Structure of Atom Test

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Structure of Atom Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

What is the name and symbol of the element with atomic number 112?

A
Ununbium, Uub

B
Unnilbium, Unb

C
Ununnillum, Uun

D
Ununtrium, Uut

Solution

The name and symbol of the element with atomic number 112 are Ununbium and Uub respectively. Alternatively, the name and symbol of the element with atomic number 112 are Copernicium and Cn respectively.
Question 2
1 / -0

If the radius of first Bohr orbit is x pm, then the radius of the third orbit would be :

A
$$(3\times x )pm$$

B
$$(6\times x )pm$$

C
$$(\frac{1}{2}\times x )pm$$

D
$$(9\times x )pm$$

Solution

The radii of the $$n^{th}$$ stationary states of Hydrogen and Hydrogen like atom is given by:
$$r_n =a_0 \frac{n^2}{Z}$$
where $$a_0 = 52.9\ pm, n=n^{th}$$Bohr orbit, Z=atomic number .
Given: The radius of the first stationary state i.e Bohr orbit=x pm
$$r_1=a_0 \frac{1^2}{Z}=x$$ pm
or $$x=a_0 \frac{1}{Z}$$
$$Z= \frac{a_0}{x}$$
Radius of third orbit: $$r_3=a_0 \frac{3^2}{Z}$$
substitute the above relation:
Radius of third orbit: $$r_3=a_0 \frac{9x}{a_0}=9x$$ pm
Question 3
1 / -0

The radius of a hydrogen atom in the ground state is $$0.53 \mathring {A}$$. What will be the radius of $$Li^{2+}$$ in the ground state?

A
$$1.06\mathring{A}$$

B
$$0.265\mathring{A}$$

C
$$0.17\mathring{A}$$

D
$$0.53\mathring{A}$$

Solution

The radii of the $$n^{th}$$ stationary state for a hydrogen-like specie is expressed as :
$$r_n = \frac{n^2 a_0}{Z}$$
where $$Z=$$atomic number and $$a_0=$$ radius of Bohr orbit.
For ground state of hydrogen atom, $$n=1$$ and $$Z=1$$
thus for Hydrogen $$r_1(H)=a_0=0.53 \mathring A$$ (given)
Thus the radius of the first stationary state of hydrogen atom i.e. in ground state, called the Bohr orbit, is $$0.53 \mathring A$$.
for $$_3Li^{2+}$$, Z=3 and ground state means $$n=1$$
thus $$r_1(Li^{2+})=\frac{1^2 \times 0.53}{3}$$
or $$r_1(Li^{2+})=0.176 \mathring A$$
Question 4
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Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?

A
Valence principal quantum number (n)

B
Nuclear charge

C
Nuclear mass

D
Number of core electrons

Solution

Nuclear mass does not affect the valence shell because nucleus consists of protons and neutrons. Protons i.e., nuclear charge affect the valence shell but neutrons do not.
Question 5
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The statement that is not correct for periodic classification of elements is :

A
the properties of elements are periodic function of their atomic numbers

B
non-metallic elements are less in number than metallic elements

C
for transition elements, the 3d orbitals are filled with electrons after 4p orbitals and before 4s orbitals

D
the first ionization enthalpies of elements generally increase with increase in atomic number as we go along a period.
Solution
Correct Option: C

Explanation

1. Periodicity of elements
- The modern periodic table is based on Moseley’s experiment and equation.
- Moseley’s law is an experimental verification that $$X-rays$$ emitted from elements are a function of their atomic number.
- It can numerically be represented as $$ \sqrt{\nu } = a(Z-b)$$ where,
  $$\nu=$$ Frequency of emitted $$X-rays$$
  $$a$$ $$\&$$ $$b=$$ Moseley’s constants
  $$Z =$$ Atomic number of element
- Therefore, Moseley proved that periodicity in elements is a function of their atomic number and he thus devised the modern periodic table on the same basis.
- Option A is correct.
2. Metallic and non-metallic elements
- Usually, elements having positive oxidation potential are considered as metals.
- Elements belonging to $$s,d$$ and $$f$$ blocs show positive oxidation potential.
- Elements belonging to $$p$$ block are mostly non-metals but elements like $$Sn, Al,etc$$ are metals even though they belong to the $$p$$ block.
- Therefore, the periodic table consists majorly of metals.
- Option B is correct
3. Electronic configuration
- The distribution of electrons of an element/molecule in atomic or molecular orbitals is known as its electronic configuration.
- Electrons are filled in orbitals in order of increasing energy.
- As energy of $$ns > (n-1)d > np$$, it is the order of filling of electrons.
- Therefore, for transition elements, the 3d orbitals are filled with electrons before 4p orbitals and after 4s orbitals.
- Option C is incorrect.
4. Trend in $$I.E.$$ along period
- Ionisation energy ($$I.E.$$) is the energy required to convert $$1$$ mole gaseous atom into cation by emission of an electron
- Along a period, $$Z_{eff}$$ increases due to an increase in atomic number, i.e., the number of protons increases.
- Therefore, the outer electrons are pulled in strongly by the nucleus.
- Thus, more energy is required to knock out an electron.
- Option D is correct
Hence, the statement ‘for transition elements, the $$3d$$ orbitals are filled with electrons after $$3p$$ orbitals and before $$4s$$ orbitals’ is incorrect.
Question 6
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Which of the following transitions will involve maximum amount of energy?

