Structure of Atom Test

Result

Structure of Atom Test - 43

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

Assertion (A) : Atoms with completely filled and half-filled subshells are stable.

Reason (R) : Completely filled and half-filled subshells have the symmetrical distribution of electrons and have maximum exchange energy.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

C
Assertion is correct but Reason is incorrect

D
Assertion is incorrect but Reason is correct

Solution

Assertion (A) : Atoms with completely filled and half filled sub-shells are stable.
Reason (R) : Completely filled and half filled sub-shells have symmetrical distribution of electrons and have maximum exchange energy.
For example, He with completely filled $$1s$$ sub-shell and $$N$$ with half filled $$2p$$ sub-shell are stable.
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
Hence, the option (A) is the correct answer.
Question 2
1 / -0

If $$n=6$$, the correct sequence for filling of electrons will be:

A
$$ns\longrightarrow \left( n-1 \right) d\longrightarrow \left( n-2 \right) f\longrightarrow np$$

B
$$ns\longrightarrow \left( n-2 \right) f\longrightarrow np\longrightarrow \left( n-1 \right) d$$

C
$$ns\longrightarrow np\longrightarrow \left( n-1 \right) d\longrightarrow \left( n-2 \right) f$$

D
$$ns\longrightarrow \left( n-2 \right) f\longrightarrow \left( n-1 \right) d\longrightarrow np$$

Solution

If $$n=6$$, the correct sequence for filling of electrons will be

$$\displaystyle ns\longrightarrow \left( n-2 \right) f\longrightarrow \left( n-1 \right) d\longrightarrow np$$.

It will be $$\displaystyle 6s\longrightarrow 4 f \longrightarrow5 d\longrightarrow 6 p$$

The energy of $$4 f$$ and $$5 d$$ orbitals are in between the energies of $$4s$$ and $$6 p$$ orbitals.

Note: Higher is the $$(n+l)$$, higher will be the energy of the orbital.

For n = 6
6s = 6+0 = 6
6p = 6+1 = 7
5d = 5+2 = 7
4f = 4+3 = 7
Hence, correct order is: $$ns\longrightarrow \left( n-2 \right) f\longrightarrow \left( n-1 \right) d\longrightarrow np$$

Hence, the correct option is $$D$$
Question 3
1 / -0

The density of water and ethanol at room temperature is $$1.0$$g $$/$$mL and $$0.789$$g$$/$$mL respectively. What volume of ethanol contains the same number of molecules as are present in $$100$$mL of water?

A
$$307.8$$mL

B
$$168.7$$mL

C
$$323.9$$mL

D
$$253.3$$mL

Solution

An equal number of moles contain equal number of molecules.

$$\displaystyle\dfrac{wt. \ of \ C_2H_5OH}{mol. \ wt. \ of \ C_2H_5OH}=\frac{wt. of H_2O}{mol. \ wt. \ of \ H_2O}$$

$$\displaystyle\frac{0.789\times \ volume \ of \ C_2H_5OH}{46}=\dfrac{1.0\times 100}{18}$$

Volume of $$C_2H_5OH=\displaystyle\frac{100}{18}\times \dfrac{46}{0.789}$$

$$=323.9$$ mL.

Hence, option $$C$$ is correct.
Question 4
1 / -0

Which of the following statement in relation to the hydrogen atom is correct?

A
$$3s,3p$$ and $$3d$$ orbitals all have the same energy

B
$$3s$$ and $$3p$$-orbitals are of lower energy than $$3d$$-orbitals

C
$$3p$$-orbital is lower in energy than$$3d$$- orbital

D
$$3s$$-orbital is lower in energy than $$3p$$-orbital

Solution

Hydrogen atom is in $$1{ s }^{ 1 }$$ and the $$3s,3p$$ and $$3d-$$ orbitals will have the same energy w.r.t $$1s$$ orbital
Question 5
1 / -0

According to the drawback of Rutherford's structure of atom electron should fall into the :

A
space

B
outer enviroment

C
shell

D
nucleus

Solution

Rutherford explained that electrons revolve around the nucleus in a circular path. By following a circular path, energy is lost by electrons, and ultimately they will fall into the nucleus.
Hence option D is correct.
Question 6
1 / -0

$$_7M^{14}+\,_2He^4\rightarrow X+\,_1H^1, X$$ is:

