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All questions of Classification Of Elements and Periodicity in Properties for NEET Exam
Lothar Meyer proposed that on arranging the elements in order of increasing atomic weights; similarities appear in which type of properties?- a)Only physical properties
- b)Only chemical properties
- c)In both physical and chemical properties
- d)thermodynamic properties
Correct answer is option 'C'. Can you explain this answer?
Lothar Meyer proposed that on arranging the elements in order of increasing atomic weights; similarities appear in which type of properties?
a)
Only physical properties
b)
Only chemical properties
c)
In both physical and chemical properties
d)
thermodynamic properties
|
Anoushka Yadav answered |
Lothar Meyer proposed the periodic table in 1864. He arranged the elements in order of increasing atomic weights. He noticed that similarities appeared in both physical and chemical properties of the elements. The correct answer is option 'C' which means similarities appear in both physical and chemical properties.
Explanation:
The periodic table arranges the elements in a way that helps in understanding their properties. The modern periodic table is based on the electronic configuration of the elements. But the original periodic table was based on the atomic weight of the elements. Lothar Meyer was the first to realize that there is a periodicity in the properties of the elements when they are arranged in order of increasing atomic weights.
Physical properties are those that can be observed or measured without changing the chemical composition of the substance. Some examples of physical properties are:
- Melting point
- Boiling point
- Density
- Electrical conductivity
- Thermal conductivity
- Atomic radius
- Ionic radius
- Electronegativity
Chemical properties are those that can be observed during a chemical reaction. Some examples of chemical properties are:
- Reactivity with acids
- Reactivity with oxygen
- Reactivity with water
- Reduction potential
- Oxidation potential
Meyer observed that when the elements are arranged in order of increasing atomic weights, there is a periodicity in their physical and chemical properties. This means that elements with similar atomic weights have similar physical and chemical properties. For example, lithium, sodium, and potassium have similar physical and chemical properties because they all belong to the same group and have similar atomic weights.
Conclusion:
Lothar Meyer proposed the periodic table in 1864. He arranged the elements in order of increasing atomic weights. He noticed that similarities appeared in both physical and chemical properties of the elements. When the elements are arranged in order of increasing atomic weights, there is a periodicity in their physical and chemical properties.
Explanation:
The periodic table arranges the elements in a way that helps in understanding their properties. The modern periodic table is based on the electronic configuration of the elements. But the original periodic table was based on the atomic weight of the elements. Lothar Meyer was the first to realize that there is a periodicity in the properties of the elements when they are arranged in order of increasing atomic weights.
Physical properties are those that can be observed or measured without changing the chemical composition of the substance. Some examples of physical properties are:
- Melting point
- Boiling point
- Density
- Electrical conductivity
- Thermal conductivity
- Atomic radius
- Ionic radius
- Electronegativity
Chemical properties are those that can be observed during a chemical reaction. Some examples of chemical properties are:
- Reactivity with acids
- Reactivity with oxygen
- Reactivity with water
- Reduction potential
- Oxidation potential
Meyer observed that when the elements are arranged in order of increasing atomic weights, there is a periodicity in their physical and chemical properties. This means that elements with similar atomic weights have similar physical and chemical properties. For example, lithium, sodium, and potassium have similar physical and chemical properties because they all belong to the same group and have similar atomic weights.
Conclusion:
Lothar Meyer proposed the periodic table in 1864. He arranged the elements in order of increasing atomic weights. He noticed that similarities appeared in both physical and chemical properties of the elements. When the elements are arranged in order of increasing atomic weights, there is a periodicity in their physical and chemical properties.
Horizontal rows in the periodic table are called:- a)Cell
- b)Table
- c)Groups
- d)Periods
Correct answer is option 'D'. Can you explain this answer?
Horizontal rows in the periodic table are called:
a)
Cell
b)
Table
c)
Groups
d)
Periods
|
Rohan Singh answered |
The Periodic Table: Families and Periods. In the periodic table of elements, there are seven horizontal rows of elements called periods. The vertical columns of elements are called groups, or families.
Can you explain the answer of this question below:In the modern periodic table, which period contains 32 elements?
- A:
Sixth
- B:
First
- C:
Seventh
- D:
Second
The answer is a.
In the modern periodic table, which period contains 32 elements?
Sixth
First
Seventh
Second
|
Preeti Iyer answered |
The answer is c.
The total number of electrons that can be accommodated in seventh period are 2 ( in 7s) + 14(in 5f) + 10(in 6d )+ 6(in 7p) = 32. The maximum number of elements present in it is 32.
