$$BC{l_3}$$  have $$s{p^2}$$  hybridisation and no lone pair of electron on central atom but $$NC{l_3}$$  have $$s{p^3}$$  hybridisation and also contains one lone pair of electron on nitrogen, so $$BC{l_3}$$  is planar.

$$N \equiv \,N\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:N \vdots \vdots N:$$

In $$B{F_3},$$  dipole moment is zero due to its symmetrical structure. Summations of all dipoles is zero.

In $${H_2}S$$  and $${H_2}O$$  due to unsymmetrical structure net $$+ve$$  dipole is there. $${H_2}O$$  has higher dipole due to higher electronegativity of oxygen than sulphur.

Electronic configuration of atom : $$1{s^2}2{s^2}2{p^5}$$
$$M.O.{\text{ configuration :}}$$
$$\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)\left( {{\sigma ^ * }2{s^2}} \right)$$      $$\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$    $$\left( {{\pi ^ * }2p_x^2 = {\pi ^ * }2p_y^2} \right)$$
$$\eqalign{ & B.O. = \frac{1}{2} \times \left( {10 - 8} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \cr} $$

Both $$Xe{F_2}\,\,{\text{and}}\,\,C{O_2}$$   have a linear structure.
$$F - Xe - F\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O = C = O$$

When a diatomic molecule lost electron, then its bond order may increase or decrease, so its bond energy may decrease or increase.

NOTE: Dipole moment is vector quantity
In trigonal planar geometry $$\left( {{\text{for}}\,s{p^2}\,{\text{hybridisation}}} \right)$$    , the vector sum of two bond moments is equal and opposite to the dipole moment of third bond.

According to Fajans’ rule,
$${\text{Covalent character}} \propto $$    $$\frac{1}{{{\text{size}}\,{\text{of}}\,{\text{cation}}}} \propto {\text{size}}\,{\text{of}}\,{\text{anion}}$$
In the given options,cation is same but anions are different. Among halogens the order of size is
$$\eqalign{ & F < Cl < Br < I \cr & \therefore {\text{Order of covalent character is}} \cr & Ml > MBr > MCl > MF \cr} $$

Molecular orbital configuration of $${B_2}\left( {10} \right)$$  as per the condition will be
$$\,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},p2p_y^2$$
Bond order of $${B_2} = \frac{{6 - 4}}{2} = 1,\,{B_2}$$    will be diamagnetic.