Structure of Atom Test

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Structure of Atom Test - 68

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Question 1
1 / -0

Directions For Questions

The electrons in a $$H$$-atom kept at rest, jumps from the $$mth$$ shell to the $$nth$$ shell $$(m>n)$$. Suppose instead of emitting electromagnetic wave, the energy released is converted into kinetic energy of the atom. Assume Bohr's model and conservation of angular momentum are valid. Answer the following question

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What principle is violated here?

A
Laws of Motion

B
Energy conservation

C
Nothing is violated

D
Cannot be decided

Solution

Laws of Motion:
Newtons second law: $$F = ma$$
Without any force, the atom cannot accelerate.
Conservation of Momentum:
The linear momentum of the atom can't change without any external force acting on it.
Question 2
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Maximum number of electrons in a subshell of an atom is determined by the following:

A
$$\text{2l+2}$$

B
$$\text{4l-1}$$

C
$$2n^2$$

D
$$\text {4l + 2}$$

Solution

Hint

Quantum number: It is defined as the set of numbers which describes the position and energy of electrons in an atom. There are four quantum numbers: principal, azimuthal, magnetic and spin quantum numbers.

Explanation

Step 1: Determine the maximum number of subshells

The principle quantum number ($$n$$) describe the distance between the nucleus and the electrons.
The azimuthal quantum number ($$l$$) is given by ($$n - l$$)

We know that the maximum number of subshells is equal to is $$\left( 2l+1 \right)$$.

Step 2: Determine the maximum number of electrons in a subshell
Maximum electrons a subshell can accommodate is $$2$$

Therefore, the total number of electrons in a subshell is

$$2\left( 2l+1 \right)=\left( 4l+2 \right)$$

Thus, maximum number of electrons in a subshell is $$( 4l+2)$$

Final answer

The correct answer is option (D).
Question 3
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Designations of some orbitals are given below. Arrange the orbitals with possible designations in the order in which they are filled with electrons $$6s,\ 8p,\ 7s,\ 4d,\ 2p,\ 3d,\ 3f,\ 4f$$.

A
$$2p < 3d < 4d < 4p < 6s < 8p < 7s<4f$$

B
$$2p < 3d < 4p < 4d < 6s < 4f < 7s < 8p$$

C
$$3d < 2p < 4p < 4d < 6s < 4f < 7s < 8p$$

D
$$2p < 3d < 4p < 4d < 4f < 6s < 7s < 8p$$

Solution

The orbitals with possible designations in the order in which they are filled with electrons is $$2p < 3d < 4p < 4d < 6s < 4f < 7s < 8p$$.
Question 4
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For the hydrogen atom, the energy of the electron is defined by the factor $$E_n=-13.58/n^2 eV$$. Here n is positive integer. The minimum quantity of energy which it can absorb in its primitive stage is:

A
$$1.00 eV$$

B
$$3.39 eV$$

C
$$6.79 eV$$

D
$$10.19 eV$$

Solution

$$E_n=\dfrac {-13.58}{n^2}$$

In primitive stage at $$n=1, E_1=-13.58 eV$$

Minimum energy in excited stage,

$$n=2,\quad E_2=-3.395 eV$$

Energy absorbed $$=-3.395+13.58=10.19eV$$

Hence, the correct option is D
Question 5
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The total number of electrons that could be held in a sub-level with azimuthal quantum number = 2 is :

A
$$2$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

The azimutual quantum number, $$'l' = 2$$ corresponds to 'd-subshell'.
A 'd-subshell' has 5 orbitals, and in each orbital there a maximum of 2 electrons can be filled.
The total number of electrons present in all 5 orbitals of d-subshell $$ = 5 \times 2 = 10$$
Question 6
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The maximum number of electrons in a shell with the principal quantum number = 4 is :

A
$$2$$

B
$$10$$

C
$$16$$

D
$$32$$

Solution

Formula to determine the number of electrons present in any shell is as follows :
Let the shell number is "n", then number of electrons present in n shell = $$2n^{2}$$.
Hence, as per formula the number of electrons iin 4th shell = $$2 \times 4^{2} = 32$$.
Question 7
1 / -0

Which letter orbital corresponds to $$l$$ = 2?

