# Periodic Table And Structure Of Atom- Combined Test

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• 1.

### Which of the following sets of quantum numbers represents the highest energy of an atom?

• A.

N=3, l=1, m=1, s=+1/2

• B.

N=3, l=2, m=1, s=+1/2

• C.

N=4, l=0, m=0, s=+1/2

• D.

N=3, l=0, m=0, s=+1/2

B. N=3, l=2, m=1, s=+1/2
Explanation
The quantum numbers n, l, m, and s represent different properties of an electron in an atom. The principal quantum number (n) represents the energy level or shell of the electron. The angular momentum quantum number (l) represents the shape of the orbital. The magnetic quantum number (m) represents the orientation of the orbital in space. The spin quantum number (s) represents the spin of the electron.

In this case, the set of quantum numbers n=3, l=2, m=1, s=+1/2 represents the highest energy of an atom because the principal quantum number (n=3) indicates that the electron is in the third energy level, which is higher than the other options. Additionally, the angular momentum quantum number (l=2) indicates that the electron is in a d orbital, which has higher energy than s or p orbitals. The combination of these quantum numbers suggests a higher energy state for the electron.

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• 2.

### In a multi-electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in absence of magnetic and electric field?     (A) n=1, l=0, m=0 (B) n=2, l=0, m=0 (C) n=2, l=1, m=1 (D) n=3, l=2, m=1 (E) n=3, l=2, m=0

• A.

D and E

• B.

C and D

• C.

B and C

• D.

A and B

A. D and E
Explanation
In a multi-electron atom, the energy of an orbital is determined by the values of the principal quantum number (n), azimuthal quantum number (l), and magnetic quantum number (m). The orbitals with the same energy are those that have the same values of n, l, and m.

Option D has n=3, l=2, and m=1. Option E has n=3, l=2, and m=0. Since both options have the same values for n and l, they will have the same energy in the absence of magnetic and electric fields.

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• 3.

### Which of the following sets of quantum numbers is correct for an electron in 4f orbital?

• A.

N=4, l=3, m=+4, s=+1/2

• B.

N=4, l=4, m=-4, s=-1/2

• C.

N=4, l=3, m=+1, s=+1/2

• D.

N=3, l=2, m=-2, s=+1/2

C. N=4, l=3, m=+1, s=+1/2
Explanation
The quantum numbers describe the energy levels, orbital shapes, orientations, and spin states of electrons in an atom. For an electron in the 4f orbital, the principal quantum number (n) is 4, indicating the energy level. The azimuthal quantum number (l) is 3, representing the orbital shape. The magnetic quantum number (m) is +1, indicating the orientation of the orbital in space. The spin quantum number (s) is +1/2, representing the spin state of the electron. Therefore, the set of quantum numbers n=4, l=3, m=+1, s=+1/2 is correct for an electron in the 4f orbital.

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• 4.

### Which one of the following sets of ions represents the collection of iso-electronic species?

• A.

K+, Ca2+, Sc3+, Cl-

• B.

Na+, Ca2+, Sc3+, F-

• C.

K+, Cl-, Mg2+, Sc3+

• D.

Na+, Mg2+, Al3+, Cl-

A. K+, Ca2+, Sc3+, Cl-
Explanation
The collection of iso-electronic species refers to a group of ions that have the same number of electrons. In this case, K+ has lost one electron, Ca2+ has lost two electrons, Sc3+ has lost three electrons, and Cl- has gained one electron. Therefore, all four ions have the same number of electrons, making them iso-electronic.

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• 5.

### Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar?

• A.

Ca < S < Ba < Se < Ar

• B.

S < Se < Ca < Ba < Ar

• C.

Ba < Ca < Se < S < Ar

• D.

Ca < Ba < S < Se < Ar

C. Ba < Ca < Se < S < Ar
Explanation
The first ionization enthalpy refers to the energy required to remove one electron from an atom in its gaseous state. As we move across a period from left to right, the atomic radius decreases, resulting in stronger attractions between the nucleus and the outermost electron. This makes it more difficult to remove an electron, thus increasing the ionization enthalpy. Among the given options, Ba has the largest atomic radius and therefore the weakest attraction to its outermost electron, making it the easiest to remove. Ca has a smaller atomic radius than Ba but larger than Se, S, and Ar, hence it has a higher ionization enthalpy than Ba but lower than the remaining elements. Se has a smaller atomic radius than Ca but larger than S and Ar, resulting in a higher ionization enthalpy than Ca but lower than S and Ar. S has a smaller atomic radius than Se but larger than Ar, giving it a higher ionization enthalpy than Se but lower than Ar. Ar has the smallest atomic radius among the given elements, resulting in the strongest attraction to its outermost electron and the highest ionization enthalpy. Therefore, the correct order of increasing first ionization enthalpy is Ba < Ca < Se < S < Ar.

