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All questions of d – and f – Block Elements for JEE Exam
Can you explain the answer of this question below:Which one of the following is a diamagnetic ion?- A:CO2+
- B:Cu2+
- C:Mn2+
- D:Sc3+
The answer is d.
Which one of the following is a diamagnetic ion?
A:
CO2+
B:
Cu2+
C:
Mn2+
D:
Sc3+
|
Learners Habitat answered |
Co+2 = [Ar] 3d7
Cu2+ = [Ar] 3d9
Mn+2 = [Ar] 3d5
Sc+3 = [Ar]
We can see that only Sc+3 has no unpaired electron, so it is a diamagnetic ion.
Cu2+ = [Ar] 3d9
Mn+2 = [Ar] 3d5
Sc+3 = [Ar]
We can see that only Sc+3 has no unpaired electron, so it is a diamagnetic ion.
Which ion will show more paramagnetic behaviour ?- a)Cu+
- b)Fe2+
- c)Ag+
- d)Fe3+
Correct answer is option 'D'. Can you explain this answer?
Which ion will show more paramagnetic behaviour ?
a)
Cu+
b)
Fe2+
c)
Ag+
d)
Fe3+
|
Lavanya Menon answered |
Since the configuration of Fe3+ ion is (argon ) 3d5,which contains maximum number of unpaired electrons, hence more will be paramagnetic behavior.
Which one of the following exists in the oxidation state other than +3?- a)B
- b)Al
- c)Ce
- d)Ga
Correct answer is option 'C'. Can you explain this answer?
Which one of the following exists in the oxidation state other than +3?
a)
B
b)
Al
c)
Ce
d)
Ga
|
|
Dino James answered |
Ce exhibits only +4oxident states
Which form of silver is colourless?- a)Ag2+
- b)Ag
- c)Ag3+
- d)Ag+
Correct answer is option 'D'. Can you explain this answer?
Which form of silver is colourless?
a)
Ag2+
b)
Ag
c)
Ag3+
d)
Ag+
|
Om Desai answered |
ilver in the form of Ag+ is colourless. For transition metal ions to exhibit color, their metal ions must have incompletely filled (n-1)d orbitals.
Ag+ =4d10,5s0
Ag+ has completely filled d orbitals hence is colourless.
Which among the following is colourless?- a)Sc2+
- b)Zn2+
- c)Ti3+
- d)V3+
Correct answer is option 'B'. Can you explain this answer?
Which among the following is colourless?
a)
Sc2+
b)
Zn2+
c)
Ti3+
d)
V3+
|
Krishna Iyer answered |
Zn2+ has completely filled d-orbitals and there are no vacant d-orbitals for the transition of electrons, hence it is colourless.
The first ionization energy of the d-block elements are?- a)Lesser than p-block elements
- b)Between s and p-block elements
- c)Lesser than s-block elements
- d)Higher than p-block elements
Correct answer is option 'B'. Can you explain this answer?
The first ionization energy of the d-block elements are?
a)
Lesser than p-block elements
b)
Between s and p-block elements
c)
Lesser than s-block elements
d)
Higher than p-block elements
|
Rahul Bansal answered |
The first ionization energy of the d-block elements are between s and p-block elements. Thus they are more electropositive than p-block elements and less electropositive than s-block elements.
Statement TypeDirection (Q. No. 25) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.Q. Statement I : The highest oxidation state of osmium is +8.Statement II : Osmium is a 5-d block element.- a)Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
- b)Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
- c)Statement I is correct but Statement II is incorrect
- d)Statement II is correct but Statement I is incorrect
Correct answer is option 'A'. Can you explain this answer?
Statement Type
Direction (Q. No. 25) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.
Q.
Statement I : The highest oxidation state of osmium is +8.
Statement II : Osmium is a 5-d block element.
a)
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
b)
Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
c)
Statement I is correct but Statement II is incorrect
d)
Statement II is correct but Statement I is incorrect
|
Vikas Saini answered |
B hona chayie
In which of the following oxidation states La achieve the noble gas configuration?- a)+7
- b)+3
- c)+5
- d)+2
Correct answer is option 'B'. Can you explain this answer?
