Density of ice is less than water due to an open-cage structure formed by hydrogen bonding.

Critical temperature of water is higher than $${O_2}$$  because $${H_2}O$$  molecule has dipole moment which is due to its V-shape.

The van der Waals’ forces are weakest forces and covalent bond is strongest, so the order of interactions is van der Waals’ < $$H$$ - bonding < dipole-dipole < covalent

$${O_2}:\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$      $$\left\{ {_{\pi 2p_z^2}^{\pi 2p_y^2},\left\{ {_{{\pi ^*}2p_z^1}^{{\pi ^*}2p_y^1}} \right.} \right.$$
Bond order $$\, = \frac{{10 - 6}}{2} = 2$$
(two unpaired electrons in antibonding molecular orbital)
$$O_2^ + :\,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},\sigma 2p_x^2,$$     $$\left\{ {_{\pi 2p_z^2}^{\pi 2p_y^2},\left\{ {_{{\pi ^*}2p_z^0}^{{\pi ^*}2p_y^1}} \right.} \right.$$
Bond order $$ = \frac{{10 - 5}}{2} = 2.5$$
(One unpaired electron in antibonding molecular orbital)
Hence $${O_2}$$ as well as $$O_2^ + $$ is paramagnetic, and bond order of $$O_2^ + $$  is greater than that of $${O_2}.$$

In $$s{p^3}$$  hybrid orbital one part, out of four orbital is $$s$$-orbital, so it have $$25\% $$  or $$\frac{1}{4}$$  $$s$$-character.

NOTE : Greater the difference between electronegativity of bonded atoms, stronger will be bond. Since $$F$$ is most electronegative hence $$F - H\,\,....\,\,F$$    is the strongest bond.

The structure of $$Ca{C_2}$$  is $$C{a^{2 + }}{\left[ {:C \equiv C:} \right]^{2 - }}$$
i.e, one $$\sigma $$ and two $$\pi $$ bonds

No explanation is given for this question. Let's discuss the answer together.

$${\text{Bond order of}}\,\,O_2^ - $$
$$O_2^ - = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},$$      $$\mathop \sigma \limits^* 2{s^2}\,\sigma 2p_z^2\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$     $$\left( {{\pi ^*}2p_x^1 = {\pi ^*}2p_y^1} \right)$$
$$\eqalign{ & {\text{Bond}}\,\,{\text{order}} \cr & = \frac{{{\text{number}}\,\,{\text{of}}\,\,{\text{electron}}\,\,{\text{in}}\,\,BMO - {\text{number}}\,\,{\text{of}}\,\,{\text{electrons}}\,\,ABMO}}{2} \cr & = \frac{{10 - 7}}{2} \cr & = \frac{3}{2} \cr & = 1.5 \cr} $$
$$O_2^ + = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},$$      $$\mathop \sigma \limits^* 2{s^2}\,\sigma 2p_z^2\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$     $$\left( {{\pi ^*}2p_x^1 = {\pi ^*}2p_y^0} \right)$$
$$\eqalign{ & BO = \frac{{10 - 5}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{5}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 2.5 \cr} $$
$${O_2} = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},$$      $$\mathop \sigma \limits^* 2{s^2}\,\sigma 2p_z^2\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$     $$\left( {{\pi ^*}2p_x^1 = {\pi ^*}2p_y^1} \right)$$
$$\eqalign{ & BO = \frac{{10 - 6}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{4}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 2 \cr & {\text{So, the correct sequence is}} \cr & O_2^ - < {O_2} < O_2^ + \cr} $$

In $$S{F_4}$$  the hybridisation is $$s{p^3}d$$  and the shape of molecule is