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• 6.

### The increasing order of the ionic radii of the given iso-electronic species is-

• A.

Cl- , Ca2+ , K+ , S2-

• B.

S2- , Cl- , Ca2+ , K+

• C.

Ca2+, K+, Cl- , S2-

• D.

K+, S2-, Ca2+, Cl-

C. Ca2+, K+, Cl- , S2-
Explanation
The ionic radii of ions are determined by the number of electrons and the effective nuclear charge. In this case, all the ions are iso-electronic, meaning they have the same number of electrons. However, the effective nuclear charge increases as you move across a period in the periodic table, resulting in a decrease in ionic radius. Therefore, the larger ions will be towards the left side of the given options. Among the given options, Ca2+ has the largest ionic radius because it has the least effective nuclear charge. K+ has the next largest ionic radius, followed by Cl- and S2- with the smallest ionic radius due to their higher effective nuclear charges.

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• 7.

### The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I, having atomic number 9, 17, 35 and 53 respectively is-

• A.

I > Br > Cl > F

• B.

F > Cl > Br > I

• C.

Cl > F > Br > I

• D.

Br > Cl > F > I

C. Cl > F > Br > I
Explanation
The electron gain enthalpy is the energy released when an atom gains an electron. It is generally observed that the electron gain enthalpy becomes less negative as we move down a group in the periodic table. This is because the atomic radius increases down the group, making it easier for the atom to accommodate an additional electron.

In this case, we are comparing the electron gain enthalpy of F, Cl, Br, and I. Since Cl has a smaller atomic number than F, Br, and I, it has a smaller atomic radius. Therefore, it has a higher electron gain enthalpy. Among F, Br, and I, F has the highest electron gain enthalpy because it has the smallest atomic radius.

So, the correct order of electron gain enthalpy with a negative sign is Cl > F > Br > I.

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• 8.

### The correct sequence which shows decreasing order of the ionic radii of the elements-

• A.

Al3+ > Mg2+ > Na+ > F+ > O2-

• B.

Na+ > Mg2+ > Al3+ > O2- > F-

• C.

Na+ > F- > Mg2+ > O2- > Al3+

• D.

O2- > F- > Na+ > Mg2+ > Al3+

D. O2- > F- > Na+ > Mg2+ > Al3+
Explanation
The correct answer is O2- > F- > Na+ > Mg2+ > Al3+. This is because as we move across a period in the periodic table, the atomic radius decreases due to the increasing effective nuclear charge. However, when an atom gains electrons to form an anion, the added electrons repel each other, causing the electron cloud to expand and the ionic radius to increase. Therefore, O2- has the largest ionic radius as it has gained two electrons, followed by F- with one added electron. Na+ has lost one electron, making it smaller than F-. Mg2+ has lost two electrons, making it smaller than Na+. Lastly, Al3+ has lost three electrons, making it the smallest ion in the sequence.

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• 9.

### Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements give the correct picture?

• A.

Reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group

• B.

In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group

• C.

Chemical reactivity increases with increase in atomic number down the group in both alkali metals and halogens

• D.

In alkali metals, the reactivity increases but in the halogens it decreases with increase in atomic number down the group

D. In alkali metals, the reactivity increases but in the halogens it decreases with increase in atomic number down the group
Explanation
The correct answer states that in alkali metals, the reactivity increases but in the halogens it decreases with an increase in atomic number down the group. This is because alkali metals have a single valence electron that is easily lost, leading to increased reactivity as the atomic number increases. On the other hand, halogens have a high electron affinity and tend to gain electrons, resulting in decreased reactivity as the atomic number increases. Therefore, this statement accurately describes the periodic trends of chemical reactivity in alkali metals and halogens.

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• 10.

### According to the periodic law of elements, the variation in properties of elements is related to their

• A.

Atomic masses

• B.

Nuclear Masses

• C.

Atomic numbers

• D.

