When $$s$$ -orbital and $$p$$ -orbital overlap each other, then the bond angle formed is $${180^ \circ }$$  as given below

The correct order of energies of molecular orbitals in $${N_2}$$  molecule is $$KK\sigma 2s < {\sigma ^ * }2s < \left( {\pi 2{p_x} = \pi 2{p_y}} \right)$$       $$ < \sigma 2{p_z} < \left( {{\pi ^ * }2{p_x} = {\pi ^ * }2{p_y}} \right) < {\sigma ^ * }2{p_z}$$

TIPS/Formulae :
(i) Non metallic oxides are more covalent (or less ionic) as compared to metallic oxides.
(ii) Higher the polarising power of cation (higher for higher oxidation state of similar size cations) more will be covalent character.
(i) $${P_2}{O_5}$$  will be more covalent than other metallic oxides.
(ii) Oxidation state of $$Mn$$  is $$+7$$  in $$M{n_2}{O_7},$$  oxidation state of $$Cr$$  in $$Cr{O_3}$$  $$+6$$  and oxidation state of $$Mn$$  is $$+2$$  in $$MnO.$$
∴ $$MnO$$  is most ionic.
NOTE : $${P_2}{O_5},$$  being anon-metallic oxide will definitely be more covalent than the other metallic oxides. Further, we know that higher the polarising power of the cation ( higher for higher oxidation state of the similar size cations ) more will be the covalent character. Here $$Mn$$  is in $$+7\,O.S$$   in $$M{n_2}{O_7},Cr$$   in $$+6$$  in $$Cr{O_3}$$  and $$Mn$$  in $$+2$$  in $$MnO.$$  So $$MnO$$  is the most ionic and $$M{n_2}{O_7}$$  is the most covalent.

$$H = \frac{1}{2}\left( {3 + 3 + 0 - 0} \right) = 3$$
∴ Boron, in $$B{F_3},$$  is $$s{p^2}$$  hybridised leading to trigonal planar shape.

Thestructure of species can be predicted on the basis of hybridisation which in turn can be known by knowing the number of hybrid orbitals $$(H)$$  in that species

$$=\frac{1}{2}\left( {6 + 4 + 0 - 0} \right) = 5$$
For $$S{F_4}$$  : $$S$$  is $$s{p^3}d$$  hybridised in $$S{F_4}.$$  Thus $$S{F_4}$$  has 5 hybrid orbitals of which only four are used by $$F,$$  leaving one lone pair of electrons on sulphur.
For $$C{F_4}:H = \frac{1}{2}\left[ {4 + 4 + 0 - 0} \right]$$     $$ = 4$$  ∴ $$s{p^3}$$  hybridisaion
Since all the four orbitals of carbon are involved in bond formation, no lone pair is present on $$C$$  having four valence electrons
For $$Xe{F_4}$$  : $$H = \frac{1}{2}\left( {8 + 4 + 0 - 0} \right) = 6,$$      ∴ $$s{p^3}{d^2}$$  hybridization of the six hybrid orbitals, four form bond with $$F,$$  leaving behind two lone pairs of electrons on $$Xe$$ .

The structure of $$IF_6^ - $$  is distorted octahedral This is due to presence of a lone pair.

$$M.O.{\text{ configuration of}}\,{O_2}:$$
$$\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)\left( {{\sigma ^ * }2{s^2}} \right)$$      $$\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$    $$\left( {{\pi ^ * }2p_x^1 = {\pi ^ * }2p_y^1} \right)$$

In ether, there is no $$H $$ - bonding while alcohols have intermolecular $$H$$ - bonding

In the given molecules nitrogen has greater electronegativity. So, it has greater dipole moment and correct order of dipole moment is $$N{H_3} > P{H_3} > As{H_3} > Sb{H_3}$$

The order of bond enthalpies of the bonds is given as, Bond order $$ \propto $$  Bond enthalpy
Hence, single bond enthalpy < double bond enthalpy < triple bond enthalpy
i.e., $$H - H < O = O < N \equiv N$$