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

For molecules such as $${B_2},{C_2},{N_2}$$   etc. the increasing order of energies of various molecular orbitals is $$\sigma 1s < {\sigma ^ * }1s < \sigma 2s < {\sigma ^ * }2s < $$      $$\left( {\pi 2{p_x} = \pi 2{p_y}} \right) < \sigma 2{p_z} < $$     $$\left( {{\pi ^ * }2{p_x} = {\pi ^ * }2{p_y}} \right) < {\sigma ^ * }2{p_z}$$

Electronic configuratios of $$Li_2^ + $$  and $$Li_2^ - $$  :
\[Li_2^ + :\,\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^1}\]
\[Li_2^ - :\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^1}\]
Now,
Bond order of \[Li_2^ + = \frac{1}{2}\left( {3 - 2} \right) = \frac{1}{2}\]
Bond order of \[Li_2^ - = \frac{1}{2}\left( {4 - 3} \right) = \frac{1}{2}\]
Here, both $$Li_2^ + $$  and $$Li_2^ - $$  have positive bond order, thus both are stable.

TIPS/Formulae : $$sp$$  type of hybridization involves the intermixing of one $$s$$ and one $$p$$ $$\left( {say\,{p_x}} \right)$$  orbitals to give two equivalent hybrid orbitals, known as $$\,sp$$  hybrid orbitals.
The two $$\,sp$$  hybrid orbitals are directed diagonally, i.e., in a straight line with an angle of $${180^ \circ }$$  (collinear orbitals). The other two $$p$$ orbitals $$\left( {{\text{say}}\,{p_y}\,{\text{and}}\,{p_z}} \right)$$   remain pure.

In $${O_2}\left( {As{F_6}} \right),O_2^ + $$   has bond order 2.5.
$${O_2}$$ has bond order 2.0.
In $$K{O_2},O_2^ - $$   has bond order 1.5.
Higher the bond order, smaller is the bond length.
Hence, the order is $${O_2}\left( {As{F_6}} \right) > {O_2} > K{O_2}.$$

For any species to have same bond order we can expect them to have same number of electrons. Calculating the number of electrons in various species.
$$O_2^ - \left( {8 + 8 + 1 = 17);\,\,C{N^ - }\left( {6 + 7 + 1 = 14} \right)} \right.$$
$$N{O^ + }\left( {7 + 8 - 1 = 14} \right);\,C{N^ + }\left( {6 + 7 - 1 = 12} \right)$$
We find $${C{N^ - }}$$   and $$N{O^ + }$$  both have $${14}$$  electrons so they have same bond order. Correct answer is (A).

All the molecules have trigonal planar structure.

$$1.\,\,NO\left( {15} \right) \to $$   one unpaired electron is present in $${\pi ^*}$$ molecular orbit hence paramagnetic.
$$2.\,CO\left( {14} \right) \to $$   $$\sigma _{1s}^2,\sigma _{1s}^{ * 2},\sigma _{2s}^2,\sigma _{2s}^{ * 2},\pi _{2{p_X}}^2,\pi _{2{p_y}}^2,\sigma _{2{p_Z}}^2$$
No unpaired electron, hence diamagnetic.
$$3.\,{O_2}\left( {16} \right) \to $$   $$\sigma _{1s}^2,\sigma _{1s}^{ * 2},\sigma _{2s}^2,\sigma _{2s}^{ * 2},\sigma _{2{p_Z}}^2,\pi _{2{p_X}}^2,\pi _{2{p_Y}}^2,\pi _{2{p_X}}^{ * 1},\pi _{2{p_Y}}^{ * 1}$$
Two unpaired electrons, hence paramagnetic.
$$4.\,{B_2}\left( {10} \right) \to $$   $$\sigma _{1s}^2,\sigma _{1s}^{ * 2},\sigma _{2s}^2,\sigma _{2s}^{ * 2},\pi _{2{p_X}}^1,\pi _{2{p_Y}}^1$$
$${B_2}$$ contains two unpaired electrons, hence paramagnetic

In nitrogen molecule, both the nitrogen atoms have same electronegativity. So it has zero polarity and hence less tendency to break away and form $$ions.$$

Intermolecular hydrogen bonding is present in concentrated acetic acid, $${H_2}{O_2}$$  and $$HCN$$  while cellulose has intra- molecular hydrogen bonding as shown below :


In above molecules, dotted lines represent hydrogen bonding.

$$\left[ {Ne} \right]3{s^2}3{p^3}$$   will have highest ionisation enthalpy due to extra stability of half-filled orbitals and smaller size.