Nuclear neutron-proton number ratios

C. Atomic numbers
Explanation
The correct answer is atomic numbers. According to the periodic law of elements, the variation in properties of elements is related to their atomic numbers. Atomic number represents the number of protons in an atom's nucleus, which determines the element's identity. As the atomic number increases, the properties of elements change in a periodic manner, leading to the organization of the periodic table. Atomic mass, nuclear masses, and neutron-proton number ratios are not directly related to the variation in properties of elements.

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• 11.

### The atomic numbers of vanadium (V) , Chromium (Cr), Manganese (Mn) and iron(Fe) are respectively 23, 24 and 25 and 26. Which one of these may be expected to have the highest second ionization enthalpy?

• A.

V

• B.

Cr

• C.

Mn

• D.

Fe

B. Cr
Explanation
Chromium (Cr) may be expected to have the highest second ionization enthalpy because it has the highest atomic number among the given elements. As the atomic number increases, the number of protons in the nucleus increases, leading to a stronger attraction between the nucleus and the electrons. This stronger attraction requires more energy to remove an electron, resulting in a higher ionization enthalpy.

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• 12.

### Which one of the following statement is incorrect for an atom having electronic configuration 2, 8, 7:

• A.

It forms diatomic molecules

• B.

It is a non-metal element

• C.

Its valency is 1

• D.

It forms basic oxide

D. It forms basic oxide
Explanation
An atom with electronic configuration 2, 8, 7 refers to an atom with 17 electrons. This corresponds to the element chlorine (Cl) in the periodic table. Chlorine is a non-metal element and does not form diatomic molecules. It forms a diatomic molecule called chlorine gas (Cl2). Its valency is 1, meaning it tends to gain one electron to achieve a stable electron configuration. Chlorine forms acidic oxides, not basic oxides, as it tends to combine with oxygen to form compounds like chlorine dioxide (ClO2) or chlorine trioxide (ClO3).

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• 13.

### Which of the following electronic configuration corresponds to atom with the lowest ionisation enthalpy?

• A.

1s2 2s2 2p3

• B.

1s2 2s2 2p6 3s1

• C.

1s2 2s2 2p6

• D.

1s2 2s2 2p5

B. 1s2 2s2 2p6 3s1
Explanation
The electronic configuration 1s2 2s2 2p6 3s1 corresponds to the atom with the lowest ionisation enthalpy because it has the highest number of valence electrons (in this case, 1) in the outermost energy level. The ionisation enthalpy is the energy required to remove an electron from an atom, and atoms with more valence electrons have a lower ionisation enthalpy because the outermost electron is less strongly attracted to the nucleus and therefore easier to remove.

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• 14.

### Lanthanide contraction relates to decrease in-

• A.

• B.

Atomic as well as M3+ radii

• C.

Valence electron

• D.

Oxidation state

B. Atomic as well as M3+ radii
Explanation
The lanthanide contraction refers to the decrease in both atomic and M3+ radii. This phenomenon occurs because as you move across the lanthanide series, the increasing positive charge in the nucleus attracts the outermost electrons more strongly, causing the electron cloud to contract. This contraction affects both the atomic radii, which is the size of the atom as a whole, and the M3+ radii, which is the size of the ion with a +3 charge. Therefore, the correct answer is atomic as well as M3+ radii.

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• 15.

### Which of the following has no unit-

• A.

Electronegativity

• B.

Electron affinity

• C.

Ionisation energy

• D.

Excitation potential

A. Electronegativity
Explanation
Electronegativity is a measure of an atom's ability to attract electrons in a chemical bond. It is a relative scale and does not have a specific unit of measurement. Electronegativity values are determined based on a comparison between different elements, rather than being measured in a specific unit. In contrast, electron affinity, ionization energy, and excitation potential are all physical properties that can be measured and have specific units associated with them.

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• 16.

### Ionization energy of nitrogen is more than oxygen due to-

• A.

Increased attraction of electron towards nucleus

• B.

Extra stability of half filled 'p' orbital

• C.

Small size of N2

• D.

None of the above

B. Extra stability of half filled 'p' orbital
Explanation
The correct answer is "Extra stability of half filled 'p' orbital". Nitrogen has a half-filled p orbital in its ground state, which provides it with extra stability. This stability makes it more difficult to remove an electron from nitrogen compared to oxygen, which does not have a half-filled p orbital. The increased attraction of electrons towards the nucleus and the small size of N2 are not the main reasons for the higher ionization energy of nitrogen.