In which of the following oxidation states La achieve the noble gas configuration?
a)
+7
b)
+3
c)
+5
d)
+2
|
Shail Chawla answered |
La stands for Lanthanum, which is a member of the lanthanide series of elements in the periodic table. It has an atomic number of 57 and an electronic configuration of [Xe]5d1 6s2.
Achieving Noble Gas Configuration
To achieve a noble gas configuration, an element must have its outermost shell filled with electrons. The noble gases are stable because they have a filled outermost shell. Lanthanum can achieve this configuration by losing three electrons from its outermost shell.
Oxidation States
The oxidation state of an element is the charge that it carries when it forms a compound or ion. Lanthanum can form compounds in various oxidation states, including +2, +3, and +4.
In order to determine in which oxidation state La achieves noble gas configuration, we need to look at its electron configuration and determine how many electrons it needs to lose to achieve a filled outermost shell.
La3+ (Oxidation State of +3)
When Lanthanum loses three electrons, it forms a +3 ion (La3+). In this oxidation state, La has a noble gas configuration of [Xe]. Therefore, option B is the correct answer.
Other Oxidation States
Lanthanum can also form compounds in other oxidation states, but they do not achieve a noble gas configuration. For example, in the +2 oxidation state, La has an electron configuration of [Xe]5d1, which is not a filled outermost shell.
Conclusion
In conclusion, Lanthanum achieves noble gas configuration in the +3 oxidation state by losing three electrons and forming a La3+ ion.
Achieving Noble Gas Configuration
To achieve a noble gas configuration, an element must have its outermost shell filled with electrons. The noble gases are stable because they have a filled outermost shell. Lanthanum can achieve this configuration by losing three electrons from its outermost shell.
Oxidation States
The oxidation state of an element is the charge that it carries when it forms a compound or ion. Lanthanum can form compounds in various oxidation states, including +2, +3, and +4.
In order to determine in which oxidation state La achieves noble gas configuration, we need to look at its electron configuration and determine how many electrons it needs to lose to achieve a filled outermost shell.
La3+ (Oxidation State of +3)
When Lanthanum loses three electrons, it forms a +3 ion (La3+). In this oxidation state, La has a noble gas configuration of [Xe]. Therefore, option B is the correct answer.
Other Oxidation States
Lanthanum can also form compounds in other oxidation states, but they do not achieve a noble gas configuration. For example, in the +2 oxidation state, La has an electron configuration of [Xe]5d1, which is not a filled outermost shell.
Conclusion
In conclusion, Lanthanum achieves noble gas configuration in the +3 oxidation state by losing three electrons and forming a La3+ ion.
Which of the following group of elements are not regarded as transition elements?- a)Sc,Y,La
- b)Cu, Ag, Au
- c)Zn, Cd, Hg
- d)Cr ,Mo,W
Correct answer is option 'C'. Can you explain this answer?
Which of the following group of elements are not regarded as transition elements?
a)
Sc,Y,La
b)
Cu, Ag, Au
c)
Zn, Cd, Hg
d)
Cr ,Mo,W
|
Neha Sharma answered |
Zn, Cd & Hg because the d-orbital of these elements are completely filled. So, they don't show the characteristics of transition elements (i.e. the d-orbital of transition elements is incomplete.
The inner transition elements are the elements in which the added electrons go to:- a)(n-1)d-orbitals
- b)(n-1)d-orbitals and (n-1)f-orbitals
- c)(n-1)d-orbitals and ns orbitals
- d)(n-2)f-orbitals
Correct answer is option 'D'. Can you explain this answer?
The inner transition elements are the elements in which the added electrons go to:
a)
(n-1)d-orbitals
b)
(n-1)d-orbitals and (n-1)f-orbitals
c)
(n-1)d-orbitals and ns orbitals
d)
(n-2)f-orbitals
|
Rohit Shah answered |
Lanthanides and actinides are called inner transition elements because they are a group of elements that are shown as the bottom two rows of the periodic table. ... Lanthanides and actinides belong to the f-block elements, which means that they have filled up their f-orbitals with electrons.
Which of the following group contain mostly radioactive elements ?- a)Lanthanoids
- b)3 d transition series
- c)4 d transition series
- d)Actinoids
Correct answer is option 'D'. Can you explain this answer?
