Structure of Atom Test

Result

Structure of Atom Test - 67

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The boher model of atoms

A
Assume that the angular momentum of electrons is quantized

B
Uses Einstein's photo electric equation

C
Predicts continuous emission spectra for atoms

D
Predicts the same emission spectra for all types of atoms

Solution

Bohr defined stable orbits of electron revolution in his second postulate. According to this postulate:

· An electron revolves around the nucleus in orbits.

· The angular momentum of revolution is an integral multiple of h/2π where h is Planck’s constant.
Hence, the angular momentum (L) of the orbiting electron is: L = nh/2π, which is quantized. (Option A is correct)
Question 2
1 / -0

In hydrogen atom which quantity is integral multiple of $$\dfrac{h}{2\pi}$$

A
Angular momentum

B
Angular velocity

C
Angular acceleration

D
Momentum

Solution
Question 3
1 / -0

Which substance is discovered by the Prafulla Chandra Ray the inventor of Indian Chemical Industry ?

A
Sodium Chloride

B
Mercurus Nitrate

C
Nitre

D
Sal Ammonia

Solution
Question 4
1 / -0

When a hydrogen atom is raised from ground to excited state.

A
The P.E increases and K.E decreases

B
The P.E decreases and K.E increases

C
Both P.E and K.E increase

D
Both P.E and K.E decreases

Solution

Potential energy of hydrogen atom is given by
$$P.E=\dfrac{-kZe}{r}$$
As atom is excited,it's radius increases. But due to negative sign or reverse sign. Its actual potential energy increases.
And kinetic energy is given by
$$K.E=\dfrac{kZe}{2r}$$
As Tom is excited, it's radius increases this resulting in decrease of kinetic energy.
The PE increases and KE decreases.
Question 5
1 / -0

If $$n = 3 , l = 0 , m = 0 $$ then atomic number is

A
$$12 , 13 $$

B
$$13 , 14 $$

C
$$10 , 11 $$

D
$$11 , 12 $$

Solution

$$n = 2 , l = 0 $$ means last shell is $$3s$$
E.C will be $$1s^{2} 2s^{2} 2p^{6} 3s^{1 - 2} $$
Atomic no. is $$11$$ or $$12$$.
Question 6
1 / -0

In the ground state, an element has $$13$$ electrons in its M-shell. The element is

A
Manganese

B
Chromium

C
Nickel

D
Iron

Solution

$$13e^{-} $$ in M $$(3rd)$$ shell means $$ 3s^{2} 3p^{6} 3d^{5} $$. hence complete configuration will be $$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} 3d^{5} $$
i.e., total $$e^{-} = 24 $$. Hence the element is chromium.
Question 7
1 / -0

The maximum number of electrons in an orbit with $$l = 2 , n = 3 $$ is

A
$$2$$

B
$$6$$

C
$$12$$

D
$$10$$

Solution

$$ n = 3 , l = 2 $$ means $$3d$$-subshell. Maximum number of electrons present in it $$ = 10 $$
Question 8
1 / -0

Any p-orbital can accommodate upto

A
four electrons

B
two electrons with parallel spin

C
six electrons

D
two electrons with anti-parallel spin.
Solution
- Any orbital can accommodate a maximum of 2 electrons with an opposite spin so a p-orbital can have 2 electrons with opposite spin.
- One p-orbital can accommodate up to two electrons with opposite spin while a p-subshell can accommodate up to six electrons.
Option D is correct.
Question 9
1 / -0

If Aufbau rule is not obeyed and the electronic filling occurs orbit by orbit till saturation is reached (for a given principal quantum number $$n$$, electrons are filled in the increasing order of azimuthal quantum number $$l$$), then the percentage change in sum of all $$(n + l)$$ values for unpaired electrons in an atom of iron is $$10x$$. What is the value of $$x$$?

A
$$5$$

B
$$6$$

C
$$7$$

D
$$3$$

Solution

When $$n+l$$ rule is obeyed, the configuration is

$$^{26}Fe= 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{2}, 3d^{6}.$$

Since, $$(n+l)$$ for each unpaired electron is $$5$$, so for $$4$$ unpaired electrons in $$^{26}Fe$$, $$(n+l)$$ is $$5\times 4= 20$$.

When $$n+l$$ rule is not obeyed, the configuration is

$$^{26}Fe= 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{8}.$$

Since, $$(n+l)$$ for each unpaired electron is $$5$$, so for $$2$$ unpaired electrons in $$^{26}Fe$$, $$(n+l)$$ is $$5\times 2= 10$$.

Now,
% change $$= \dfrac{20-10}{20} \times 100 = 50 = 10x$$.

$$\therefore x=5.$$
Question 10
1 / -0

The orbital diagram in which aufbau principle is violated is:

A

B

C

D

Solution

Hint: Aufbau principle states that electrons getting filled in the orbitals in the increasing order of energy.
Step 1: The energy of an orbital is directly proportional to the $$(n + l)$$ value.
The $$(n + l)$$ value for $$2s$$ level:
For $$2s$$ energy level-
$$n = 2$$
Where $$n$$ is the principal quantum number and $$l$$ is the azimuthal quantum number.
For $$s$$ orbital, $$l=0$$
So, $$(n + l) = 2 + 0$$
$$(n + l) = 2$$
Step 2:
The $$(n + l)$$ value for $$2p$$ level:
For $$2p$$ energy level-
$$n = 2$$
Where $$n$$ is the principal quantum number and $$l$$ is the azimuthal quantum number.
For $$p$$ orbital, $$l=1$$
So, $$(n + l) = 2 + 1$$
$$(n + l) = 3$$
Step 3:
Comparison of $$(n + l)$$ value of two energy levels:
As the value of $$(n + l)$$ for $$2s$$ orbital is less than that of $$2p$$ orbital so $$2s$$ orbital has lower energy than $$2p$$ orbital. Therefore $$2s$$ orbital will get filled first then $$2p$$ orbital will starts filling.
Final answer:
So the orbital diagram in which Aufbau principle is violated is option (B).
So the correct option is (B).