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Byju's Answer

Standard XII

Chemistry

Borax

solution of t...

Question

solution of the cpmpound $(A) + g a s B - --- \to (C) δ \to (A) + (D)$

conc. solution of $(C) c o n c . H_{2} S O_{4} - ------ \to (E)$ (weak acid)

$(D) +$ cobalt oxide $Δ - \to (F)$ (coloured bead)

$(E) + C_{2} H_{5} O H Δ - \to$ green edged flame $(G)$

A to G based on the following reactions are:

A: ${N a B O}_{2}$
B: ${C O}_{2}$
C: ${N a}_{2} B_{4} O_{7}$
D: $B_{2} O_{3}$
E: $H_{3} {B O}_{3}$
F: ${C o ({B O}_{2})}_{2}$
G: $(C_{2} H_{5} O)_{3} B$

If the statement is true enter 1, else enter 0.

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Solution

The compound A is ${N a B O}_{2}$
The compound B is ${C O}_{2}$
The compound Cis ${N a}_{2} B_{4} O_{7}$
The compound D is $B_{2} O_{3}$
The compound E is $H_{3} {B O}_{3}$
The compound Fis ${C o ({B O}_{2})}_{2}$
The compound G is $(C_{2} H_{5} O)_{3} B$
${N a B O}_{2}$ reacts with ${C O}_{2}$ gas to form ${N a}_{2} B_{4} O_{7}$

$4 {N a B O}_{2} + {C O}_{2} \to {N a}_{2} B_{4} O_{7} + N a_{2} C O - 3$

${N a}_{2} B_{4} O_{7}$ on heating produces ${N a B O}_{2}$ and
$B_{2} O_{3}$ (glassy bead)

${N a}_{2} B_{4} O_{7} h e a t - - \to 2 {N a B O}_{2} + B_{2} O_{3}$

Conc. solution of ${N a}_{2} B_{4} O_{7}$ reacts with conc. sulphuric acid to form weak acid

$H_{3} {B O}_{3}$

${N a}_{2} B_{4} O_{7} + H_{2} S O_{4} + 5 H_{2} O \to 4 H_{3} {B O}_{3} + N a_{S} O_{4}$

$B_{2} O_{3}$ on heating with cobalt oxide gives a coloured bead of ${C o ({B O}_{2})}_{2}$
$B_{2} O_{3} + C o O \to {C o ({B O}_{2})}_{2}$

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Similar questions

(A), (B), (C), (D), (E) and (F) base on the reactions are

A :

Z n S

B :

H_{2} S

C :

{Z n S O}_{4}

D :

P b S

E :

{Z n (O H)}_{2}

F :

{N a}_{2} {Z n O}_{2}

{N a}_{2} [{Z n (O H)}_{4}]

If the given statement is true enter 1, else enter 0.

Q. Classify the following into Lewis acids and Lewis bases:
(a)

H_{2} O

(b)

N H_{3}

(c)

H^{+}

(d)

B F_{3}

(e)

A I H_{3}

(f)

R O H

(g)

C H_{3} N H_{2}

(h)

C N^{-}

(i)

^{+} C H_{3}

Lewis acids : (c), (d), (e), (i)
Lewis bases : (a), (b), (f), (g), (h)
If true enter 1, else enter 0.

A(orange)

Δ - \to C (g) + r e s i d u e (c) (g r e e n) + H_{2} O

↓ N a O H

D(g)

H C l - - \to

white fumes

B(g)

A l - \to Δ (E) H_{2} O - - \to g a s (B)

Residue (C)

c o n c . H C l - ---- \to (F) N a O h, H_{2} O_{2} - ------ \to Δ G

(yellow)

C H_{3} C O O H - -------- \to (C H_{3} C O O)_{2} P b

H (yellow

If true enter 1, else enter 0.

Compounda (A) to (H) are identified as (A)

({N H}_{4})_{2} {C r}_{2} O_{7}

, (B)

N_{2}

, (C)

{C r}_{2} O_{3}

, (D)

{N H}_{3}

, (E)

A I N

, (F)

{C r C l}_{3}

, (G)

{N a}_{2} {C r O}_{4}

and (H)

{P b C r O}_{4}

Q. If true enter 1, else enter 0.
(i) A white precipitate (B) is formed, when a mineral (A) is boiled with

{N a}_{2} {C O}_{3}

solution.
(ii) The precipitate is filtered an the filtrate contains two compounds (C) and (D). The compound (C) is removed by crystallisation and when

{C O}_{2}

is passed through the mother liquor left, (D) changes to (C).
(iii) The compound (C) on strong heating gives two compound (D) and (E).
(iv) E on heating with

C o O

forms blue colour bead (F).
(A) to (F) are identified as (A) :

C a_{2} B_{6} O_{11}

C a C O_{3}

N a_{2} B_{4} O_{7}

N a B O_{2}

B_{2} O_{3}

and

C o (B O_{2})_{2}

A, B, C, D, E and F in the following are A:

N a_{2} S_{2} O_{3}

; B:

A g_{2} S_{2} O_{3}

; C:

N a_{3} [A g (S_{2} O_{3})_{2}]

; D :

A g_{2} S

; E :

H_{2} S O_{4}

; F:

B a S O_{4}

If the statement is true enter 1, else enter 0.

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