Structure of Atom Test

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Structure of Atom Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

An element has three electrons in the 4th shell, the atomic number of the element is:

A
13

B
21

C
27

D
31

Solution

The configuration is $$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^1$$.

Three electrons in the 4th shell is the element Gallium (Ga), atomic number 31.

Hence, option D is correct.
Question 2
1 / -0

$$12\ g$$ of a trivalent element combines with $$4\ g$$ of oxygen. Its atomic mass is:

A
$$48$$

B
$$72$$

C
$$3$$

D
$$36$$

Solution

Equivalent mass of metal $$= \dfrac{\text{Weight of metal}}{\text{Weight of oxygen}} \times 8$$

$$= \dfrac{12}{8} \times 8$$

$$ = 12 $$

Therefore, atomic mass = Equivalent weight $$\times$$ valency

$$ = 12 \times 3 $$

$$ = 36 gm$$
Question 3
1 / -0

Radius of $$2nd$$ shell of $$He^+$$ (where $$a_0$$- Bohr radius)

A
$$3a_0$$

B
$$a_0$$

C
$$\dfrac{3}{2}a_0$$

D
$$2a_0$$

Solution

Radius of the Bohr's orbit is given by,
$$r=\dfrac{h^2n^2}{4\pi^2 m Z e^2}$$
Substituting values of h, m for helium ion and the electronic charge r may be calculated as
$$r=\dfrac{0.529n^2}{Z}A^0$$
Where $$0.529A^0$$ is is the Bohr's radius $$a_0=0.529$$
Substituting $$Z=2$$, $$n=2$$
$$r=2a_0A^0$$
Question 4
1 / -0

The number of orbitals in n = 3 are:

A
1

B
4

C
9

D
16

Solution

There are nine orbitals in the n = 3 shell. There is one orbital in the 3s subshell and three orbitals in the 3p subshell. n = 3 shell, however, also includes 3d orbitals.
Also, we have to note that the number of orbitals is given by $$n^2$$
Question 5
1 / -0

The active mass of $$7.0\ g$$ of nitrogen in a $$2.0\ L$$ container would be:

A
$$0.25$$

B
$$0.125$$

C
$$0.5$$

D
$$14.0$$

Solution

Moles of $$N_2=\dfrac{7}{28} = 0.25\ mol$$

Active mass = conc. in $$ mol/L$$

= $$0.25 mol / 2 L$$

= $$0.125$$
So, the correct option is $$B$$
Question 6
1 / -0

The key feature Of Bohr's theory Of spectrum Of hydrogen atom is the quantization Of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy Of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.A diatomic molecule has moment Of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n=0 is not allowed) is

A
$$\cfrac{h^2}{(n^2)(8\pi^2I)}$$

B
$$\cfrac{h^2}{(n)(8\pi^2I)}$$

C
$$\cfrac{n h^2}{8\pi^2I}$$

D
$$\cfrac{n^2 h^2}{(8\pi^2I)}$$

Solution

$$\begin{array}{l}\text { Given - * A Diatomic molecule- } \\\text { let, the structure of the molecule be like- }\end{array}$$

$$\Rightarrow I=2 m r^{2}$$

$$\begin{array}{l}\text { we howe Rotational encrgy- } \\\qquad E=\frac{1}{2} I \omega^{2}=m \omega^{2} r^{2}-(1) \\\text { By Bohr quantization of angular momentum } \\\qquad 2 \text { mwr}^2=\frac{n h}{2 \pi} \quad-(2) \\\text { From } e q(1) \text { and (2) we get ,}\\\qquad E=\frac{n^{2} h^{2}}{16 \pi^{2} m r^{2}}=\frac{n^{2} h^{2}}{8\pi^{2} I}\end{array}$$
Question 7
1 / -0

The electronic configuration of an element is $$1s^2, 2s^2, 2p^6, 3s^23p^3$$. What is the atomic number of the element which is just below the given element in the periodic table?