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• 17.

### In which of the following arrangements the order is NOT according to the property indicated against it?

• A.

Al3+ < Mg2+ < Na+ < F- : Increasing ionic size

• B.

B < C < N < O : increasing first ionization energy

• C.

I < Br < F < Cl - increasing electron gain enthalpy (With negative sign)

• D.

Li < Na < K < Rb - increasing metallic radius

B. B < C < N < O : increasing first ionization energy
Explanation
The correct answer is B < C < N < O because the trend for first ionization energy is typically to increase from left to right across a period. However, in this arrangement, the order is reversed, with B having a higher first ionization energy than C, N, and O. This indicates that there may be an error in the arrangement or a different property is being indicated.

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• 18.

### Which one of the following is isoelectronic-

• A.

Fe3+, Co3+

• B.

Fe3+, Mn2+

• C.

Co3+, Sr3+

• D.

Sc3+, Ti3+

B. Fe3+, Mn2+
Explanation
Fe3+ and Mn2+ are isoelectronic because they have the same number of electrons. Fe3+ has lost 3 electrons to become positively charged, while Mn2+ has lost 2 electrons to become positively charged. Therefore, both ions have the same electron configuration and are isoelectronic.

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• 19.

### Similarity between the characteristics of Li and Mg is termed as-

• A.

Direct relation

• B.

Group relation

• C.

Diagonal relation

• D.

Linear relation

C. Diagonal relation
Explanation
The similarity between the characteristics of Li and Mg is referred to as a diagonal relation. This term is used to describe elements that are located diagonally from each other in the periodic table and share similar properties. Li and Mg are both located in Group 2 of the periodic table and have similar characteristics such as being lightweight, having low melting and boiling points, and forming ionic compounds with similar properties. This diagonal relationship is due to the similarities in the electronic configurations of these elements.

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• 20.

### In alkali metals, correct order of 1st ionisation energy is-

• A.

Li > Na > K > Rb

• B.

K > Rb > Li > Na

• C.

Li > Na > Rb > K

• D.

Rb > Li > K > Na

A. Li > Na > K > Rb
Explanation
The correct order of 1st ionization energy in alkali metals is Li > Na > K > Rb. This means that it requires the most energy to remove an electron from a lithium atom, followed by sodium, potassium, and finally rubidium. This order can be explained by the trend in atomic radius and effective nuclear charge. As you move down the group, the atomic radius increases, resulting in a weaker attraction between the nucleus and the outermost electron. Additionally, the effective nuclear charge decreases due to the shielding effect of inner electrons. These factors make it easier to remove an electron, leading to a decrease in ionization energy.

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• 21.

### Elements with similar chemical properties-

• A.

Occur only within the same period

• B.

Have identical atomic mass(weight)

• C.

Have identical number of neutrons

• D.

Have the same number of electrons in the outer shell

D. Have the same number of electrons in the outer shell
Explanation
Elements with similar chemical properties have the same number of electrons in their outer shell. The outer shell is the valence shell, which determines the element's reactivity and ability to form bonds with other elements. Elements with the same number of electrons in their outer shell have similar chemical behaviors because they have similar valence electron configurations. This similarity allows them to exhibit similar chemical reactions and form similar compounds.

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• 22.

### Zinc and Copper have similarity in their-

• A.

Group number

• B.

Electronic configuration

• C.

Number of electron present in their d subshells

• D.

Both are transition metals

A. Group number
Explanation
Zinc and Copper have similarity in their group number because they both belong to Group 11 of the periodic table. Group number refers to the vertical columns in the periodic table which indicate the number of valence electrons an element has. Zinc and Copper both have one valence electron, making them part of the same group.

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• 23.

### Which of the following statements is false?

• A.

Screening effect increases down the group

• B.

Zeff increases down the group

• C.

Zeff. increases in a period

• D.

All of the above

B. Zeff increases down the group
Explanation
The given statement "Zeff increases down the group" is false. Zeff, also known as effective nuclear charge, refers to the net positive charge experienced by an electron in an atom. It is determined by the number of protons in the nucleus and the shielding effect of inner electrons. As we move down a group in the periodic table, the number of protons and inner electrons both increase, resulting in an increase in screening effect and a decrease in Zeff. Therefore, Zeff decreases down the group, making the statement false.

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• 24.

### According to Pauli's exclusion principle, spin quantum number of two electrons in any subshell can be -

• A.

