(a) On bolining a minerak `(A)` with `NaCO_(3)` solution, a white precipirtate `(B)` is formed.
(b) The precipitate is fitered and the filtrate contains two conpounds `( C)` and `(D)`. The componds `( C)` is removed by crystallisation and when `CO_(2)` is passed through the mother liquor left, `(D)` changes to `( C)`.
( c) On strongly heating `( C)`, two compounds `(D)` and `( E)` are formed.
(d) `(E)` on heating with cobalt oxide provides blue-coloured substance `(F)`.
Identify `(A)` to `(F)` and give chemical reaction for `(a)` to `(d)`.

Explanation

(a) The mineral `(A)` is colemanite, `Ca_(2)B_(6) O_(11). 5H_(2)O`
`Ca_(2) B_(6) underset((A))(O_(11)).5H_(2) O + 2Na_(2)CO_(3(aq)) overset("Boiled")rarr`
`underbrace(2CaCO_(3) darr + underset((C))(Na_(2)B_(4)O_(7(aq))) + underset((D))(2NaBO_(2(aq))))`
`(B)`, i.e white ppt. is of calcium carbonate.
(b) The filterate formed in step `(a)` contains borax `Na_(2)B_(4)O_(7) (C)` and sodium metaborate, `NaBO_(2) (D)`.
On crystallisation, borax gets separated out in the form of white crystals. Filterate now consists of `NaBO_(2)` only. On passing `CO_(2)` though the filterate, sodium metaborate `NaBO_(2) (D)` reacts to form borax `Na_(2)B_(4) O_(7)`, i.e. `(C)`.
`underset((D))(NaBO_(2)) + CO_(2) rarr underset((C))(Na_(2)B_(4)O_(7))+ Na_(2)CO_(3)`
( c) On strongly heating `(C)`, sodium metaborate `(D)` and boro tioxide `(E)` are formed.
`Na_(2)B_(4)O_(7).10 H_(2) O overset(Delta)rarr Na_(2)B_(4)O_(7) + 10 H_(2) O`
`Na_(2)B_(4)O_(7) overset(Delta)rarr underset((D))(2NaBO_(2))+ underset(underset("Boron trioxide")((E)))(B_(2)O_(3))`
(d) On heating `(E)` with cobalt oxide `(CoO)`, blue-coloured cobalt metaborate, `Co(BO_(2))_(2) (F)` is formed.
`B_(2)O_(3) + CoO rarr underset(underset(("blue coloured"))("Cobalt metaborate"))(Co"(BO_(2))_(2))`
Hence, `(A)` is `Ca_(2)B_(6)O_(11). 5H_(2) O , (B)` is `CaCO_(3) , ( C)` is `Na_(2) B_(4)O_(7) . 10 H_(2) O , (D)` is `NaBO_(2)` and `(E)` is `Co(BO_(2))_(2)`.