A
$$34$$

B
$$49$$

C
$$33$$

D
$$31$$

Solution

The element with the given electronic configuration is P (Phosphorus) and the atomic number is 15 below this is atomic number 33 As (Arsenic).
The electronic configuration of the Arsenic is given as follow:
$$[Ar]3d^{10}4s^24p^3$$
Question 8
1 / -0

$$7.8$$ grams of metal in water and liberated $$2.24L$$ of hydrogen at standard conditions. Calculate the atomic number of the metal.

A
$$23$$

B
$$24$$

C
$$7$$

D
$$19$$

E
$$35$$
Question 9
1 / -0

If an electron jumps from orbit $$A$$ to orbit $$B$$ it loses energy while when it jumps from $$C$$ to $$B$$ it gains energy. Arrange the orbits in the increasing order of distance from the nucleus.

A
$$A < B < C$$

B
$$C < A < B$$

C
$$C < B < A$$

D
$$A < C < B$$

Solution

When an electrons goes from lower debit to higher orbit it gains energy while when it comes to lower orbit from higher orbit it releases energy.So,the correct order will be.
$$C<B<A$$
Hence, option $$C$$ is correct answer.
Question 10
1 / -0

The orbit having Bohr radius equal to $$1^{st}$$ Bohr orbit of $$H-$$atom is:

A
$$n=2$$ of $$He^{+}$$

B
$$n=2$$ of $$B^{+4}$$

C
$$n=3$$ of $$Li^{+2}$$

D
$$n=2$$ of $$Be^{+3}$$

Solution

r$$=\dfrac { { n }^{ 2 }{ h }^{ 2 } }{ 4{ \pi }^{ 2 }{ m }{ e }^{ 2 } } \times \dfrac { 1 }{ z } $$
=$$0.529\times \dfrac { { n }^{ 2 } }{ z } $$
where n=Orbit number
z=atomic number
n=1 of H atom
r$$=\dfrac{1}{1}=1$$
For $$n=2 $$ of $${B}^{+4}$$
r$$=\dfrac{4}{5}$$
For $$n=2$$ of $${He}^{+}$$
r$$=\dfrac{4}{2}=2$$
For $$n=3$$ of $${Li}^{+2}$$
r$$=\dfrac{9}{3}=3$$
For $$n=2$$ of $${Be}^{+2}$$
r$$=\dfrac{4}{4}=1$$
So radius of 1 st orbit of H atom is equal to $$n=2$$ of $${Be}^{+3}$$ atom

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Re-Attempt Weekly Quiz Competition

Structure of Atom Test - 52

Result

Structure of Atom Test - 52

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SHARING IS CARING

Question 1

13

21

27

31

Solution

Question 2

$$48$$

$$72$$

$$3$$

$$36$$

Solution

Question 3

$$3a_0$$

$$a_0$$

$$\dfrac{3}{2}a_0$$

$$2a_0$$

Solution

Question 4

1

4

9

16

Solution

Question 5

$$0.25$$

$$0.125$$

$$0.5$$

$$14.0$$

Solution

Question 6

$$\cfrac{h^2}{(n^2)(8\pi^2I)}$$

$$\cfrac{h^2}{(n)(8\pi^2I)}$$

$$\cfrac{n h^2}{8\pi^2I}$$

$$\cfrac{n^2 h^2}{(8\pi^2I)}$$

Solution

Question 7

$$34$$

$$49$$

$$33$$

$$31$$

Solution

Question 8

$$23$$

$$24$$

$$7$$

$$19$$

$$35$$

Question 9

$$A < B < C$$

$$C < A < B$$

$$C < B < A$$

$$A < C < B$$

Solution

Question 10

$$n=2$$ of $$He^{+}$$

$$n=2$$ of $$B^{+4}$$

$$n=3$$ of $$Li^{+2}$$

$$n=2$$ of $$Be^{+3}$$

Solution

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