Structure of Atom Test

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Structure of Atom Test - 69

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Question 1
1 / -0

If the principal quantum number of a shell is 2, what types of orbitals will be present?

A
$$s$$

B
$$s$$ and $$ p$$

C
$$s,\ p$$ and $$d$$

D
$$s,\ p,\ d$$ and $$f$$

Solution

For a shell with the principal quantum number "$$n$$", the types of orbitals are determined by the value of the magnetic quantum number "$$l$$".

For a shell with the principal quantum number "$$n$$", the '$$l$$' values ranges from $$0$$ to $$n-1$$.

When $$ n = 2$$ then '$$l$$' has values $$= 0 ,\ 1$$.

$$l = 0$$ means type of the orbitals is s-orbital.
$$l = 1$$ means type of orbitals is p-orbital.
Hence, there are $$s$$ and $$p$$ orbitals for a shell with the principal quantum number "$$n$$".
Question 2
1 / -0

Statement I : $$\displaystyle ^{ 12 }{ C }\ $$ is an isotope of $$\displaystyle ^{ 14 }{ C }\ $$.
Statement II: The nuclei of both atoms have the same number of neutrons.

A
true, false

B
false, true

C
false, false

D
true, true, correct explanation

Solution

$$^{12}C$$ is an isotope of $$^{14}C$$.

As isotopes are variants of a particular chemical element that differ in neutron number but have the same number of protons.

$$^{12}C$$ has $$12 - 6 = 6$$ neutrons.

$$^{14}$$ has $$14 - 6 = 8$$ neutrons.
Hence, the statement $$I$$ is true but the statement $$II$$ is false.
Question 3
1 / -0

The predominant yellow line in the spectrum of a sodium vapour lamp has a wavelength of $$590 nm$$. What minimum accelerated potential is needed to excite this line in an electron tube having sodium vapours?

A
$$8.44$$ volt.

B
$$2.11$$ volt.

C
$$6.11$$ volt.

D
$$4.22$$ volt.

Solution

Energy$$=\cfrac{hc}{\lambda}=\cfrac{6.626\times 10^{-34}\times 3\times 10^8}{590\times 10^{-9}}=3.369\times 10^{-19}J$$
Ex-potential $$=$$ Energy.
$$\therefore$$ Potential required (min)$$=\cfrac{3.369\times 10^{-19}}{1.6\times 10^{-19}}v\approx 2.11v$$
Question 4
1 / -0

In ground state of chromium atom $$(Z=24)$$ the total number of orbitals populated by one or more electrons is?

A
$$15$$

B
$$16$$

C
$$20$$

D
$$14$$

Solution

The configuration of $$_{24}Cr$$ is $$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$$
$$\therefore$$ total s-orbitals$$=4$$
total p-orbitals$$=6$$
total d-orbitals$$=5$$ and thus
Thus, total orbitals $$=4+6+5=15$$.
Question 5
1 / -0

The mass of photon having wavelength $$1 nm$$ is:

A
$$2.21\times { 10 }^{ -35 }kg$$

B
$$2.21\times { 10 }^{ -33 }g$$

C
$$2.21\times { 10 }^{ -33 }kg$$

D
$$2.21\times { 10 }^{ -26 }kg$$

Solution

Photons travel with the speed of light = $$ 3\times 10^8 m/s$$
De Broglie wavelength $$ \lambda = \cfrac{h}{mv}$$
$$ h = 6.626 \times 10^{-34} Js$$
Now, $$1 \times 10^{-9} = \cfrac{6.626 \times 10^{-34}}{m \times 3 \times 10^8}$$
$$ m = 2.21 \times 10^{-33}kg$$
The mass of photon having wavelength 1nm is $$ 2.21 \times 10^{-33}kg$$
Question 6
1 / -0

Two trials of a reaction were run in a lab where absorbance data was collected over time.
In trial $$2$$, the absorbance values for the reaction for each unit of time were higher than they were for trial $$1$$.
Also, the rate of trial $$2$$ was higher than that of trial $$1$$.
Which of the following is most likely the case?

A
The concentration of one or more reactants was increased in trial $$2$$.

B
The concentration of the reactants in trial $$1$$ was higher than that in trial $$2$$.

C
The concentration of reactants was decreased for trial $$2$$.

D
Something interfered with the reaction.

Solution

Absorbance$$=\varepsilon \times concentration \times path length\dots [Beer \quad law]$$

$$\therefore $$ Absorbance $$\propto$$ concentration.

Also rate of reaction $$\propto$$ concentration of reactant

$$\therefore$$ It is possible that the concentration of one or more reactant was increased in trial $$2$$, giving higher values in trial $$2$$.
Question 7
1 / -0

Gas A is a colorless gas and Gas B is orange in color. Consider the absorbance data for gases A and B when they were analyzed through spectroscopy.
Which of the following statements best explains the data?

A
More energy is required to make Gas A molecules vibrate than Gas B.

B
The ionization energy of Gas B atom is higher than Gas A.

C
Gas A has more electrons compared to Gas B.

D
There are lower energy transitions available in Gas B compared to Gas A.

Solution

Gas A absorbs only UV light whereas gas B absorbs both UV and visible light. Thus transitions in Gas B can be carried out with visible light as well, i.e., they require lesser energy, as UV light is more energetic than visible light. Thus the correct answer is option D
Question 8
1 / -0

Beer's Law states that $$A = abc$$, where A is absorbance, a is the molar absorptivity constant, b is the path length, and c is the molar concentration.
Which of the following would be an incorrect application of Beer's Law?