Which of the following group contain mostly radioactive elements ?
a)
Lanthanoids
b)
3 d transition series
c)
4 d transition series
d)
Actinoids
|
Parth Sharma answered |
The actinoids are said to be most radioactive metals because in this group all the metals have very oxidation number and are very rare metals found on earth and it has all the elements radioactive in its group and it has most radioactive metal uranium in it's group.
Mischmetal is an alloy of:
- a)Lanthanoid metal and Iron
- b)Actinoid metal and Iron
- c)Manganese, nickel and iron
- d)Zinc ,cobalt and iron
Correct answer is option 'A'. Can you explain this answer?
Mischmetal is an alloy of:
a)
Lanthanoid metal and Iron
b)
Actinoid metal and Iron
c)
Manganese, nickel and iron
d)
Zinc ,cobalt and iron
|
Snehal Iyer answered |
Explanation:
Mischmetal is an alloy of Lanthanoid Metal and Iron. It is a rare earth alloy, which is primarily used in the production of alloy steels, stainless steels, and other high-performance alloys.
Composition:
Mischmetal typically contains 50-55% cerium, 20-25% lanthanum, and small amounts of other rare earth metals such as neodymium, praseodymium, and gadolinium. The remaining portion of the alloy is typically iron.
Properties:
Mischmetal has a number of unique properties that make it an attractive alloy for use in a variety of applications. These properties include:
- High heat resistance
- High strength and durability
- Corrosion resistance
- Good machinability
Applications:
Mischmetal is primarily used in the production of alloy steels, stainless steels, and other high-performance alloys. It is also used in the production of magnesium alloys, which are used in a variety of applications, including aerospace, automotive, and sporting goods.
Some other applications of Mischmetal are:
- In the production of lighter flints
- As a pyrophoric material for the ignition of torches and lighters
- In the production of certain types of electrodes for welding and cutting
- In the production of certain types of batteries
Conclusion:
In conclusion, Mischmetal is an alloy of lanthanoid metal and iron, primarily used in the production of alloy steels, stainless steels, and other high-performance alloys. It has a number of unique properties that make it an attractive alloy for use in a variety of applications.
Mischmetal is an alloy of Lanthanoid Metal and Iron. It is a rare earth alloy, which is primarily used in the production of alloy steels, stainless steels, and other high-performance alloys.
Composition:
Mischmetal typically contains 50-55% cerium, 20-25% lanthanum, and small amounts of other rare earth metals such as neodymium, praseodymium, and gadolinium. The remaining portion of the alloy is typically iron.
Properties:
Mischmetal has a number of unique properties that make it an attractive alloy for use in a variety of applications. These properties include:
- High heat resistance
- High strength and durability
- Corrosion resistance
- Good machinability
Applications:
Mischmetal is primarily used in the production of alloy steels, stainless steels, and other high-performance alloys. It is also used in the production of magnesium alloys, which are used in a variety of applications, including aerospace, automotive, and sporting goods.
Some other applications of Mischmetal are:
- In the production of lighter flints
- As a pyrophoric material for the ignition of torches and lighters
- In the production of certain types of electrodes for welding and cutting
- In the production of certain types of batteries
Conclusion:
In conclusion, Mischmetal is an alloy of lanthanoid metal and iron, primarily used in the production of alloy steels, stainless steels, and other high-performance alloys. It has a number of unique properties that make it an attractive alloy for use in a variety of applications.
Melting point of d block elements across a period:- a)Increases from left to right
- b)Deceases from left to right
- c)Increases to a maximum at d5 and then decreases with increase of atomic number.
- d)Does not change on moving from left to right.
Correct answer is option 'C'. Can you explain this answer?
Melting point of d block elements across a period:
a)
Increases from left to right
b)
Deceases from left to right
c)
Increases to a maximum at d5 and then decreases with increase of atomic number.
d)
Does not change on moving from left to right.
|
Jayant Mishra answered |
The melting and boiling points first increase, reaches maximum and then steadily decrease across any transition series. ... The low melting points of Zn, Cd, and Hg are due to the absence of unpaired d-electrons in their atoms and thus low metallic bonding.
Number of electrons with l = 2 and s = 1/2 in zinc atom are
Correct answer is '5'. Can you explain this answer?
Number of electrons with l = 2 and s = 1/2 in zinc atom are
|
|
Krishna Iyer answered |
l = 2 means d-subshell. Zinc has 10 electrons ind-subshell in which 5 electrons are + 1/2 spin and 5 electrons are - 1/2.