Among the following statements the one that is not true about Mendeleev’s Periodic Table is:- a)Elements of group 7 and 8 were arranged on the basis of equivalent weights
- b)Some elements in the same group differ in their properties
- c)Elements were arranged on the basis of atomic weights
- d)The position of hydrogen was not justified
Correct answer is option 'A'. Can you explain this answer?
Among the following statements the one that is not true about Mendeleev’s Periodic Table is:
a)
Elements of group 7 and 8 were arranged on the basis of equivalent weights
b)
Some elements in the same group differ in their properties
c)
Elements were arranged on the basis of atomic weights
d)
The position of hydrogen was not justified
|
Om Desai answered |
In Mendeleev’s periodic table, elements were arranged on the basis of atomic weights.
An element with atomic number will form a basic oxide________ - a)7
- b)17
- c)14
- d)11
Correct answer is option 'D'. Can you explain this answer?
An element with atomic number will form a basic oxide________
a)
7
b)
17
c)
14
d)
11
|
EduRev NEET answered |
Sodium forms a basic oxide because it is a metal. Metals generally form basic oxides because they tend to donate electrons to oxygen atoms. In the case of sodium, when it reacts with oxygen, it forms sodium oxide, in which each sodium atom donates one electron to oxygen. This results in the formation of sodium ions (Na+) and oxide ions (O2-), which can combine to form the ionic compound sodium oxide. Sodium oxide is basic because it can react with water to form a solution of sodium hydroxide, a strong base.
Which electronic configuration of an element has abnormally high difference between second and third ionization energy ? [1993]
- a)1s2, 2s2, 2p6, 3s1
- b)1s2, 2s2, 2p6, 3s1 3p1
- c)1s2, 2s2, 2p6, 3s2 3p2
- d)1s2, 2s2, 2p6, 3s2
Correct answer is option 'D'. Can you explain this answer?
Which electronic configuration of an element has abnormally high difference between second and third ionization energy ? [1993]
a)
1s2, 2s2, 2p6, 3s1
b)
1s2, 2s2, 2p6, 3s1 3p1
c)
1s2, 2s2, 2p6, 3s2 3p2
d)
1s2, 2s2, 2p6, 3s2
|
Shivani Tiwari answered |
In option D, after removal of second valence electron from 3s orbital, the ion formed achieves noble gas configuration. Therefore, to remove the third electron from 2p orbital, a lot of energy is required. Thus, there is an abnormally high difference between second and third ionization enthalpies.
Eka silicon predicted by Mendeleev is which element:- a)Germanium
- b)Aluminium
- c)Gallium
- d)Sodium
Correct answer is option 'A'. Can you explain this answer?
Eka silicon predicted by Mendeleev is which element:
a)
Germanium
b)
Aluminium
c)
Gallium
d)
Sodium
|
Suresh Iyer answered |
Mendeleev predicted the existence of 'eka-silicon', which would fit into a gap next to silicon. The element germanium was discovered later. Its properties were found to be similar to the predicted ones and confirmed Mendeleev's periodic table.
Which scientist proposed that atomic number is more fundamental property of an element than its atomic mass?- a)Newlands
- b)Lothar Meyer
- c)Henry Moseley
- d)Meyer
Correct answer is option 'C'. Can you explain this answer?
Which scientist proposed that atomic number is more fundamental property of an element than its atomic mass?
a)
Newlands
b)
Lothar Meyer
c)
Henry Moseley
d)
Meyer
|
|
Pooja Shah answered |
The scientist who first of all showed that the atomic number is more fundamental property of an element than its atomic mass was Henry Moseley.
The lanthanide series starts with:
- a)Thorium
- b)Lutetium
- c)Lawrencium
- d)cerium
Correct answer is option 'D'. Can you explain this answer?
The lanthanide series starts with:
a)
Thorium
b)
Lutetium
c)
Lawrencium
d)
cerium
|
|
Suresh Reddy answered |
The lanthanide series starts from cerium and ends in lutetium
What is the relationship between the number of elements in each period and the number of atomic orbitals available in the energy level that is being filled?- a)Number of elements is twice the number of atomic orbitals.
- b)Number of elements is three times the number of atomic orbitals.
- c)Number of elements is half the number of atomic orbitals.
- d)Number of elements is same as the number of atomic orbitals.
Correct answer is 'A'. Can you explain this answer?
What is the relationship between the number of elements in each period and the number of atomic orbitals available in the energy level that is being filled?
a)
Number of elements is twice the number of atomic orbitals.
b)
Number of elements is three times the number of atomic orbitals.
c)
Number of elements is half the number of atomic orbitals.
d)
Number of elements is same as the number of atomic orbitals.
|
Krishna Iyer answered |
The number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled.
