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

$$I{F_3}$$  has bent-$$T$$ geometry

$$2lp + 3bp = s{p^3}d$$    hybridisation
$$PC{l_3}$$  has pyramidal geometry

$$1lp + 3bp = s{p^3}$$   hybridisation
$$N{H_3}$$  has trigonal pyramidal geometry

$$1lp + 3bp = s{p^3}$$   hybridisation
$$B{F_3}$$  has trigonal planar geometry

$$3bp\,\,{\text{only}} = s{p^2}$$   (hydridisation)

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$${F_2}\,{\text{and}}\,O_2^{2 - }\,{\text{are}}\,\,{\text{isoelectronic}}{\text{.}}$$
$${F_2}:\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$     $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$      $$\left( {{\pi ^ * }2p_x^2 = {\pi ^ * }2p_y^2} \right)$$
$$B.O. = \frac{1}{2} \times \left( {10 - 8} \right) = 1$$
$$O_2^{2 - }:\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$      $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$      $$\left( {{\pi ^ * }2p_x^2 = {\pi ^ * }2p_y^2} \right)$$
$$B.O. = \frac{{10 - 8}}{2} = 1$$

$$RbCl$$  is an electrovalent compound hence the bond involves is non-directional.

$$O_2^ - :\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$     $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$      $$\left( {{\pi ^ * }2p_x^2 = {\pi ^ * }2p_y^1} \right)$$
$${N_2},{F_2}$$  and $$O_2^{2 - }$$  have paired electrons.

Only $${H_2}O$$  forms $$H$$ - bonding while other substances $$HCl,HI$$   and $${H_2}S$$  do not form $$H$$ - bonding.

$$\eqalign{ & 1\,e = 1.602 \times {10^{ - 19}}C \cr & 1\,esu = 3.33 \times {10^{ - 10}}C \cr & \frac{{1\,e}}{{1\,esu}} = \frac{{1.602 \times {{10}^{ - 19}}C}}{{3.33 \times {{10}^{ - 10}}C}} \cr & 1\,e = 4.802 \times {10^{ - 10}}esu \cr & {\text{Dipole moment = }}q \times {\text{distance}} \cr & \Rightarrow 1\,D \approx {\text{1}}{{\text{0}}^{ - 18}}esu\,cm \cr & 0.38 \times {10^{ - 18}}esu\,cm = q \times \left( {1.61 \times {{10}^{ - 8}}cm} \right) \cr & q = 2.36 \times {10^{ - 11}}esu \cr & q = \frac{{2.36 \times {{10}^{ - 11}}\,esu}}{{4.802 \times {{10}^{ - 10}}esu}} \cr & q = 0.049 \cr & q \approx 0.05\,{\text{fractional charge}} \cr} $$

$$Be{F_2} - sp$$   hybridisation
$$B{F_3} - s{p^2}$$   hybridisation
$${C_2}{H_2} - sp$$   hybridisation
$$N{H_3} - s{p^3}$$   hybridisation

The molecular orbital configuration of
$$O_2^ - \left( {8 + 8 + 1 = 17} \right) = $$     $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},\sigma 2p_x^2,$$      $$\pi 2p_y^2 \approx \pi 2p_z^2,\mathop \pi \limits^* 2p_y^2 \approx \mathop \pi \limits^* 2p_z^1$$
$$\eqalign{ & {\text{Bond}}\,{\text{order}}\left( {{\text{BO}}} \right) \cr & = \frac{{{N_b} - {N_a}}}{2} \cr & = \frac{{10 - 7}}{2} \cr & = 1.5 \cr} $$
$$NO\left( {7 + 8 = 15} \right) = $$     $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},\sigma 2p_x^2,$$      $$\pi 2p_y^2 \approx \pi 2p_z^2,\mathop \pi \limits^* 2p_y^1 \approx \mathop \pi \limits^* 2p_z^0$$
$$\eqalign{ & {\text{BO}} = \frac{{10 - 5}}{2} \cr & \,\,\,\,\,\,\,\,\, = 2.5 \cr} $$
$$C_2^{2 - }\left( {6 + 6 + 2 = 14} \right) = $$     $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$     $$\pi 2p_y^2 \approx \pi 2p_z^2,\sigma 2p_x^2$$
$$\eqalign{ & {\text{BO}} = \frac{{10 - 4}}{2} \cr & \,\,\,\,\,\,\,\,\, = 3 \cr} $$
$$\eqalign{ & He_2^ + \left( {2 + 2 - 1 = 3} \right) = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^1} \cr & {\text{BO}} = \frac{{2 - 1}}{2} \cr & \,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\, = 0.5 \cr} $$
Hence, order of increasing bond order is $$He_2^ + < O_2^ - < NO < C_2^{2 - }$$

The values of dipole moments of methyl halides :
$$\eqalign{ & C{H_3}F - 1.84\,D;C{H_3}Cl - 1.86\,D \cr & C{H_3}Br - 1.83\,D;C{H_3}I - 1.63\,D \cr & \therefore C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I \cr} $$