A
The path length is irrelevant, as long as it is inversely proportional to the molar concentration when determining the concentration of an unknown as compared to a standard curve.

B
The molar concentration is directly related to absorption of light by the sample and is used to develop a standard curve.

C
The concentration of an unknown sample can be easily determined from a standard curve of known concentrations plotted against absorbance for a fixed path length.

D
The use of Beer's Law and a spectrophotometer for a colored sample is a better method of determining molar concentration than visual comparison.

Solution

The standard curve is the graph of obserbance versus concentration. The path length is relevant, because it is known, we can easily find the molar absorptinity constant with the help of slope of the graph and Beer's law.
Question 9
1 / -0

The wave number of a particular spectral line in the atomic spectrum of a hydrogen like species increases $${9}/{4}$$ times when deuterium nucleus is introduced into its nucleus, then which of the following will be the initial hydrogen like species?

A
$${ Li }^{ 2+ }$$

B
$${ Li }^{ + }$$

C
$${ He }^{ + }$$

D
$${ Be }^{ 3+ }$$

Solution

For hydrogen-like atoms spectra,

$$\lambda \propto \cfrac{n^{2}}{Z}$$

where,
$$\lambda =$$wavelength
$$n=$$ The number which element excited
$$Z=$$Atomic number

Given that for hydrogen

$$\cfrac{n^{2}}{Z}=\cfrac{9}{8}$$

check which ion's $$\lambda \propto \cfrac{9}{8}$$

$$A)$$ for $$Li^{+2}$$

$$\lambda \propto \cfrac{n^{2}}{Z} \Rightarrow \lambda \propto \cfrac{2^{2}}{3}=\cfrac{4}{3}$$

$$B)$$ for $$Li^{+}$$

$$\lambda \propto \cfrac{n^{2}}{Z} \Rightarrow \lambda \propto \cfrac{1^{2}}{3}=\cfrac{1}{3}$$

$$c)$$ for $$He^{+}$$

$$\lambda \propto \cfrac{n^{2}}{Z} \Rightarrow \lambda \propto \cfrac{1^{2}}{2}=\cfrac{1}{2}$$

$$c)$$ for $$Be^{3+}$$

$$\lambda \propto \cfrac{n^{2}}{Z} \Rightarrow \lambda \propto \cfrac{3^{2}}{4}=\cfrac{9}{4}$$

$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (D).}$$
Question 10
1 / -0

A man writes his biodata with carbon pencil on the plane paper having mass 150 mg. After writing his biodata, he weighs the written paper and find its mass 152 mg. What is the number of carbon atoms present in the paper?

A
$$1.0036 \times 10^{20}$$

B
$$5.02 \times 10^{20}$$

C
$$1.0036 \times 10^{23}$$

D
$$0.502 \times 10^{20}$$

Solution

Mass of C-atoms $$= 152 - 150 = 2$$ mg

Molar mass of C-atoms $$= 12 g = 12000$$ mg

12000 mg of carbon contains $$6.022 \times 10^{23} $$ atoms carbon

$$\therefore$$ 2 mg of carbon contains $$= \dfrac{6.022 \times 10^{23}}{12000} \times 2$$

$$= 0.0010036 \times 10^{23}$$

$$= 1.0036 \times 10^{20}$$ atoms

Therefore, option A is correct.

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Structure of Atom Test - 69

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Structure of Atom Test - 69

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Question 1

$$s$$

$$s$$ and $$ p$$

$$s,\ p$$ and $$d$$

$$s,\ p,\ d$$ and $$f$$

Solution

Question 2

true, false

false, true

false, false

true, true, correct explanation

Solution

Question 3

$$8.44$$ volt.

$$2.11$$ volt.

$$6.11$$ volt.

$$4.22$$ volt.

Solution

Question 4

$$15$$

$$16$$

$$20$$

$$14$$

Solution

Question 5

$$2.21\times { 10 }^{ -35 }kg$$

$$2.21\times { 10 }^{ -33 }g$$

$$2.21\times { 10 }^{ -33 }kg$$

$$2.21\times { 10 }^{ -26 }kg$$

Solution

Question 6

The concentration of one or more reactants was increased in trial $$2$$.

The concentration of the reactants in trial $$1$$ was higher than that in trial $$2$$.

The concentration of reactants was decreased for trial $$2$$.

Something interfered with the reaction.

Solution

Question 7

More energy is required to make Gas A molecules vibrate than Gas B.

The ionization energy of Gas B atom is higher than Gas A.

Gas A has more electrons compared to Gas B.

There are lower energy transitions available in Gas B compared to Gas A.

Solution

Question 8

The path length is irrelevant, as long as it is inversely proportional to the molar concentration when determining the concentration of an unknown as compared to a standard curve.

The molar concentration is directly related to absorption of light by the sample and is used to develop a standard curve.

The concentration of an unknown sample can be easily determined from a standard curve of known concentrations plotted against absorbance for a fixed path length.

The use of Beer's Law and a spectrophotometer for a colored sample is a better method of determining molar concentration than visual comparison.

Solution

Question 9

$${ Li }^{ 2+ }$$

$${ Li }^{ + }$$

$${ He }^{ + }$$

$${ Be }^{ 3+ }$$

Solution

Question 10

$$1.0036 \times 10^{20}$$

$$5.02 \times 10^{20}$$

$$1.0036 \times 10^{23}$$

$$0.502 \times 10^{20}$$

Solution

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