Let's understand it by taking the elements of the 4th period.
In the fourth period, there are 18 elements. The types of subshells used in the 4th period are s, p, and d.
Number of orbitals of s subshell = 1
Number of orbitals of p subshell = 3
Number of orbitals of d subshell = 5
Total number of orbitals = 9
The total number of elements is 18 i.e. twice the total number of orbitals.
In the fourth period, there are 18 elements. The types of subshells used in the 4th period are s, p, and d.
Number of orbitals of s subshell = 1
Number of orbitals of p subshell = 3
Number of orbitals of d subshell = 5
Total number of orbitals = 9
The total number of elements is 18 i.e. twice the total number of orbitals.
Elements in the same group have same:- a)Number of valence electrons
- b)Density
- c)Atomic radius
- d)Nuclear charge
Correct answer is option 'A'. Can you explain this answer?
Elements in the same group have same:
a)
Number of valence electrons
b)
Density
c)
Atomic radius
d)
Nuclear charge
|
Lavanya Menon answered |
- Elements in the same group have the same number of valence electrons in their outermost shell, hence, have similar properties.
- Elements in the same period don't have same number of valence electrons, hence, have different properties. But, elements in the same period have same number of shells.
If the atomic number of an element is 33, it will be placed in the periodic table in the- a)First group
- b)Third group
- c)Fifteenth group
- d)Seventh group.
Correct answer is option 'C'. Can you explain this answer?
If the atomic number of an element is 33, it will be placed in the periodic table in the
a)
First group
b)
Third group
c)
Fifteenth group
d)
Seventh group.
|
|
Preeti Iyer answered |
The electronic configuration of element with atomic number 33 is 1s22s22p63s23p63d104s24p3
As, its last shell have five electrons
and hence, its group is 10 + 5 = 15th or V A.
Hence,it lies in the 15th group.
As, its last shell have five electrons
and hence, its group is 10 + 5 = 15th or V A.
Hence,it lies in the 15th group.
The symbol for element with atomic number 111 and name Unununnium is- a)Uun
- b)Uuu
- c)UUU
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
The symbol for element with atomic number 111 and name Unununnium is
a)
Uun
b)
Uuu
c)
UUU
d)
None of the above
|
Gaurav Kumar answered |
- The element was to be called unununium (with the corresponding symbol of Uuu),a systematic element name as a placeholder, until the element was discovered (and the discovery then confirmed) and a permanent name was decided on.
- Although widely used in the chemical community on all levels, from chemistry classrooms to advanced textbooks, the recommendations were mostly ignored among scientists in the field, who called it element 111, with the symbol of E111, (111) or even simply 111.
Among the elements with atomic numbers 9, 12, 16 and 36 which is highly electropositive?- a)Element with atomic number 9
- b)Element \vith atomic number 12
- c)Element with atomic number 16
- d)Element with atomic number 36
Correct answer is option 'B'. Can you explain this answer?
Among the elements with atomic numbers 9, 12, 16 and 36 which is highly electropositive?
a)
Element with atomic number 9
b)
Element \vith atomic number 12
c)
Element with atomic number 16
d)
Element with atomic number 36
|
Pooja Deshpande answered |
Introduction:
In order to determine which element is highly electropositive among the elements with atomic numbers 9, 12, 16, and 36, we need to understand the concept of electronegativity and how it relates to electropositivity. Electronegativity is a measure of an atom's ability to attract electrons towards itself in a chemical bond. Electropositivity, on the other hand, is the tendency of an atom to lose electrons and form positive ions. Therefore, an element with low electronegativity and high electropositivity will readily lose electrons and form positive ions.
Explanation:
Let's analyze each element and its electropositivity based on their atomic numbers:
Element with atomic number 9 (Fluorine):
Fluorine is a halogen and has a high electronegativity. It readily gains an electron to achieve a stable electron configuration. Therefore, it is not highly electropositive.
Element with atomic number 12 (Magnesium):
Magnesium is an alkaline earth metal and has a low electronegativity. It readily loses its two valence electrons to achieve a stable electron configuration. Therefore, it is highly electropositive.
Element with atomic number 16 (Sulfur):
Sulfur is a nonmetal and has a moderate electronegativity. It can gain or lose electrons depending on the reaction conditions. However, it is not highly electropositive as it tends to gain electrons to achieve a stable electron configuration.
Element with atomic number 36 (Krypton):
Krypton is a noble gas and has a very high electronegativity. It has a stable electron configuration and tends not to lose or gain electrons. Therefore, it is not highly electropositive.
