The stability of + 5 oxidation state decreases down the group due to inert pair effect. The only well characterised $$Bi\left( V \right)$$  compound is $$Bi{F_5}$$  as fluorine being most electronegative element is able to unpair $$ns$$  electrons.

\[{{H}_{2}}S+C{{l}_{2}}\xrightarrow{\text{water}}2HCl+S\]
Due to formation of free sulphur the solution becomes turbid.

$${B_2}{H_6} + \mathop {2L}\limits_{\left( {{\text{Lewis base}}} \right)} \to \mathop {2B{H_3} \cdot L}\limits_{{\text{Adduct}}} $$

The ease of formation and stability of hydrides decreases rapidly from $$N{H_3}\,{\text{to}}\,Bi{H_3}.$$   This is evident from their dissociation temperature which decreases from $$N{H_3}\,{\text{to}}\,Bi{H_3}.$$    As we go down the group the size of central atom increases and thus metal-hydrogen bond becomes weaker due to decreased overlap between the large central atom and small hydrogen atom.
$$\mathop {N{H_3} > P{H_3}}\limits_{{\text{(most stable)}}} > As{H_3} > Sb{H_3} > \mathop {Bi{H_3}}\limits_{{\text{(least stable}})} $$

Species	Hybridisation	Bond angle
$$N{O_2}$$	$$sp$$	less than $${120^ \circ }$$
$$NO_2^ - $$	$$s{p^2}$$	$${115.4^ \circ }$$
$$NO_2^ + $$	$$sp$$  (linear)	$${180^ \circ }$$
$$NO_3^ - $$	$$s{p^2}$$	$${120^ \circ }$$

So, $$NO_2^ + $$  has maximum bond angle.

The following reaction occurs
$$N{a_2}{S_2}{O_3} + 4C{l_2} + 5{H_2}O\,\,\, \to \,\,2NaHS{O_4} + 8HCl.\,\,$$

$$F$$ and $$Cl$$  are more oxidising in nature and can achieve $$Pb$$  in $$\left( {IV} \right)O.S.$$   but $$B{r_2}$$  and $${I_2}$$  can not achieve $$Pb$$  in $$\left( {IV} \right)O.S.$$   secondly $$P{b^{4 + }}$$  is strong in oxidising nature and in its presence, $$B{r^ - }$$  and $${I^ - }$$  can not exist.

The thermal stability of tetrahalides decreases in order $$C{X_4} > Si{X_4} > Ge{X_4} > Sn{X_4}$$      and in terms of same metal with different halides is in order of $$M{F_4} > MC{l_4} > MB{r_4} > M{I_4}.$$

$${\text{Borax is}}\,N{a_2}{B_4}{O_7} \cdot 10{H_2}O.$$

$$C{O_2}$$  forms carbonic acid $${H_2}C{O_3},$$  when dissolved in water, $$CO$$  is neutral, whereas other two $$Ge{O_2}$$  and $$Sn{O_2}$$  are solids.