Question

(a) Complete the following chemical equations:
(i) $Xe{F}_4+Sb{F}_5\to$
(ii) $C{l}_2+F_2\left(excess\right)\to$
(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down group 15.
(iii) The bond angles ($O-N-O$) are not of the same value in $N{O}_2^{-}$ and $N{O}_2^{+}$.

Answer 1

a) i) $Xe{F}_4$ or xenon tetrafluoride reacts with $Sb{F}_5$ which is an Lewis acid and forms adduct-

$Xe{F}_4+Sb{F}_5\to [Xe{F}_3{]}^{+}+[Sb{F}_6{]}^{-}$

ii) $C{l}_2+3F_2\to 2Cl{F}_3$

B) i) Nitrogen is much less reactive than phosporous as it exists as $N_2$ in which 2 nitrogen atoms are bonded with triple bond and are stable.

ii) The stability of +5 oxidation state decreases down the group due to inert pair effect.

iii) $N{O}_2^{+}$ has more bond angle (180 degrees) than $N{O}_2^{-}$ because $N{O}_2^{+}$ has no lone pairs which do not disturb the regular bond angle of 180 degree but where as in $N{O}_2^{-}$ there is one lone pairs so the l.p and b.p repulsion leads to decrease in bond angle from regular bond angle of ${180}^o$.