Conclusion:
Based on the analysis, the element with atomic number 12 (Magnesium) is highly electropositive among the elements with atomic numbers 9, 12, 16, and 36. It has a low electronegativity and readily loses its two valence electrons to form a stable 2+ ion.
In order to determine which element is highly electropositive among the elements with atomic numbers 9, 12, 16, and 36, we need to understand the concept of electronegativity and how it relates to electropositivity. Electronegativity is a measure of an atom's ability to attract electrons towards itself in a chemical bond. Electropositivity, on the other hand, is the tendency of an atom to lose electrons and form positive ions. Therefore, an element with low electronegativity and high electropositivity will readily lose electrons and form positive ions.
Explanation:
Let's analyze each element and its electropositivity based on their atomic numbers:
Element with atomic number 9 (Fluorine):
Fluorine is a halogen and has a high electronegativity. It readily gains an electron to achieve a stable electron configuration. Therefore, it is not highly electropositive.
Element with atomic number 12 (Magnesium):
Magnesium is an alkaline earth metal and has a low electronegativity. It readily loses its two valence electrons to achieve a stable electron configuration. Therefore, it is highly electropositive.
Element with atomic number 16 (Sulfur):
Sulfur is a nonmetal and has a moderate electronegativity. It can gain or lose electrons depending on the reaction conditions. However, it is not highly electropositive as it tends to gain electrons to achieve a stable electron configuration.
Element with atomic number 36 (Krypton):
Krypton is a noble gas and has a very high electronegativity. It has a stable electron configuration and tends not to lose or gain electrons. Therefore, it is not highly electropositive.
Conclusion:
Based on the analysis, the element with atomic number 12 (Magnesium) is highly electropositive among the elements with atomic numbers 9, 12, 16, and 36. It has a low electronegativity and readily loses its two valence electrons to form a stable 2+ ion.
Eka aluminium predicted by Mendeleev is which element?- a)Germanium
- b)Magnesium
- c)Gallium
- d)Sodium
Correct answer is option 'C'. Can you explain this answer?
Eka aluminium predicted by Mendeleev is which element?
a)
Germanium
b)
Magnesium
c)
Gallium
d)
Sodium
|
|
Gaurav Kumar answered |
Eka aluminium predicted by Mendeleev is Gallium. Eka-aluminium and gallium are the two names of the same element as Eka -Aluminum has almost exactly the same properties as the actual properties of the gallium element.
According to Dobereiner’s law of triads the number of elements present in each group is:
- a)2
- b)4
- c)5
- d)3
Correct answer is option 'D'. Can you explain this answer?
According to Dobereiner’s law of triads the number of elements present in each group is:
a)
2
b)
4
c)
5
d)
3
|
|
Geetika Shah answered |
- According to Dobereiner's law of triads each triad contains three elements.
- He also noticed that the middle element of each of the triads had an atomic weight about halfway between the atomic weights of the other two.
Moseley performed experiments and studied the frequencies of which radiations emitted from the elements?- a)Gamma rays
- b)X-rays
- c)UV rays
- d)Infra red rays
Correct answer is option 'B'. Can you explain this answer?
Moseley performed experiments and studied the frequencies of which radiations emitted from the elements?
a)
Gamma rays
b)
X-rays
c)
UV rays
d)
Infra red rays
|
|
Naina Bansal answered |
Moseley performed experiments and studied the frequencies of the X-rays emitted from the elements.
One would expect proton to have very large- a)Charge
- b)Ionization potential
- c)Hydration energy
- d)Radius. [1993]
Correct answer is option 'C'. Can you explain this answer?
One would expect proton to have very large
a)
Charge
b)
Ionization potential
c)
Hydration energy
d)
Radius. [1993]
|
Harshitha Dey answered |
Proton (H+) being very small in size would have very large hydration energy.
The first period has how many orbitals?- a)1
- b)3
- c)2
- d)4
Correct answer is option 'A'. Can you explain this answer?
The first period has how many orbitals?
a)
1
b)
3
c)
2
d)
4
|
|
Raghav Bansal answered |
1st period contains only 2 elements hydrogen and helium having atomic no. 1 and 2 respectively.
The electronic configuration of hydrogen is 1s1 while that of helium is 1s2. Both contain only s subshell. S subshell that only contains 1 orbital i.e. z.
Thus, 1st period has 1 orbital.
Johann Dobereiner classified elements in group of three elements called as- a)Trinity
- b)Trials
- c)Triads
- d)Diads
Correct answer is option 'C'. Can you explain this answer?
