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

Each boron atom in diborane $$\left( {{B_2}{H_6}} \right)$$   is $$s{p^3}$$ hybridised. In the structure of diborane four $$H - {\text{atoms,}}$$   two on the left and two on the right, known as terminal hydrogens, are in different environments from the other two hydrogen atoms which are known as bridging atoms. The two boron atoms and the four terminal hydrogen atoms lie in the same plane while the two boron atoms and the two bridging hydrogen atoms, one above and the other below, lie in a plane perpendicular to this plane.

When two hybridised orbitals of two atoms undergoes linear combination, they form sigma bond.

\[H-C\equiv C-\underset{\begin{smallmatrix} \left. {} \right| \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ \left| {} \right. \end{smallmatrix}}{\mathop{C}}}\,-H;\]
No. of $$sigma$$  bonds = 6, No. of $$Pi$$  bonds = 2

The highest lattice energy is given by one with the highest ionic charges and smallest ionic sizes.

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

In a coordinate or dative bond, electrons required for sharing are contribution by one atom and shared by both the atoms.

According to octet rule, the central atom must have 8 electrons but in some compounds the number of electrons is more than 8, or less than 8 or an odd number of electrons is left on the central atom e.g., $$P{F_5},BC{l_3},NO.$$

The bond order of $${N_2},{O_2},$$   and $$O_2^ - $$  are respectively 3, 2 and 1.5
Since higher bond order implies higher bond dissociation energy hence the correct order will be $${N_2} > {O_2} > O_2^ - $$

\[{{\left[ O-\underset{\begin{smallmatrix} \left| {} \right. \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ \parallel \end{smallmatrix}}{\mathop{P}}}\,-O \right]}^{3-}}\leftrightarrow {{\left[ O-\underset{\begin{smallmatrix} \left| {} \right. \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ \left. {} \right| \end{smallmatrix}}{\mathop{P}}}\,=O \right]}^{3-}}\]       \[\leftrightarrow {{\left[ O-\underset{\begin{smallmatrix} \parallel \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ \left| {} \right. \end{smallmatrix}}{\mathop{P}}}\,-O \right]}^{3-}}\leftrightarrow {{\left[ O=\underset{\begin{smallmatrix} \left| {} \right. \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ \left| {} \right. \end{smallmatrix}}{\mathop{P}}}\,-O \right]}^{3-}}\]

$$\eqalign{ & {\text{Formal charge on each }}O{\text{ - atom}} \cr & {\text{ = }}\frac{{{\text{Total charge}}}}{{{\text{No}}{\text{. of }}O{\text{ - atoms}}}} \cr & = \frac{{ - 3}}{4} \cr & = - 0.75 \cr & \left( {P - {\text{ }}O} \right){\text{ Bond order}} \cr & {\text{ = }}\frac{{{\text{Total no}}{\text{. of bonds}}}}{{{\text{Total no}}{\text{. of resonating structures}}}} \cr & = \frac{5}{4} \cr & = 1.25 \cr} $$

Key concept The species that have same number of electrons have same bond order.

Thus, both $$C{N^ - }$$  and $$CO$$  have equal number of electrons. So, their bond order will be same.