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Byju's Answer

Standard XII

Chemistry

Bond Length

In the given ...

Question

In the given reaction:
${[C o {({N H}_{3})}_{5} H_{2} O]}^{2 +} + {N H}_{3} ⟶ c o m p o u n d (Y) + e^{-} + H_{2} O$
Compound (X)
Number of unpaired electron in compound X and compound Y are respectively:

(1, 0)

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(3, 1)

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(1, 3)

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(3, 3)

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Solution

The correct option is B $(1, 0)$
${[C o {({N H}_{3})}_{5} H_{2} O]}^{2 +} \to d^{7}$
$(X)$ $\to$ low spin
$\to$ No. of unpaired $e^{-} = 1$
${[({C o (N H}_{3})_{6})]}^{3 +} \to d^{6}$
$(Y)$ $\to$ low spin
$\to$ No. of unpaired $e^{-} = 0$

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Similar questions

Q. Consider the following two reactions:
I.

Z n + c o n c . H N O_{3} (h o t) \to Z n (N O_{3})_{2} + X + H_{2} O

II.

Z n + d i l . H N O_{3} (c o l d) \to Z n (N O_{3})_{2} + Y + H_{2} O

Compounds

X

and

Y

are respectively :

Q. The following two reactions of

H {N O}_{3}

with

Z n

are given as (equations are not balanced)

Z n + c o n c . H {N O}_{3} ⟶ Z n {({N O}_{3})}_{2} + X + H_{2} O . . . . . . . . . . . . . . . . . . (A)

Z n + d i l . H {N O}_{3} ⟶ Z n {({N O}_{3})}_{2} + Y + H_{2} O . . . . . . . . . . . . . . . . . . . . (B)

In reactions

A

and

B

, the compounds

X

and

Y

respectively are

Question 04

Hydrogen bonds are formed in many compounds e.g., $H_{2} O, H F, N H_{3}$ . The boiling point of such compounds depends to a extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points above compounds is

(a) $H F > H_{2} O > N H_{3}$

(b) $H_{2} O > H F > N H_{3}$

(d) $N H_{3} > H_{2} O > H F$

Q. In the following reaction, X and Y are respectively _________.

C H_{3} C O O H + N H_{3} \to X Δ - \to Y + H_{2} O

Q. The correct order of magnitude of bond angles among the compounds

C H_{4}, N H_{3} a n d H_{2} O

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In the given reaction:[Co(NH3)5H2O]2++NH3⟶compound(Y)+e−+H2OCompound (X)Number of unpaired electron in compound X and compound Y are respectively:

The correct option is B (1,0)[Co(NH3)5H2O]2+→d7 (X) → low spin → No. of unpaired e−=1[(Co(NH3)6)]3+→d6 (Y) → low spin → No. of unpaired e−=0

In the given reaction:
${[C o {({N H}_{3})}_{5} H_{2} O]}^{2 +} + {N H}_{3} ⟶ c o m p o u n d (Y) + e^{-} + H_{2} O$
Compound (X)
Number of unpaired electron in compound X and compound Y are respectively:

The correct option is B $(1, 0)$
${[C o {({N H}_{3})}_{5} H_{2} O]}^{2 +} \to d^{7}$
$(X)$ $\to$ low spin
$\to$ No. of unpaired $e^{-} = 1$
${[({C o (N H}_{3})_{6})]}^{3 +} \to d^{6}$
$(Y)$ $\to$ low spin
$\to$ No. of unpaired $e^{-} = 0$