Observe the following statements. Which is correct statement regarding lanthanides?- a)Cerium shows + 4 and + 3 oxidation states
- b)Element with atomic numbers 66 and 99 belong to f-block
- c)Lanthanides belong to 3rd group and 7th period
- d)Monazite is the mineral that constitutes the major source of the lanthanides
Correct answer is option 'B'. Can you explain this answer?
Observe the following statements. Which is correct statement regarding lanthanides?
a)
Cerium shows + 4 and + 3 oxidation states
b)
Element with atomic numbers 66 and 99 belong to f-block
c)
Lanthanides belong to 3rd group and 7th period
d)
Monazite is the mineral that constitutes the major source of the lanthanides
|
Suresh Iyer answered |
The correct answer is option B
Ce shows +4 oxidation state in aq.solution
Ce (58) to lu (71)- lanthanide
Th (90) to Lr (103)- actinoids.
Ce shows +4 oxidation state in aq.solution
Ce (58) to lu (71)- lanthanide
Th (90) to Lr (103)- actinoids.
One Integer Value Correct TypeDirection (Q. Nos. 20-24) This section contains 5 questions. When worked out will result in an integer from 0 to 9 (both inclusive).Q. The number of elements that do not have completely filled d subshell among the elements Ni, Pd, Pt, Cu, Ag, Au, Zn, Cd and Hg are
Correct answer is '2'. Can you explain this answer?
One Integer Value Correct Type
Direction (Q. Nos. 20-24) This section contains 5 questions. When worked out will result in an integer from 0 to 9 (both inclusive).
Q.
The number of elements that do not have completely filled d subshell among the elements Ni, Pd, Pt, Cu, Ag, Au, Zn, Cd and Hg are
|
Preeti Iyer answered |
Ni has 3d84s2 and Pt has 5d96s1.
Only One Option Correct TypeDirection (Q. Nos. 1- 10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.Q. Set of continuous atomic numbers of elements are present in the same group as well as same period.- a)56,57,58,59
- b)88,89,90,91
- c)101,102,103,104
- d)68,69,70,71
Correct answer is option 'D'. Can you explain this answer?
Only One Option Correct Type
Direction (Q. Nos. 1- 10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
Set of continuous atomic numbers of elements are present in the same group as well as same period.
a)
56,57,58,59
b)
88,89,90,91
c)
101,102,103,104
d)
68,69,70,71
|
|
Krishna Iyer answered |
Er (68), Tm (69), Yb (70) and Lu (71) elements are present in same group as well as same period.
Which of the following is not an interstitial compound?- a)Mn4N
- b)TiC
- c)Fe3H
- d)FeCl3
Correct answer is option 'D'. Can you explain this answer?
Which of the following is not an interstitial compound?
a)
Mn4N
b)
TiC
c)
Fe3H
d)
FeCl3
|
|
Lavanya Menon answered |
The correct answer is option D
An interstitial compound is a compound that is formed when an atom with a small enough radius sits a hole in a metal lattice
They are usually transitional elements
Eg: Tic, Mn4N, Fe3H, TiH2 etc.
Main Characteristics:-
· High melting points than metals
· They are much harder
· They become chemically inert
· They become less malleable
So, the answer is (a), (b), (c) .
An interstitial compound is a compound that is formed when an atom with a small enough radius sits a hole in a metal lattice
They are usually transitional elements
Eg: Tic, Mn4N, Fe3H, TiH2 etc.
Main Characteristics:-
· High melting points than metals
· They are much harder
· They become chemically inert
· They become less malleable
So, the answer is (a), (b), (c) .
Identify the incorrect statement among the following -[AIEEE-07]- a)d-block elements show irregular and erratic chemical properties among themselves
- b)La and Lu have partially filled d-orbitals and no other partially filled orbitals
- c)The chemistry of various lanthanoids is very similar
- d)4f and 5f-orbitals are equally shielded
Correct answer is option 'D'. Can you explain this answer?
Identify the incorrect statement among the following -
[AIEEE-07]
a)
d-block elements show irregular and erratic chemical properties among themselves
b)
La and Lu have partially filled d-orbitals and no other partially filled orbitals
c)
The chemistry of various lanthanoids is very similar
d)
4f and 5f-orbitals are equally shielded
|
|
Nikita Singh answered |
The correct answer is option D
4f and 5f belongs to different energy levels, hence the shielding effect on them is not the same. Shielding of 4f is more than 5f.