A
$$_8O^{18}$$

B
$$_8O^{17}$$

C
$$_8N^{14}$$

D
$$_7N^{15}$$

Solution

$$_7M^{14}+\ _2He^4\rightarrow\ _8O^{17}\ +_1H^1$$

Balancing mass number: $$14+4=17+1=18$$

Balancing atomic number: $$7+2=8+1=9$$
Question 7
1 / -0

The ionization energy of $${ Li }^{ 2+ }$$ is equal to

A
$$9 hcR$$

B
$$6 hcR$$

C
$$2 hcR$$

D
$$hcR$$

Solution

Ionization energy of an atom $$=hc R{ Z }^{ 2 }$$
where $$R$$ is the Rydberg constant and $$Z$$ is the atomic number.
We know, $$Z=3$$ for $${ Li }^{ 2+ }$$
Ionization energy of $$Li^{2+}$$ $$={ \left( 3 \right) }^{ 2 }hcR$$ $$=9 hcR$$
Question 8
1 / -0

The energy liberated when an excited electron returns to its ground state can have:

A
any value from zero to infinity

B
only negative values

C
only specified positive values

D
none of the above

Solution

The energy liberated when an excited electron returns to its ground state can have only specified positive values which corresponds to the energy difference between excited state and ground state. Such value cannot be negative.
Question 9
1 / -0

According to classical theory, if an electron is moving in a circular orbit around the nucleus:

A
it will continue to do so for sometime

B
its orbit will continuously shrink

C
its orbit will continuously enlarge

D
it will continue to do so for all the time

Solution

According to the classical theory if an electron is moving in a circular orbit around the nucleus its orbit will continuously shrink.

The electron must emit radiation and gradually lose energy, its motion would slow down and it would not be able to withstand the attraction of the nucleus. The orbit would become smaller and smaller and the electron would fall into the nucleus.
Question 10
1 / -0

On Bohr stationary orbits:

A
electrons do not move

B
electrons move emitting radiations

C
energy of the electron remains constant

D
angular momentum of the electron is $$\dfrac{h}{2 \pi}$$
Solution
Hint: Bohr orbit is the hypothetical path of an electron around the nucleus of the Bohr atom.
Explanation:
- Electrons revolve in fixed stationary orbits where they do not emit radiation.
- The angular momentum of electrons is $$nh/2\pi$$.
- Each stationary state is associated with a definite amount of energy.
- The energy of an electron remains constant.
Hence, the correct answer is option $$C.$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Structure of Atom Test - 43

Result

Structure of Atom Test - 43

Score

-

Rank

-

SHARING IS CARING

Question 1

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

Solution

Question 2

$$ns\longrightarrow \left( n-1 \right) d\longrightarrow \left( n-2 \right) f\longrightarrow np$$

$$ns\longrightarrow \left( n-2 \right) f\longrightarrow np\longrightarrow \left( n-1 \right) d$$

$$ns\longrightarrow np\longrightarrow \left( n-1 \right) d\longrightarrow \left( n-2 \right) f$$

$$ns\longrightarrow \left( n-2 \right) f\longrightarrow \left( n-1 \right) d\longrightarrow np$$

Solution

Question 3

$$307.8$$mL

$$168.7$$mL

$$323.9$$mL

$$253.3$$mL

Solution

Question 4

$$3s,3p$$ and $$3d$$ orbitals all have the same energy

$$3s$$ and $$3p$$-orbitals are of lower energy than $$3d$$-orbitals

$$3p$$-orbital is lower in energy than$$3d$$- orbital

$$3s$$-orbital is lower in energy than $$3p$$-orbital

Solution

Question 5

space

outer enviroment

shell

nucleus

Solution

Question 6

$$_8O^{18}$$

$$_8O^{17}$$

$$_8N^{14}$$

$$_7N^{15}$$

Solution

Question 7

$$9 hcR$$

$$6 hcR$$

$$2 hcR$$

$$hcR$$

Solution

Question 8

any value from zero to infinity

only negative values

only specified positive values

none of the above

Solution

Question 9

it will continue to do so for sometime

its orbit will continuously shrink

its orbit will continuously enlarge

it will continue to do so for all the time

Solution

Question 10

electrons do not move

electrons move emitting radiations

energy of the electron remains constant

angular momentum of the electron is $$\dfrac{h}{2 \pi}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.