Using molecular orbital theory, compare the bond energy and magnetic character of $\mathrm{O}_2^{+}$and $\mathrm{O}_2^{-}$species.
According to molecular orbital theory electronic configurations of $\mathrm{O}_2^{+}$ and $\mathrm{O}_2^{-}$ species are as follows
$$\mathrm{O}_2^{+}:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x^2, \pi 2 p_y^2\right)\left(\pi^* 2 p_x^1\right)$$
Bond order of $\mathrm{O}_2^{+}=\frac{10-5}{2}=\frac{5}{2}=2.5$
$$\mathrm{O}_2^{-}:(\sigma 1 s)^2\left(\sigma^{\star} 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^{\star} 2 s^2\right)\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x^2, \pi 2 p_y^2\right)\left(\pi^{\star} 2 p_x^2, \pi^* 2 p_y^1\right)$$
Bond order of $\mathrm{O}_2^{-}=\frac{10-7}{2}=\frac{3}{2}=1.5$
Higher bond order of $\mathrm{O}_2^{+}$shows that it is more stable than $\mathrm{O}_2^{-}$. Both the species have unpaired electrons. So, both are paramagnetic in nature.
Explain the shape of BrF$$_5$$.
The central atom Br has seven electrons in the valence shell. Five of these will form bonds with five fluorine atoms and the remaining two electrons are present as one lone pair.
Hence, total pairs of electrons are six ( 5 bond pairs and 1 lone pair). To minimize repulsion between lone pairs and bond pairs, the shape becomes square pyramidal.
Structures of molecules of two compounds are given below.
(a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding?
(b) The melting point of compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point?
(c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with easily and be more soluble in it?
(a) Compound (I) will form intramolecular H -bonding. Intramolecular H -bonding is formed when H -atom, in between the two highly electronegative atoms, is present within the same molecule. In ortho-nitrophenol (compound I), H -atom is in between the two oxygen atoms.
Compound (II) forms intermolecular H-bonding. In para-nitrophenol (II) there is a gap between $\mathrm{NO}_2$ and OH group. So, H -bond exists between H -atom of one molecule and O-atom of another molecule as depicted below.
(b) Compound (II) will have higher melting point because large number of molecules are joined together by H-bonds.
(c) Due to intramolecular H -bonding, compound ( I ) is not able to form H -bond with water, so it is less soluble in water. While molecules of compound II form H -bonding with $\mathrm{H}_2 \mathrm{O}$ easily, so it is soluble in water.
Why does type of overlap given in the following figure not result in bond formation?
In the figure (I), area of ++ overlap is equal to + - overlap, so net overlap is zero, while in figure (II), there is no overlap due to different symmetry.
Explain why $\mathrm{PCl}_5$ is trigonal bipyramidal whereas $\mathrm{IF}_5$ is square pyramidal.
$\mathrm{PCl}_5-$ The ground state and the excited state outer electronic configurations of phosphorus $(Z=15)$ are represented below
In $\mathrm{PCl}_5$, P is $s p^3 d$ hybridised, therefore, its shape is trigonal bipyramidal. $\mathrm{IF}_5$-The ground state and the excited state outer electronic configurations of iodine $(Z=53)$ are represented below.
In $\mathrm{IF}_5, \mathrm{I}$ is $s p^3 d^2$ hybridised, therefore, shape of $\mathrm{IF}_5$ is square pyramidal.