A
S

B
p

C
d

D
f

E
n

Solution

$$l =0$$, means s
$$l = 1 $$ means p
$$l = 2 $$ means d
$$l = 3 $$ means f
Question 8
1 / -0

Which color of light has the highest energy?

A
Violet

B
Green

C
Yellow

D
Orange

E
Red

Solution

Violet has highest frequency.
$$E = hv$$
$$v = frequency, $$
Violet has highest energy.
Question 9
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The ratio of the energy of photon of $$2000 A^0$$ wavelength to that of $$4000 A^0$$ wavelength is:

A
1 : 4

B
4 : 1

C
1 : 2

D
2 : 1

Solution

$$E=\cfrac { hc }{ \lambda } \Rightarrow E\propto \cfrac { 1 }{ \lambda } $$

$$\therefore \cfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\cfrac { { \lambda }_{ 2 } }{ { \lambda }_{ 1 } } =\cfrac { 4000 }{ 2000 } =2:1$$

Hence, option $$D$$ is correct.
Question 10
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An element $$X$$ combines with oxygen to form compounds $$P$$ and $$Q$$. If the ratio of the valency of element $$X$$ in $$P$$ to element $$X$$ in $$Q$$ is $$3 : 5$$ respectively, what could be the probable compounds $$P$$ and $$Q$$?

A
$$\begin{matrix} P & Q \\ CO & C{ O }_{ 2 } \end{matrix}$$

B
$$\begin{matrix} P & Q \\ { N }_{ 2 }{ O }_{ 3 } & { N }_{ 2 }{ O }_{ 5 } \end{matrix}$$

C
$$\begin{matrix} P & Q \\ { H }_{ 2 }O & { H }_{ 2 }{ O }_{ 2 } \end{matrix}$$

D
$$\begin{matrix} P & Q \\ C{ H }_{ 4 } & { C }_{ 3 }{ H }_{ 6 } \end{matrix}$$

Solution

Let $$P = X_2O_a$$ where a is the valency of X in P.

and let $$Q = X_2O_b$$ where b is the valency of X in Q

According to the question:
The ratio of valency of X in P to element X in Q is $$3 : 5$$ respectively.

Thus, $$\dfrac{a}{b} = \dfrac{3}{5}$$

Therefore, $$a = 3$$ and $$b = 5$$

$$\therefore$$ compound $$P = X_2 O_3$$

and compound $$Q = X_2 O_5$$

According to the options, option (B) is satisfied, So, $$P = N_2O_3$$ and $$Q = N_2 O_5$$

Hence, the correct option is (B)

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Structure of Atom Test - 68

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Structure of Atom Test - 68

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SHARING IS CARING

Question 1

Laws of Motion

Energy conservation

Nothing is violated

Cannot be decided

Solution

Question 2

$$\text{2l+2}$$

$$\text{4l-1}$$

$$2n^2$$

$$\text {4l + 2}$$

Solution

Question 3

$$2p < 3d < 4d < 4p < 6s < 8p < 7s<4f$$

$$2p < 3d < 4p < 4d < 6s < 4f < 7s < 8p$$

$$3d < 2p < 4p < 4d < 6s < 4f < 7s < 8p$$

$$2p < 3d < 4p < 4d < 4f < 6s < 7s < 8p$$

Solution

Question 4

$$1.00 eV$$

$$3.39 eV$$

$$6.79 eV$$

$$10.19 eV$$

Solution

Question 5

$$2$$

$$6$$

$$8$$

$$10$$

Solution

Question 6

$$2$$

$$10$$

$$16$$

$$32$$

Solution

Question 7

S

p

d

f

n

Solution

Question 8

Violet

Green

Yellow

Orange

Red

Solution

Question 9

1 : 4

4 : 1

1 : 2

2 : 1

Solution

Question 10

$$\begin{matrix} P & Q \\ CO & C{ O }_{ 2 } \end{matrix}$$

$$\begin{matrix} P & Q \\ { N }_{ 2 }{ O }_{ 3 } & { N }_{ 2 }{ O }_{ 5 } \end{matrix}$$

$$\begin{matrix} P & Q \\ { H }_{ 2 }O & { H }_{ 2 }{ O }_{ 2 } \end{matrix}$$

$$\begin{matrix} P & Q \\ C{ H }_{ 4 } & { C }_{ 3 }{ H }_{ 6 } \end{matrix}$$

Solution

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