4f and 5f belongs to different energy levels, hence the shielding effect on them is not the same. Shielding of 4f is more than 5f.
The electronic configuration of Eu 3+ ion is:- a)[Xe] 4f4
- b)[Xe] 4f7
- c)[Xe] 4f2
- d)[Xe] 4f6
Correct answer is option 'D'. Can you explain this answer?
The electronic configuration of Eu 3+ ion is:
a)
[Xe] 4f4
b)
[Xe] 4f7
c)
[Xe] 4f2
d)
[Xe] 4f6
|
Aditi Azade answered |
Atomic number of Eu is 63...so electronic configuration of Eu is, Eu=[xe] 4f^9 Hence,Eu^+3=[xe]4f^6
N2(g) + 3H2 (g) 
2NH3(g) ; Haber's process, Mo is used as -- a)A catalyst
- b)A catalytic promoter
- c)An oxidising agent
- d)As a catalytic poison
Correct answer is option 'B'. Can you explain this answer?
N2(g) + 3H2 (g) 2NH3(g) ; Haber's process, Mo is used as -
2NH3(g) ; Haber's process, Mo is used as -a)
A catalyst
b)
A catalytic promoter
c)
An oxidising agent
d)
As a catalytic poison
|
Prashanth Banerjee answered |
MO is catalytic promoter & it improves catalytic properties as catalyst
What is the most common oxidation state for actinoids?- a)+3
- b)+7
- c)+2
- d)+5
Correct answer is option 'A'. Can you explain this answer?
What is the most common oxidation state for actinoids?
a)
+3
b)
+7
c)
+2
d)
+5
|
Mamali . answered |
Unlike lanthanides which show the +3, oxidation States,actinides show a variety of Oxidation State from +3to+6.However+3&+4 are the principal Oxidation State.The+3 Oxidation state is the most stable in AC and all the other elements of the series.Thats it.
In context with the transition elements, which of the following statements is incorrect?[AIEEE-09]- a)In the highest oxidation states, the transition metal show basic character and form cationic complexes.
- b)In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and electrons are used for bonding.
- c)Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in
bonding decreases. - d) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.
Correct answer is option 'A'. Can you explain this answer?
In context with the transition elements, which of the following statements is incorrect?
[AIEEE-09]
a)
In the highest oxidation states, the transition metal show basic character and form cationic complexes.
b)
In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and electrons are used for bonding.
c)
Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in
bonding decreases.
bonding decreases.
d)
In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.
|
Keerthana Mehta answered |
As oxidation state increase, electronegativity increase thus acidic characteristic increases not basic.
In neutral solution, potassium permanganate forms:- a)Manganese dioxide
- b)Manganous sulphate
- c)Pyrolusite
- d)Potassium manganite
Correct answer is option 'A'. Can you explain this answer?
In neutral solution, potassium permanganate forms:
a)
Manganese dioxide
b)
Manganous sulphate
c)
Pyrolusite
d)
Potassium manganite
|
Prakash answered |
KMno₄ is an oxidising agent (here Mn⁺⁷)in nuetral medium there is a change of 3 in it's oxidation state M⁺⁷-------Mn⁺⁴
The actinoids exhibits more number of oxidation states in general than the lanthanoids. This is because [AIEEE-07]- a)The 5f-orbitals are more buried than the 4f-orbitals
- b)There is a similarly between 4f-and-5f in the their angular part of the wave function
- c)The actinoids are more reactive than the lanthanoids
- d)The 5f-orbitals extend further from the nucleus than the 4f-orbitals
Correct answer is option 'D'. Can you explain this answer?
The actinoids exhibits more number of oxidation states in general than the lanthanoids. This is because
[AIEEE-07]
a)
The 5f-orbitals are more buried than the 4f-orbitals
b)
There is a similarly between 4f-and-5f in the their angular part of the wave function
c)
The actinoids are more reactive than the lanthanoids
d)
The 5f-orbitals extend further from the nucleus than the 4f-orbitals
|
|
T.ttttt answered |
Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states.
