Sodium borohydride reacts with iodine in the presence of diglyme to give A. A heated at 388 K give B. A heated at 373 K in sealed tube to form C. A further heated at red hot condition to give element D. Find out A,B, C and D. Give the reactions.

Explanation

(i) Sodium borohydride reacts with iodine in the presence of diglyme to give diborane (A).
`2NaBH_(4)+I_(2)overset("diglyme")(to)underset((A))(B_(2)H_(6))+2NaI+H_(2)`
(ii) Diborane (A) heated at 388 K gives pentaborane (11)(B)
`underset((A))underset(("Diborane"))(5B_(2)H_(6))underset((U-"tube"))overset(388K)(to)underset((B))underset(("pentaborane-11"))(2B_(5)H_(11))+4H_(2)`
(iii) Diborane (A) heated at 373 K in the sealed tube to form decaborane (14)(C)
`underset((A))underset(("Diborane"))(5B_(2)H_(6))underset("sealed tube")overset(373K)(to)underset((C))underset(("decaborane"-14))(B_(10)H_(14)+8H_(2)`
(iv) Diborane (A) heated at red hot conditions to give element boron (D).
`underset((A))underset(("Diborane"))(B_(2)H_(6))overset("Red hot")(to)underset((D))underset(("Boron"))(2B)+3H_(2)`