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Cyclotrimetaphosphoric acid is $${\left( {HP{O_3}} \right)_3}.$$

$${N_2}O = + 1,NO = + 2,N{O_2} = + 4,$$       $$NO_3^ - = + 5,NH_4^ + = - 3$$
Increasing order of oxidation state will be
$$NH_4^ + < {N_2}O < NO < N{O_2} < NO_3^ - $$

In sulphur trioxide trimer $${{S_3}{O_9}}$$  ( also called $$\gamma - $$  sulphur trioxide ) two sulphur atoms are linked to each other via $$O\,$$ atoms, hence there is no $$S - S$$   bond.

$$\eqalign{ & 2A{g_2}O\left( s \right) \to 4Ag\left( s \right) + {O_2}\left( g \right) \cr & 2P{b_3}{O_4}\left( s \right) \to 6PbO\left( s \right) + {O_2}\left( g \right) \cr & 2Pb{O_2}\left( s \right) \to 2PbO\left( s \right) + {O_2}\left( g \right) \cr} $$

$${B_2}{H_6} + 6{H_2}O \to 2{H_3}B{O_3} + 6{H_2} \uparrow $$

As the size of halogen atom increases, the acidic strength of boron halides increases. Thus, $$B{F_3}$$  is the weakest Lewis acid. This is because of the $$p\pi - p\pi $$   back bonding between the fully filled unutilised $$2p$$ -orbitals of $$F$$  and vacant $$2p$$ -orbitals of boron which makes $$B{F_3}$$  less electron deficient. Such back donation is not possible in case of $$BC{l_3}$$  or $$BB{r_3}$$  due to larger energy difference between their orbitals. Thus, these are more electron deficient. Since on moving down the group the energy difference increases, the Lewis acid character also increases. Thus, the tendency to behave as Lewis acid follows the order
$$BB{r_3} > BC{l_3} > B{F_3}$$

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

Valence shell electronic configuration of oxygen is $$2{s^2}2{p^4}.$$

Acidic nature of oxides decreases down a group.
So, $${N_2}{O_5}$$  is most acidic.
Another reason of acidic strength of $${N_2}{O_5}$$  is that the electronegativity of $$N$$  is maximum in the given $${V^{th}}$$ group elements. As we know that on increasing the electronegative character, acidic nature increases.