Transition metals with highest melting point is- a)Cr
- b)W
- c)Hg
- d)Sc
Correct answer is option 'B'. Can you explain this answer?
Transition metals with highest melting point is
a)
Cr
b)
W
c)
Hg
d)
Sc
|
Pragati Choudhury answered |
W belongs to 5d series and also it have lot of unpaired electrons thus it forms strong metallic bonding.
The elements which lie between s and p block elements in the long form periodic table are called as:- a)Actinides
- b)d-block elements
- c)Lanthanides
- d)Electropositive elements
Correct answer is option 'B'. Can you explain this answer?
The elements which lie between s and p block elements in the long form periodic table are called as:
a)
Actinides
b)
d-block elements
c)
Lanthanides
d)
Electropositive elements
|
|
Harsh Desai answered |
D-block elements
The long form periodic table consists of four blocks: s, p, d, and f. The s and p blocks are located on the left and right sides of the periodic table, respectively. The d-block elements are located in the middle of the periodic table, between the s and p blocks. These elements are also known as transition elements.
Explanation:
The d-block elements are characterized by the presence of partially filled d-orbitals in their valence shells. These elements are often referred to as transition elements because they exhibit a transition between the highly reactive s-block elements and the relatively inert p-block elements. The d-block elements are known for their unique chemical and physical properties, such as their ability to form complex ions and their high melting and boiling points.
Examples of d-block elements include titanium, iron, copper, and zinc. These elements are widely used in industry and technology due to their unique properties, such as their strength, durability, and conductivity.
In summary, the elements which lie between s and p block elements in the long form periodic table are called d-block elements or transition elements. These elements exhibit unique chemical and physical properties and are widely used in industry and technology.
The long form periodic table consists of four blocks: s, p, d, and f. The s and p blocks are located on the left and right sides of the periodic table, respectively. The d-block elements are located in the middle of the periodic table, between the s and p blocks. These elements are also known as transition elements.
Explanation:
The d-block elements are characterized by the presence of partially filled d-orbitals in their valence shells. These elements are often referred to as transition elements because they exhibit a transition between the highly reactive s-block elements and the relatively inert p-block elements. The d-block elements are known for their unique chemical and physical properties, such as their ability to form complex ions and their high melting and boiling points.
Examples of d-block elements include titanium, iron, copper, and zinc. These elements are widely used in industry and technology due to their unique properties, such as their strength, durability, and conductivity.
In summary, the elements which lie between s and p block elements in the long form periodic table are called d-block elements or transition elements. These elements exhibit unique chemical and physical properties and are widely used in industry and technology.
Brass is an alloy of:- a)Copper and molybdenum
- b)Copper and zinc
- c)Copper and tungsten
- d)Copper and tin
Correct answer is option 'B'. Can you explain this answer?
Brass is an alloy of:
a)
Copper and molybdenum
b)
Copper and zinc
c)
Copper and tungsten
d)
Copper and tin
|
|
Krishna Iyer answered |
Brass is an alloy of copper and zinc, in proportions which can be varied to achieve varying mechanical and electrical properties.It is a substitutional alloy: atoms of the two constituents may replace each other within the same crystal structure.
Brass is similar to bronze, another alloy containing copper that uses tin in place of zinc; both bronze and brass may include small proportions of a range of other elements including arsenic, lead, phosphorus, aluminum, manganese, and silicon.
Brass is similar to bronze, another alloy containing copper that uses tin in place of zinc; both bronze and brass may include small proportions of a range of other elements including arsenic, lead, phosphorus, aluminum, manganese, and silicon.
Which one is of the following is the lightest transition element?- a)Ti
- b)Sc
- c)Fe
- d)Hg
Correct answer is option 'B'. Can you explain this answer?
Which one is of the following is the lightest transition element?
a)
Ti
b)
Sc
c)
Fe
d)
Hg
|
|
T.ttttt answered |
Scandium (Sc) is the lightest transition element. It is the first element in the 3d transition series. Among the transition elements, Sc has the lowest density.
A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is - a)(NH4)2Cr2O7
- b)HgI2
- c)HgO
- d)Pb3O4
Correct answer is option 'B'. Can you explain this answer?
A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is
a)
(NH4)2Cr2O7
b)
HgI2
c)
HgO
d)
Pb3O4
|
Infinity Academy answered |
Mohr’s salt is a- a)Acidic salt
- b)Double salt
- c)Basic Acidic salt
- d)Normal salt
Correct answer is option 'B'. Can you explain this answer?