+1/2, +1/2

• B.

-1/2 , +1/2

• C.

-1/2 , -1/2

• D.

0,1

B. -1/2 , +1/2
Explanation
According to Pauli's exclusion principle, no two electrons in an atom can have the same set of four quantum numbers. One of these quantum numbers is the spin quantum number, which can have two possible values: +1/2 or -1/2. This means that in any subshell, the spin quantum number of two electrons must be different. Therefore, the correct answer is -1/2 , +1/2.

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• 25.

### If Hund's rule is violated, what is the minimum number of electrons that can be present in a 2p orbital of Nitrogen atom?

• A.

0

• B.

1

• C.

2

• D.

3

A. 0
Explanation
Hund's rule states that electrons will occupy empty orbitals of the same energy level before pairing up. In the case of a 2p orbital of a Nitrogen atom, there are three orbitals (2px, 2py, 2pz) that are degenerate, meaning they have the same energy. According to Hund's rule, each orbital must be singly occupied before any of them can start pairing up. Therefore, if Hund's rule is violated, it means that there are no electrons present in the 2p orbital, resulting in a minimum of 0 electrons.

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• 26.

### Number of unpaired electrons in O2- atom is

• A.

2

• B.

1

• C.

3

• D.

0

D. 0
Explanation
O2- is an oxygen ion with a charge of -2. Oxygen normally has 6 electrons, but since it has a charge of -2, it gains 2 extra electrons, resulting in a total of 8 electrons. In order to determine the number of unpaired electrons, we need to look at the electron configuration. The electron configuration of oxygen is 1s2 2s2 2p4. In O2-, the two extra electrons will occupy the 2p orbital, filling it completely. Since all the electrons are paired, there are no unpaired electrons in O2-, so the answer is 0.

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• 27.

### Which of the following has maximum number of unpaired electron

• A.

Fe

• B.

Fe+2

• C.

Fe+3

• D.

Fe+4

C. Fe+3
Explanation
Fe+3 has the maximum number of unpaired electrons. When Fe loses three electrons to become Fe+3, it loses them from its 3d orbitals. Fe has 6 electrons in its 3d orbitals, and when it loses three electrons, it leaves behind 3 unpaired electrons. Fe, Fe+2, and Fe+4 do not have any unpaired electrons.

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• 28.

### Maximum number of electrons present in n=5 is

• A.

25

• B.

50

• C.

30

• D.

10

B. 50
Explanation
The maximum number of electrons present in an energy level can be calculated using the formula 2n^2, where n is the principal quantum number. For n=5, the maximum number of electrons would be 2(5^2) = 50.

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• 29.

### Choose the answers appropriately.  A: If both Assertion and Reason are True and Reason is the correct explanation of Assertion. B : If both Assertion and Reason are True but Reason is the wrong explanation of Assertion. C : If Assertion is True but Reason is False. D : If both Assertion and Reason are False. ASSERTION: Na, K and Rb is Dobereiner's triad. REASON: Atomic weight of K is twice the atomic weight of Rb.

• A.

A

• B.

B

• C.

C

• D.

D

D. D
Explanation
The assertion states that Na, K, and Rb form Dobereiner's triad. However, this is incorrect as Dobereiner's triad consists of elements with similar chemical properties and atomic weights. Na, K, and Rb do not fulfill this criteria. The reason states that the atomic weight of K is twice the atomic weight of Rb, which is also incorrect. Therefore, both the assertion and reason are false.

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• 30.

### Choose the answers appropriately.  A: If both Assertion and Reason are True and Reason is the correct explanation of Assertion. B : If both Assertion and Reason are True but Reason is the wrong explanation of Assertion. C : If Assertion is True but Reason is False. D : If both Assertion and Reason are False. ASSERTION: Na, K and Rb is Dobereiner's triad. REASON: Atomic weight of K is twice the atomic weight of Rb.

• A.

A

• B.

B

• C.

C

• D.

D

C. C
Explanation
The assertion states that Na, K, and Rb form Dobereiner's triad. This means that these three elements have similar chemical properties and can be grouped together. The reason given is that the atomic weight of K is twice the atomic weight of Rb. However, this is incorrect as the atomic weight of K is actually less than twice the atomic weight of Rb. Therefore, the reason is false, but the assertion is true. Hence, the correct answer is C.

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• Mar 21, 2023
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• Jun 26, 2020
Quiz Created by
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