Johann Dobereiner classified elements in group of three elements called as
a)
Trinity
b)
Trials
c)
Triads
d)
Diads
|
|
Om Desai answered |
In 1829, Johann Dobereiner, a German scientist made some groups of three elements each and called them triads.
He observed that the atomic mass of the middle element of a triad was nearly equal to the arithmetic mean of the atomic masses of the other two elements. All three elements of a triad were similar in their properties.
Mendeleev predicted the existence of which element/elements in the periodic table?- a)Gallium
- b)Sodium and germanium
- c)Gallium and germanium
- d)Germanium and Gold
Correct answer is option 'C'. Can you explain this answer?
Mendeleev predicted the existence of which element/elements in the periodic table?
a)
Gallium
b)
Sodium and germanium
c)
Gallium and germanium
d)
Germanium and Gold
|
|
Neha Joshi answered |
Gallium and Germanium were the elements not discovered at that time and Mendeleev put gaps in the periodic table.
Gallium was called as Eka aluminium
Germanium was called as Eka silicon
Gallium was called as Eka aluminium
Germanium was called as Eka silicon
For alkali metals, which one of the following trends is INCORRECT?
- a)Hydration energy: Li > Na > K > Rb
- b)Ionization energy: Li > Na > K > Rb
- c)Atomic size: Li < Na < K < Rb
- d)Density: Li < Na < K < Rb
Correct answer is option 'D'. Can you explain this answer?
For alkali metals, which one of the following trends is INCORRECT?
a)
Hydration energy: Li > Na > K > Rb
b)
Ionization energy: Li > Na > K > Rb
c)
Atomic size: Li < Na < K < Rb
d)
Density: Li < Na < K < Rb
|
|
Om Desai answered |
Density: Li < Na < K < Rb.
The density of K is lower than that of Na. Thus, option D is incorrect. The correct trend is
Li < K < Na < Rb
The density of K is lower than that of Na. Thus, option D is incorrect. The correct trend is
Li < K < Na < Rb
Na+, Mg2+, Al3+ and Si4+ are isoelectronic. The order of their ionic size is [1993]- a)Na+ > Mg2+ < Al3 + < Si4+
- b)Na+ < Mg2+ > Al3+ > Si4+
- c)Na+ > Mg2+ > Al3+ > Si4+
- d)Na+ < Mg2+ > Al3+ < Si4+
Correct answer is option 'C'. Can you explain this answer?
Na+, Mg2+, Al3+ and Si4+ are isoelectronic. The order of their ionic size is [1993]
a)
Na+ > Mg2+ < Al3 + < Si4+
b)
Na+ < Mg2+ > Al3+ > Si4+
c)
Na+ > Mg2+ > Al3+ > Si4+
d)
Na+ < Mg2+ > Al3+ < Si4+
|
|
Neha Sharma answered |
Amongst isoelectronic ions, the size of the cation decreases as the magnitude of the charge increases.
Which one of the following oxides is expected to exhibit paramagn etic beh aviour? [2 00 5]- a)CO2
- b)SiO2
- c)SO2
- d)ClO2
Correct answer is option 'D'. Can you explain this answer?
Which one of the following oxides is expected to exhibit paramagn etic beh aviour? [2 00 5]
a)
CO2
b)
SiO2
c)
SO2
d)
ClO2
|
|
Nidhi Tamak answered |
Paramagnetic behavior is seen by those who has unpaired electrons that's why option d is correct because it has unpaired electron.
The basis of long form of periodic table is:- a)Mass number
- b)Atomic weight
- c)Atomic number
- d)Ionic radius
Correct answer is option 'C'. Can you explain this answer?
The basis of long form of periodic table is:
a)
Mass number
b)
Atomic weight
c)
Atomic number
d)
Ionic radius
|
|
Nandini Patel answered |
In long form of periodic table elements are arranged in increasing order of their Atomic number.
The maximum number of elements present in seventh period of the modern periodic table is:- a)32
- b)8
- c)2
- d)18
Correct answer is option 'A'. Can you explain this answer?
The maximum number of elements present in seventh period of the modern periodic table is:
a)
32
b)
8
c)
2
d)
18
|
|
Anjali Iyer answered |
The total number of electrons that can be accommodated in seventh period are 2 ( in 7s) + 14(in 5f) + 10(in 6d )+ 6(in 7p) = 32. The maximum number of elements present in it is 32.
What is the maximum number of electrons that can be filled in 3d subshell?- a)2
- b)14
- c)6
- d)10
Correct answer is option 'D'. Can you explain this answer?
