will first find the hybridization of the central atom as the hybridization of the central atom will help us in determining the bond angle and geometry of the molecule.
-When 2 atoms or molecules are attached to the central atom, the hybridization of the central atom will be ‘sp’ and the geometry of the molecule will be linear. The bond angles in this case are $180{}^\circ $.
-When 3 atoms or molecules are attached to the central atom, the hybridization of the central atom will be $'s{{p}^{2}}'$ and the geometry of the molecule will be trigonal planar. The bond angles in this case are $120{}^\circ $.
-When 4 atoms or molecules are attached to the central atom, the hybridization of the central atom will be $'s{{p}^{3}}'$ and the geometry of the molecule will be tetrahedral. The bond angles in this case are $109.5{}^\circ $.
-When 5 atom or molecules are attached to the central atom, the hybridization of the central atom will be $'s{{p}^{3}}d'$ and the geometry of the molecule will be trigonal bipyramidal. The bond angles in this case are $90{}^\circ ,120{}^\circ ,180{}^\circ $.
-When 6 atom or molecules are attached to the central atom, the hybridization of the central atom will be $'s{{p}^{3}}{{d}^{2}}'$ and the geometry of the molecule will be octahedral. The bond angles in this case are $90{}^\circ ,180{}^\circ $.
-Let us now consider the sets of molecules given to us as options and try predicting the geometry and bond angles of them from the knowledge so far we gained.
-In option A, $C{{O}_{2}} > B{{F}_{3}} > C{{H}_{4}}$ is given to us.
(i) \[C{{O}_{2}}\]have 2 oxygen atoms, therefore the carbon is in ‘sp’ hybridization and the geometry of the molecule will be linear. The bond angle in this case is $180{}^\circ $.
(ii) In $B{{F}_{3}}$, 3 fluorine atoms are attached to boron atom, therefore boron is in $'s{{p}^{2}}'$and the geometry of the molecule is trigonal planar. The bond angle in this case is $120{}^\circ $.
(iii) In $C{{H}_{4}}$, 4 hydrogen atoms are attached to the carbon atom, therefore the carbon is in $'s{{p}^{3}}'$ hybridization and the geometry of the molecule is tetrahedral. The bond angle in this case is $109.5{}^\circ $.

So, the correct order of bond angle is $C{{O}_{2}} > B{{F}_{3}} > C{{H}_{4}}$.
-In the second option, $N{{O}_{2}}^{+}>N{{O}_{3}}^{-}>N{{O}_{2}}^{-}$is given to us.
(i) In the nitronium ion, the nitrogen is attached to two oxygen atoms and has 0 lone pairs. The nitrogen in this is in ‘sp’ hybridization hence the predicted bond angle is $180{}^\circ $.
(ii) In the nitrate ion, the nitrogen has 1 lone pair and is in $'s{{p}^{2}}'$hybridization. The predicted bond angle in this case is $120{}^\circ $.
(iii) In the nitrite ion, the nitrogen has 1 lone pair and the geometry hence becomes bent rather than being linear. Due to bent geometry, the bond angle of the molecule is $<120{}^\circ $.
So, the correct order of bond angle is $N{{O}_{2}}^{+} > N{{O}_{2}}^{-} > N{{O}_{3}}^{-}$, which is contradictory to the sequence given to us.
-In option C, $Xe{{F}_{2}} > Xe{{F}_{4}} > Xe{{O}_{4}}$ is given to us.
(i) \[Xe{{F}_{2}}\]have 2 fluorine atoms, therefore the xenon is in ‘sp’ hybridization and the geometry of the molecule will be linear. The bond angle in this case is $180{}^\circ $.
(ii) In $Xe{{F}_{4}}$, 4 fluorine atoms are attached to xenon atoms, therefore the xenon has two lone pairs. Due to the presence of lone pairs, the hybridization of xenon is, hence the geometry of the molecule is square planar. The bond angle in this case is $90{}^\circ ,180{}^\circ $.
 (iii) In $Xe{{O}_{4}}$, 4 oxygen atoms are attached to a carbon atom and xenon here have 0 lone pairs. Xenon here is in $'s{{p}^{3}}'$hybridization and the geometry of the molecule is tetrahedral. The bond angle in this case is $109.5{}^\circ $.
So, the correct order of bond angle is $Xe{{F}_{2}} > Xe{{F}_{4}} > Xe{{O}_{4}}$.
-In option D, $P{{H}_{3}} > As{{H}_{3}} > Sb{{H}_{3}}$ is given to us. Here the number of atoms attached is the same, one the central atom is changing. As we move from top to bottom in a group, the atomic size of the element increases, hence the bond angle decreases.
So, the correct bond angle order is $P{{H}_{3}} > As{{H}_{3}} > Sb{{H}_{3}}$.

Hence, the correct answer is option B.

Note:
A lone pair refers to a pair of valence electron(s) which are not shared by the atom with another atom in the formation of a covalent bond. Lone pair occupies the space and hence the presence of lone pair decreases the bond angle.