Chemistry Test

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Chemistry Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

The uncertainty in position of an electron is equal to its de Broglie wavelength. The minimum percentage error in its measurement of velocity under this circumstance will be approximately.

A
4

B
8

C
2

D
18

Solution

$\lambda=\frac{h}{p}$
or $\Delta x=\frac{h}{p}$
By uncertainty principle
$\Delta x \cdot \Delta p=\frac{h}{4 \pi}$
or $\frac{h}{p} \times \Delta p=\frac{h}{4 \pi}$
$\frac{\Delta p }{ p }=\frac{1}{4 \pi}$
$\frac{m \times \Delta v}{m \times v}=\frac{1}{4 \pi}$
$\frac{\Delta v}{v}=\frac{1}{4 \pi}$
uncertainty in velocity $=\frac{\Delta v}{v} \times 100=\frac{1}{4 \pi} \times 100=8 \%$
Hence, the correct option is (B)
Question 2
1 / -0

A

B

C

D

Solution

—OH is at position (1) is above the plane, so –OH addition will take above the plane.

In this compound, attack on double bond will be in syn manner and from below the plane. This will result in dash configuration which would further avoid steric hindrance.
Question 3
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When conc. H₂SO₄ is treated with K₄[Fe(CN)₆], Co gas is evolved. By mistake, somebody used dil.H₂SO₄, then which gas will be released?

A
CO

B
HCN

C
N₂

D
CO₂

Solution

$K_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+6 \mathrm{H}_{2} \mathrm{SO}_{4}+6 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{FeSO}_{4}+3\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}+6 \mathrm{CO}$

$\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{FeSO}_{4}+6 \mathrm{HCN} \uparrow$

Hence option (B) is correct.
Question 4
1 / -0

\begin{equation}XCl _{2}( excess )+ YCl _{2} \rightarrow XCl _{4}+ Y \downarrow\end{equation}\begin{equation}Y O \longrightarrow \frac{\Delta}{>400^{\circ} C} \rightarrow \frac{1}{2} O_{2}+Y\end{equation} ore of Y would be

A
Siderite

B
Cinnabar

C
Malachite

D
Horn silver

Solution

\begin{equation}\begin{array}{l}
SnCl _{2}+ HgCl _{2} \longrightarrow SnCl _{4}+ Hg _{} \\
\left( XCl _{2}\right) \quad\left( Y Cl _{2}\right) \quad\left( X Cl _{4}\right) \quad\left( Y\right)
\end{array}\end{equation}\begin{equation}HgO \frac{\Delta}{>400^{\circ} C } \rightarrow \underset{( Y )}{ Hg }+\frac{1}{2} O _{2}\end{equation} Cinnabar (HgS) is Ore of Hg.
Siderite is $FeCO _{3}$
Malachite is $Cu _{2} CO _{3}( OH )_{2}$
Hornsilver is AgCl.
Hence, option (B) is correct.
Question 5
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$S a l t (A) + S \to B \overset{B a C l_{2}}{\to} W h i t e p p t .$
A is paramagnetic. Thus A is

A
K₂O

B
K₂O₂

C
KO₂

D
K₂SO₄

Solution

is paramagnetic due to (superoxide ion)
$2 K o_{2} + S \to \overset{K_{2} S O_{4}}{(B)} \overset{B a C l_{2}}{\to} \overset{B a S O_{4}}{w h i t e p p t .} ↓ + 2 K C l$
Hence, (C) is correct options.
Question 6
1 / -0

[X₃B ← NH₃], in which of the following boric halide, tendency to accept electrons from nitrogen of ammonia will be least (X = halogens)

A
BI₃

B
BCl₃

C
BBr₃

D
BF₃

Solution

In BF₃, due to F back bonding to B, least Lewis acid character is observed. So out of above four, BF₃ will have least tendency to accept electron from NH₃.
Hence, option (D) is correct.
Question 7
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The enolate ion that reacts with 3–buten–2–one to form (Y) is:

A

B

C

D

Solution
Question 8
1 / -0

Major product A. is:

A

B

C

D

Solution
Question 9
1 / -0

Product A is;

A

B

C

D

Solution
Question 10
1 / -0

E₁, E₂ and E₃ are activation energies then, which of the following is correct.

A
E₁ > E₃ > E₂

B
E₃ > E₁ > E₂

C
E₃ > E₂ > E₁

D
E₁ > E₂ > E₃

Solution

Here, aromaticity is being ruptured hence activation energy is highest.

Here double bonds are in conjugation, so it will be difficult to hydrogenate then. E2 will also be high but not so high as E1.

Here, no conjugation or aromaticity present in reaction so process will be easy. E3 will be lowest.

E1 > E2 > E3.

Hence, option D is correct.