What is the maximum number of electrons that can be filled in 3d subshell?
a)
2
b)
14
c)
6
d)
10
|
|
Shreya Gupta answered |
There are 5 orbitals in d subshell and each orbital can accommodate 2 electrons. Hence maximum number of electrons that can be filled in 3d subshell is 10.
How many orbitals are filled in second period?- a)6
- b)3
- c)1
- d)4
Correct answer is option 'D'. Can you explain this answer?
How many orbitals are filled in second period?
a)
6
b)
3
c)
1
d)
4
|
|
Lavanya Menon answered |
The 4 orbitals filled in second period are one 2s (with 2 electrons) and three 2p (with 6 electrons).
What is the principal quantum number for second period?a)1b)3c)2d)4Correct answer is option 'C'. Can you explain this answer?
|
|
Krishna Iyer answered |
Principle quantum no is defined as the no of shells an element possesses. Since every element in the 2nd period has 2 shells, therefore, principle quantum no for 2nd period is 2.
Newland arranged elements in increasing order of atomic weights and noted that every eighth element had properties similar to:- a)Third element
- b)Fourth element
- c)Second element
- d)First element
Correct answer is option 'D'. Can you explain this answer?
Newland arranged elements in increasing order of atomic weights and noted that every eighth element had properties similar to:
a)
Third element
b)
Fourth element
c)
Second element
d)
First element
|
Hansa Sharma answered |
According to Newlands' law of octaves when the elements are arranged in order of increasing atomic weights then every eighth element has properties similar to that of the first element.
14 elements after actinium is called
a) Lanthanides
b) Actinides
c) D-block elements
d) P block elements
Correct answer is option 'B'. Can you explain this answer?
|
|
Neha Sharma answered |
The correct answer is Option B.
14 elements after actinium is called Actinides.
The 3d transition series starts from which atomic number?- a)19
- b)21
- c)20
- d)22
Correct answer is option 'B'. Can you explain this answer?
The 3d transition series starts from which atomic number?
a)
19
b)
21
c)
20
d)
22
|
Ishani Pillai answered |
The 3d transition series starts from atomic number 21, which is the element Scandium (Sc).
Explanation:
- The transition elements or transition metals are a group of elements in the periodic table that have partially filled d orbitals in their valence shells.
- The 3d transition series refers to the elements that have their d orbitals filled progressively, from scandium (Sc) to zinc (Zn), as the atomic number increases.
- Scandium has an atomic number of 21, which means it has 21 protons in its nucleus and 21 electrons surrounding it.
- Scandium is the first element in the 3d transition series and has one electron in its 3d orbital.
- The other elements in the 3d transition series are titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu), and zinc (Zn).
- These elements have their d orbitals filled progressively, from 1 electron in Sc to 10 electrons in Zn.
- The 4d and 5d transition series follow the same pattern, with the d orbitals being progressively filled from the first element in each series.
Therefore, the correct answer is option B, 21.
Explanation:
- The transition elements or transition metals are a group of elements in the periodic table that have partially filled d orbitals in their valence shells.
- The 3d transition series refers to the elements that have their d orbitals filled progressively, from scandium (Sc) to zinc (Zn), as the atomic number increases.
- Scandium has an atomic number of 21, which means it has 21 protons in its nucleus and 21 electrons surrounding it.
- Scandium is the first element in the 3d transition series and has one electron in its 3d orbital.
- The other elements in the 3d transition series are titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu), and zinc (Zn).
- These elements have their d orbitals filled progressively, from 1 electron in Sc to 10 electrons in Zn.
- The 4d and 5d transition series follow the same pattern, with the d orbitals being progressively filled from the first element in each series.
Therefore, the correct answer is option B, 21.
Which one of the following ions will be the smallest in size? [1996]- a)Na+
- b)Mg2+
- c)F–
- d)O2–
Correct answer is option 'B'. Can you explain this answer?
Which one of the following ions will be the smallest in size? [1996]
a)
Na+
b)
Mg2+
c)
F–
d)
O2–
|
Prisha Singh answered |
Greater is the positive charge on atom, large is effective nuclear charge. Hence smaller is the size
Elements of which of the following groups will form anions most readily ? [1992]- a)Oxygen family
- b)Nitrogen family
- c)Halogens
- d)Alkali metals
Correct answer is option 'C'. Can you explain this answer?
Elements of which of the following groups will form anions most readily ? [1992]
a)
Oxygen family
b)
Nitrogen family
c)
Halogens
d)
Alkali metals
|
Kunal Rane answered |
Electron affinity values are high in case of halogen because halogens have seven electrons (ns2np5) in the valence shell, they have a strong tendency to acquire the nearest inert gas configuration by gaining an electron from the metallic atom and form halide ions easily.