A
$$M\rightarrow M^{+} + e^{-}$$

B
$$M^{-}\rightarrow M^{+} + 2e^{-}$$

C
$$M^{2+}\rightarrow M^{3+} +e^{-}$$

D
$$M^{+}\rightarrow M^{2+} + e^{-}$$

Solution

The ionization energy of a multivalent atom increases with the consecutive removal of the electron. This is due to an increase in effective nuclear charge on the valence electron that makes it difficult to ionize a cation.
Therefore $$IE_1<IE_2<IE_3$$ and so on
thus maximum amount of energy is required for $$M^{2+} \rightarrow M^{3+}+e^-$$
option C
Question 7
1 / -0

The radius of the stationary state which is also called Bohr radius is given by the expression $$r_n= n^2a_0$$ where the value of $$a_0$$ is:

A
52.9 pm

B
5.29 pm

C
529 pm

D
0.529 pm

Solution

The radii of the $$n^{th}$$ stationary states of Hydrogen and Hydrogen like atom is given by:
$$r_n =a_0 \frac{n^2}{Z}$$
where $$a_0 = 52.9\ pm, n=n^{th}$$Bohr orbit, $$Z$$=atomic number .
The radius of the first stationary state of Hydrogen atom $$(n=1)$$, is called as Bohr radius that has a value of $$52.9\ pm$$. As n increases the value of r will increase. In other words, the electron will be present away from the nucleus.
Question 8
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Which of the sequences given below shows the correct increasing order of energy?

A
$$3s, 3p, 4s, 3d, 5s, 5p, 4d$$

B
$$3s, 3p, 4s, 4p, 4d, 5s, 5p$$

C
$$3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p$$

D
$$3s, 3p, 4s, 5s, 3d, 4d, 5p$$

Solution

The energy of an orbital depends on the (n+l) value. Higher is the (n+l) higher is the energy of the orbital.
Value of n majorly decides the energy of an orbital. For two orbitals with the same value for (n+l), the orbital with a higher value of n will have higher energy.
Thus increasing order of energy for orbitals is:
$$1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p$$
For the given set of orbitals the order is:
$$3s<3p<4s<3d<4p<5s<4d<5p$$
Question 9
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The electron in Bohr's model of hydrogen atom is pictured as revolving around the nucleus in order for it to :

A
emit photons

B
keep from being pulled into the nucleus

C
keep from being repelled by the nucleus

D
possess energy

Solution

The electron in Bohr's model of a hydrogen atom is pictured as revolving around the nucleus in order for it to possess energy as each state has a fixed energy.
option D
Question 10
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The atoms in a molecule do not remain stationary. They vibrate around their mean position, by stretching or bending out of plane. We gain useful knowledge about these vibrations and the energy they carry by studying:

A
$$X-$$ray spectra

B
ultraviolet spectra

C
infrared spectra

D
visible spectra

Solution

Infrared spectroscopy involves the interaction of infrared radiation with matter.
IR spectroscopy is concerned with the study of absorption of infrared radiation, which causes vibrational transition in the molecule.

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Structure of Atom Test - 49

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Structure of Atom Test - 49

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Question 1

Ununbium, Uub

Unnilbium, Unb

Ununnillum, Uun

Ununtrium, Uut

Solution

Question 2

$$(3\times x )pm$$

$$(6\times x )pm$$

$$(\frac{1}{2}\times x )pm$$

$$(9\times x )pm$$

Solution

Question 3

$$1.06\mathring{A}$$

$$0.265\mathring{A}$$

$$0.17\mathring{A}$$

$$0.53\mathring{A}$$

Solution

Question 4

Valence principal quantum number (n)

Nuclear charge

Nuclear mass

Number of core electrons

Solution

Question 5

the properties of elements are periodic function of their atomic numbers

non-metallic elements are less in number than metallic elements

for transition elements, the 3d orbitals are filled with electrons after 4p orbitals and before 4s orbitals

the first ionization enthalpies of elements generally increase with increase in atomic number as we go along a period.

Solution

Question 6

$$M\rightarrow M^{+} + e^{-}$$

$$M^{-}\rightarrow M^{+} + 2e^{-}$$

$$M^{2+}\rightarrow M^{3+} +e^{-}$$

$$M^{+}\rightarrow M^{2+} + e^{-}$$

Solution

Question 7

52.9 pm

5.29 pm

529 pm

0.529 pm

Solution

Question 8

$$3s, 3p, 4s, 3d, 5s, 5p, 4d$$

$$3s, 3p, 4s, 4p, 4d, 5s, 5p$$

$$3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p$$

$$3s, 3p, 4s, 5s, 3d, 4d, 5p$$

Solution

Question 9

emit photons

keep from being pulled into the nucleus

keep from being repelled by the nucleus

possess energy

Solution

Question 10

$$X-$$ray spectra

ultraviolet spectra

infrared spectra

visible spectra

Solution

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