Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry

$${H_2}C{O_3}$$  and $$B{F_3}$$  both have triangular planar structure
$$B{F_3}$$  is symmetrical
$$\therefore \,\mu = 0$$   hence non polar, $${H_2}C{O_3}$$  is not symmetrical hence polar in nature.

When $${p_x}$$  and $${p_y}$$  form bond, then the bond angle is minimum and it is only $${90^ \circ }.$$
NOTE
Bond angle in $$s{p^3}$$  bonds, $$H - O - H$$    in water and in $$sp$$  bonds are $${109^ \circ }28,{180^ \circ }$$   and $${180^ \circ }$$ respectively.

Thus, $${H_2}O$$  has highest dipole moment.

NOTE : The delocalised $$p\pi - p\pi \,$$   bonding between filled $$p - {\text{orbital}}$$   of $$F$$  and vacant $$p - {\text{orbital}}$$   of $$B$$  leads to shortening of $$B - F$$  bond length which results in higher bond dissociation energy of the $$B - F$$  bond.

The structures represent canonical or resonating structures of carbonate ion.

The number of lone pairs of electrons on central atom in various given species are
$$\eqalign{ & {\rm{Species}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Number of lone}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{pairs on central}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{atom}} \cr & I{F_7}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,nil \cr & I{F_5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cr & Cl{F_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr & Xe{F_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3 \cr & {\rm{Thus the correct increasing order is}} \cr & \mathop {I{F_7}}\limits_0 < \mathop {I{F_5}}\limits_1 < \mathop {Cl{F_3}}\limits_2 < \mathop {Xe{F_2}}\limits_3 \cr} $$

\[{{H}_{2}}S{{O}_{4}};H-O-\underset{\begin{smallmatrix} \downarrow \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ \uparrow \end{smallmatrix}}{\mathop{S}}}\,-O-H\]

In acetylene molecule, triple bond between two carbon atoms is formed by one $$sigma$$  and two $$pi$$ bonds.

Since the dipoles are equal and acting in opposite directions. Therefore, the net dipole moment is zero.

$$\mu = 1.85\,D = 1.85 \times {10^{ - 18}}esu\,cm = q \times d$$


$$\eqalign{ & \cos \,\,{52.5^ \circ } = \frac{d}{{0.94\mathop {\text{A}}\limits^{\text{o}} }} \Rightarrow d = 0.609 \times 0.94\mathop {\text{A}}\limits^{\text{o}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.572\mathop {\text{A}}\limits^{\text{o}} \cr & \therefore \mu = q \times d \cr & {q_1} = \frac{\mu }{d} = \frac{{1.85D}}{{0.572\mathop {\text{A}}\limits^{\text{o}} }} \cr & = \frac{{1.85 \times {{10}^{ - 18}}esu\,cm}}{{0.572 \times {{10}^{ - 8}}cm}} \cr & = 3.2 \times {10^{ - 10}}esu \cr} $$