B2H6 - NH3 - Addition compound X X 450K- Y - ZgIn the above sequence Y and Z are respectively-

The correct option is C borazine and hydrogen B_2H_6+2NH_3 ⟶(X)B_2H_62NH_3 3B_2H_62NH_3(Y)2B_3N_3H_6+(Z)3H_2 (Y) amp; (Z) are borazine ...

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Byju's Answer

Standard XII

Chemistry

Diborane

B2H6 + NH3 → ...

Question

$B_{2} H_{6} + N H_{3} \to$ Addition compound $(X)$
$(X) 450 K \to Y + Z (g)$
In the above sequence $Y$ and $Z$ are respectively:

borazine,

H_{2} O

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boron,

H_{2}

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boron nitride,

H_{2}

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borazine and hydrogen

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Solution

The correct option is C borazine and hydrogen
$B_{2} H_{6} + 2 N H_{3} ⟶ B_{2} H_{6} 2 N H_{3} (X)$
$3 B_{2} H_{6} 2 N H_{3} 450 K - -- \to 2 B_{3} N_{3} H_{6} (Y) + 3 H_{2} (Z)$
$(Y)$ & $(Z)$ are borazine and $H_{2}$ respectively.

Suggest Corrections

Similar questions

Consider the following reactions:

B C l_{3} + L i A l H_{4} \to X + L i C l + A l C l_{3}

X + H_{2} O \to H_{2} + Y R e d h e a t \to Z

In the above reaction sequence,

X, Y

and

Z

are:

Q. A compound of boron X reacts at

200^{\circ} C

temperature with

N H_{3}

to give another compound

Y

which is called as inorganic benzene. The compound

Y

is a colourless liquid and is highly sensitive. Its melting point is

- 57^{\circ} C

. The compound

X

with excess

N H_{3}

and at a still higher temperature gives boron nitride

{(B N)}_{n}

. The compounds

X

and

Y

respectively are:

Q. In the given reaction,

B_{2} H_{6} + 2 K O H + 2 X \to 2 Y + 6 H_{2},

X and Y are respectively

Q. A compound of boron X reacts at

200^{\circ} C

temperature with

N H_{3}

to give another compound

Y

which is called as inorganic benzene. The compound

Y

is a colourless liquid and is highly sensitive. Its melting point is

- 57^{\circ} C

. The compound

X

with excess

N H_{3}

and at a still higher temperature gives boron nitride

{(B N)}_{n}

. The compounds

X

and

Y

respectively are:

N a B H_{4} + I_{2} \to X + Y + Z

B F_{3} + N a H 450 K - -- \to X + P

B F_{3} + L i A l H_{4} \to X + Q

Identify

X, Y, Z, P and Q .

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B2H6+NH3→ Addition compound (X)(X)450K→Y+Z(g)In the above sequence Y and Z are respectively:

The correct option is C borazine and hydrogenB2H6+2NH3⟶B2H62NH3(X)3B2H62NH3450K−−−→2B3N3H6(Y)+3H2(Z)(Y) & (Z) are borazine and H2 respectively.

$B_{2} H_{6} + N H_{3} \to$ Addition compound $(X)$
$(X) 450 K \to Y + Z (g)$
In the above sequence $Y$ and $Z$ are respectively:

The correct option is C borazine and hydrogen
$B_{2} H_{6} + 2 N H_{3} ⟶ B_{2} H_{6} 2 N H_{3} (X)$
$3 B_{2} H_{6} 2 N H_{3} 450 K - -- \to 2 B_{3} N_{3} H_{6} (Y) + 3 H_{2} (Z)$
$(Y)$ & $(Z)$ are borazine and $H_{2}$ respectively.