Boron reacts with oxygen at `700^@C` to give `(A)`. Compound `(A)` reacts with carbon and dry chloride to give `(B)` and carbon monoxide. `(B)` on reduction with `LiAlH_(4)` gives `( C)` along with `LiCl` and `AlCl_(3). ( C)` on reaction with ammonia gives `(D)`. Which on heating gives `(E).(C)` on reaction with `NaH` gives `(F)`.
Compound `(F)` behaves as.

Options

A. Reducing agent
B. Oxidising agent
C. Complexon
D. Buffer

Answer

A

Explanation

`B +O_(2) overset(700^@C)rarr (A)`
`(A) +C+Cl_(2) rarr(B) + CO`
`(B) +LiAlH_(4) rarr (C) + LiCl + AlCl_(3)`
`(C) + NH_(3) overset(-120^@C)rarr (D)`
`(D) overset(Delta) underset(200^@ C)rarr (E)`
`(C) + NaH rarr (F)`
`4B +3O_(2) overset(700^@ C)rarr underset((A))(2B_(2)O_(3))`
`B_(2)O_(3) +3C +3Cl_(2) rarr underset((B))(2BCl_(3))+ 3CO`
`underset((B)) (4BCl_(3)) + 3LiAlH_(4) rarr underset((C)) (2B_(2) H_(6)) + 3LiAl + 3AlCl_(3)`
`B_(2)H_(6) +2NH_(3) overset (-120^@ C)rarr underset((D))([H_(2)B(NH_(3))_(2)]^(oplus) [BH_(4)]^(Ө))`
`3[H_(2)B(NH_(3))_(2)]^(oplus) [BH_(4)]^(Ө) overset(200^@ C)rarr underset((E)) (2B_(3)N_(3)H_(6) + 8H_(2))`
`B_(2)H_(6) + 2NaH rarr underset((F))(2NaBH_(4))`.
Compound `(F)` is `NaBH_(4)`, which behaves as a reducing agent.