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

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Thus here bond angles between
$$\eqalign{ & {X_4} - M - {X_2} = {180^ \circ } \cr & {X_1} - M - {X_3} = {180^ \circ } \cr & {X_5} - M - {X_6} = {180^ \circ } \cr} $$

Molecular orbital configuration of
$$O_2^ + \left( {8 + 8 - 1 = 15} \right)$$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$      $$\sigma 2p_z^2,\pi 2p_x^2 \approx \pi 2p_y^2,\mathop \pi \limits^* 2p_x^1 \approx \mathop \pi \limits^* 2p_y^0$$
$${\text{Bond}}\,\,{\text{order}}\left( {BO} \right) = \frac{{{N_b} - {N_a}}}{2}$$
( where, $${{N_b}}$$ = number of electrons in bonding molecular orbital, $${{N_a}}$$ = number of electrons in antibonding molecular orbital )
$$\eqalign{ & \therefore \,\,BO = \frac{{10 - 5}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2.5 \cr & {\text{Similarly,}} \cr & \left( {\text{B}} \right)O_2^ - \left( {8 + 8 + 1 = 17} \right) \cr & {\text{So,}}\,\,BO = \frac{{{N_b} - {N_a}}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10 - 7}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.5 \cr & \left( {\text{C}} \right)O_2^{2 - }\left( {8 + 8 + 2 = 18} \right) \cr & BO = \frac{{{N_b} - {N_a}}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{10 - 8}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 1 \cr & \left( {\text{D}} \right){O_2}\left( {8 + 8 = 16} \right) \cr & BO = \frac{{10 - 6}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 2 \cr} $$
Thus, $$O_2^ - $$  shows the bond order 1.5.

$${H_{2\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,H - H$$

The geometry of $$I{F_5}$$  is square Pyramidal with an unsymmetric charge distribution therefore this molecule is polar.

Number of electrons involved in bonding is 6.

$$\left( {\text{A}} \right)S{F_4} = $$   irregular tetrahedral ( $$s{p^3}d,$$  one lone pair )
    $$Xe{F_4} = $$   square planar ( $$s{p^3}{d^2},$$  two lone pairs )
$$\left( {\text{B}} \right)SO_3^{2 - } = $$   pyramidal ( $$s{p^3},$$  one lone pair )
       $$NO_3^ - = $$  trigonal planar ( $$s{p^2}$$ )
$$\left( {\text{C}} \right)B{F_3} = $$   trigonal planar ( $$s{p^2}$$ )
      $$N{F_3} = $$  pyramidal ( $$s{p^3}$$ )
$$\left( {\text{D}} \right)BrO_3^ - = $$   pyramidal ( $$s{p^3},$$  one lone pair )
       $$Xe{O_3} = $$  pyramidal ( $$s{p^3},$$  one lone pair )

Species having equal number of electrons are known as isoelectronic species.
Number of electrons,
$$\eqalign{ & {\text{In}}\,\,CO = 6 + 8 = 14 \cr & {\text{In}}\,\,N{O^ + } = 7 + 8 - 1 = 14 \cr & {\text{In}}\,\,C{N^ - } = 6 + 7 + 1 = 14 \cr & {\text{In}}\,\,C_2^{2 - } = 12 + 2 = 14 \cr} $$
Hence, all have 14 electrons, so they are isoelectronic species.

Hybridization $$\left( H \right) = \frac{1}{2}$$   [ no. of valence electrons of central atom + no. of Monovalent atoms attached to it + ( - ve charge if any ) - ( + ve charge if any ) ]
$$NO_2^ + = \frac{1}{2}\left[ {5 + 0 + 0 - 1} \right] = 2$$       i.e. $$sp$$  hybridisation
$$NO_2^ - = \frac{1}{2}\left[ {5 + 0 + 1 - 0} \right] = 3$$       i.e. $$s{p^2}$$ hybridisation
$$NO_3^ - = \frac{1}{2}\left[ {5 + 0 + 1 - 0} \right] = 3$$       i.e. $$s{p^2}$$ hybridisation
The lewis structure of $$N{O_2}$$ shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridization will be $$s{p^2}$$

For all elements which have atomic number more than 7 ( beyond nitrogen ) the energy of $$\sigma 2{p_z}$$  is lower than $$\pi 2{p_x}$$  and $$\pi 2{p_y}$$  - orbitals.