An atom has electronic configuration 1s2 2s2 2p6 3s2 3p6 3d3 4s2, you will place it in which group?- a)Fifth
- b)Fifteenth [2002]
- c)Secon d
- d)Third
Correct answer is option 'A'. Can you explain this answer?
An atom has electronic configuration 1s2 2s2 2p6 3s2 3p6 3d3 4s2, you will place it in which group?
a)
Fifth
b)
Fifteenth [2002]
c)
Secon d
d)
Third
|
Prasenjit Pillai answered |
The electronic configuration clearly suggest that it is a d-block element (having contiguration (n – 1) d1– 10 ns0 – 2) which starts from III B and goes till II B. Hence with d3 configuration it would be classified in the group.
The last element of actinoid series is:- a)Lutetium
- b)Thorium
- c)Cerium
- d)Lawrencium
Correct answer is option 'D'. Can you explain this answer?
The last element of actinoid series is:
a)
Lutetium
b)
Thorium
c)
Cerium
d)
Lawrencium
|
|
Pooja Mehta answered |
The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and its electronic configuration is[Rn]5 f146d17s2. The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f14 configuration.
Which one of the following ionic species has the greatest proton affinity to form stable compound?- a)NH 2-
- b)F– [2007]
- c)I–
- d)HS–
Correct answer is option 'A'. Can you explain this answer?
Which one of the following ionic species has the greatest proton affinity to form stable compound?
a)
NH 2-
b)
F– [2007]
c)
I–
d)
HS–
|
Smrity answered |
Correct option is a bcoz the conjugate base NH2- is weak and it is also known as strong base .
The stability of + 1 oxidation state increases in the sequence: [2009]
- a)Tl < In < Ga < Al
- b)In < Tl < Ga < Al
- c)Ga < In < Al < Tl
- d)Al < Ga < In < Tl
Correct answer is option 'D'. Can you explain this answer?
The stability of + 1 oxidation state increases in the sequence: [2009]
a)
Tl < In < Ga < Al
b)
In < Tl < Ga < Al
c)
Ga < In < Al < Tl
d)
Al < Ga < In < Tl
|
Deepak Joshi answered |
The stability of +1 oxidation state increases from aluminium to thallium i.e.
Al < Ga < In < Tl
Al < Ga < In < Tl
In the periodic table from left to right in a period, the atomic volume [1993]- a)Decr eases
- b)In creases
- c)Remains same
- d)First decrease then increases
Correct answer is option 'D'. Can you explain this answer?
In the periodic table from left to right in a period, the atomic volume [1993]
a)
Decr eases
b)
In creases
c)
Remains same
d)
First decrease then increases
|
Anu Bajaj answered |
Atomic volume is the volume occupied by one gram of an element. Within a period from left to right, atomic volume first decreases and then increases due to increases of nuclear charge and increase in the number of electrons in the valence shell.
Which of the following electronic configuration an atom has the lowest ionisation enthalpy? [2007]- a)1s2 2s2 sp3
- b)1s2 2s2 2p5 3s1
- c)1s2 2s2 2p6
- d)1s2 2s2 2p5
Correct answer is option 'B'. Can you explain this answer?
Which of the following electronic configuration an atom has the lowest ionisation enthalpy? [2007]
a)
1s2 2s2 sp3
b)
1s2 2s2 2p5 3s1
c)
1s2 2s2 2p6
d)
1s2 2s2 2p5
|
Devansh Mehra answered |
The electronic configuration 1s2 2s22p5 3s1 shows lowest ionisation energy because this configuration is unstable due to the presence of one electron is s- orbital. Hence, less energy is required to remove the electron.
For IUPAC nomenclature of elements with atomic number greater than 100, the abbreviation used for digit zero is:- a)o
- b)MN
- c)n
- d)N
Correct answer is 'C'. Can you explain this answer?
For IUPAC nomenclature of elements with atomic number greater than 100, the abbreviation used for digit zero is:
a)
o
b)
MN
c)
n
d)
N
|
Mira Sharma answered |
For IUPAC nomenclature of elements with atomic number greater than 100, the abbreviation used for digit zero is n and name is nil.
Which of the following statements is true? [2002]- a)Silicon exhibits 4 coordination number in its compound
- b)Bond energy of F2 is less than Cl2
- c)Mn(III) oxidation state is more stable than Mn(II) in aqueous state
- d)Elements of 15th group shows only +3 and +5 oxidation states
Correct answer is option 'B'. Can you explain this answer?
Which of the following statements is true? [2002]
a)
Silicon exhibits 4 coordination number in its compound
b)
Bond energy of F2 is less than Cl2
c)
Mn(III) oxidation state is more stable than Mn(II) in aqueous state
d)
Elements of 15th group shows only +3 and +5 oxidation states
|
Ashwini Khanna answered |
Due to small size and high electronegativity of fluorine, there exists repulsion between the nucleus of two fluorine atoms which ultimately results in the weakening of the F-F bond. Hence, the bond dissociation enthalpy of F
2
is less than that of Cl
2
The distribution of electrons into orbitals of an atom is called its:- a)Electronic configuration
- b)Atomic configuration
- c)Molecular configuration
- d)Configuration
Correct answer is option 'A'. Can you explain this answer?
The distribution of electrons into orbitals of an atom is called its:
a)
Electronic configuration
b)
Atomic configuration
c)
Molecular configuration
d)
Configuration
|
Vaishnavi Bajaj answered |
The arrangement of electrons in different shells and sub-shells is known as the electronic configuration of a particular element. The electronic configuration diagram represents an element in its ground state or stable state. There are a set of rules to remember while distribution off electrons in different orbits.
Correct order of first IP among following elements Be, B, C, N, O is [2001]- a)B < Be < C < O < N
- b)B < Be < C < N < O
- c)Be < B < C < N < O
- d)Be < B < C < O < N
Correct answer is option 'A'. Can you explain this answer?
Correct order of first IP among following elements Be, B, C, N, O is [2001]
a)
B < Be < C < O < N
b)
B < Be < C < N < O
c)
Be < B < C < N < O
d)
Be < B < C < O < N
|
Subhankar Datta answered |
Be – 1s22s2; B – 1s22s22p1; C – 1s22s22p2; N – 1s22s22p3; O – 1s22s22p4. IP increases along the period. But IP of Be > B. Further IP of O < N because atoms with fully or partly filled orbitals are most stable and hence have high ionisation energy.
The element, with atomic number 118, will be- a)alkali
- b)noble gas
- c)lanthanide
- d)transition element
Correct answer is option 'B'. Can you explain this answer?
The element, with atomic number 118, will be
a)
alkali
b)
noble gas
c)
lanthanide
d)
transition element
|
Krish Patel answered |
Electron ic con figuration of element with atomic number 118 will be [Rn]5f146d10 7s27p6. Since its elctronic configuration in the outer most orbit (ns2np6) resemble with that of inert or noble gases, therefore it will be noble gas element.
K+ and Cl- ions are isoelectronic. Which of the statements is not correct?
- a)Both K+ and Cl- ions contain 18 electrons
- b)Both K+ and Cl- ions have same configuration
- c)K+ ion is bigger than Cl- ion in ionic size
- d)Cl- ion is bigger than K+ ion in size
Correct answer is option 'C'. Can you explain this answer?
K+ and Cl- ions are isoelectronic. Which of the statements is not correct?
a)
Both K+ and Cl- ions contain 18 electrons
b)
Both K+ and Cl- ions have same configuration
c)
K+ ion is bigger than Cl- ion in ionic size
d)
Cl- ion is bigger than K+ ion in size
|
|
Ajay Yadav answered |
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for C, Ca, Al, F and O? [NEET Kar. 2013]- a)Ca < Al < C < O < F
- b)Al < Ca < O < C < F
- c)Al < O < C < Ca < F
- d)C < F < O < Al < Ca
Correct answer is option 'A'. Can you explain this answer?
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for C, Ca, Al, F and O? [NEET Kar. 2013]
a)
Ca < Al < C < O < F
b)
Al < Ca < O < C < F
c)
Al < O < C < Ca < F
d)
C < F < O < Al < Ca
|
Ritika Khanna answered |
As the nuclear charge increases, the force of attraction between the nucleus and the incoming electron increses and hence the elecron gain enthalpy becomes more negative, hence the correct order is Ca < Al < C < O < F
Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy? [2009]- a)Ne [3s23p2]
- b)Ar [3d104s24p3 ]
- c)Ne [3s23p1]
- d)Ne [3s23p3]
Correct answer is option 'D'. Can you explain this answer?
Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy? [2009]
a)
Ne [3s23p2]
b)
Ar [3d104s24p3 ]
c)
Ne [3s23p1]
d)
Ne [3s23p3]
|
Pankaj Banerjee answered |
The smaller the atomic size, larger is the value of ionisation potential. Further the atoms having half filled or fully filled orbitals are comparitively more stable, hence more energy is required to remove the electron from such atoms.
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