Mohr’s salt is a
a)
Acidic salt
b)
Double salt
c)
Basic Acidic salt
d)
Normal salt
|
Amar Pillai answered |
Mohr salt is an example of double salt. Mohr Salt is
FeSO4 . (NH4)2 SO4.6H2O
Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statement about cerium is incorrect -[AIEEE-04]- a)Cerium (IV) acts as an oxidising agent
- b)The +3 oxidation state of cerium is more stable than the +4 oxidation state
- c)The +4 oxidation state of cerium is not known in solutions
- d) The common oxidation states of cerium are +3 and +4
Correct answer is option 'C'. Can you explain this answer?
Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statement about cerium is incorrect -
[AIEEE-04]
a)
Cerium (IV) acts as an oxidising agent
b)
The +3 oxidation state of cerium is more stable than the +4 oxidation state
c)
The +4 oxidation state of cerium is not known in solutions
d)
The common oxidation states of cerium are +3 and +4
|
|
Ritu Singh answered |
The incorrect statement about the Cerium is only a C option.
C: The +4 oxidation state of cerium is not known in solutions [ Incorrect ]
As +4 oxidation state is common for the Cerium.
The correct statement about the Cerium are as follows:
A: The common oxidation states of Cerium are +3, and +4.
B : The +3 oxidation state of the Cerium is more stable than the +4 oxidation state.
D: Cerium (IV) acts as an oxidising agent
C: The +4 oxidation state of cerium is not known in solutions [ Incorrect ]
As +4 oxidation state is common for the Cerium.
The correct statement about the Cerium are as follows:
A: The common oxidation states of Cerium are +3, and +4.
B : The +3 oxidation state of the Cerium is more stable than the +4 oxidation state.
D: Cerium (IV) acts as an oxidising agent
The observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.
Q. Which is correct statement?
- a)Ce4+ is more stable than Ce3+ in solution
- b)Ce3+ is more sable than Ce4+ in solution
- c)Ce4+ and Ce3+ are equally stable
- d)Ce4+ has xenon configuration
Correct answer is option 'B'. Can you explain this answer?
The observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.
Q. Which is correct statement?
a)
Ce4+ is more stable than Ce3+ in solution
b)
Ce3+ is more sable than Ce4+ in solution
c)
Ce4+ and Ce3+ are equally stable
d)
Ce4+ has xenon configuration
|
|
Nabanita Sen answered |
Ce4+ is [Xe] 4f0, it tends to revert to more stable oxidation state of +3 by gain of an electron. That is why, in aqueous solution, Ce4+ is good oxidising agent.
If molarity of K2Cr2O7 solution is 0.5 M, its normality value is
Correct answer is '3'. Can you explain this answer?
If molarity of K2Cr2O7 solution is 0.5 M, its normality value is
|
Ishani Patel answered |
Normality = molarity x number of electrons gained per mole of dichromate in a redox reaction. Hence, Normality = Molarity x 6
Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect ?[AIEEE-09]- a)The ionic sizes of Ln (III) decrease in general with increasing atomic number.
- b) Ln (III) compounds are generally colourless.
- c) Ln (III) hydroxides are mainly basic in character.
- d)Because of the large size of the Ln (III)ions the bonding in its compounds is predominantly ionic in character.
Correct answer is option 'B'. Can you explain this answer?
Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect ?
[AIEEE-09]
a)
The ionic sizes of Ln (III) decrease in general with increasing atomic number.
b)
Ln (III) compounds are generally colourless.
c)
Ln (III) hydroxides are mainly basic in character.
d)
Because of the large size of the Ln (III)ions the bonding in its compounds is predominantly ionic in character.
|
|
Riya Banerjee answered |
The correct answer is option B
Ln(III) compounds are coloured due to the f−f transitions in the lanthanide but the colour is fainter as compared to the d−d transitions in transition metals.
Ln(III) compounds are coloured due to the f−f transitions in the lanthanide but the colour is fainter as compared to the d−d transitions in transition metals.
Which one of the following is a diamagnetic ion?- a)CO2+
- b)Cu2+
- c)Mn2+
- d)Sc3+
Correct answer is option 'D'. Can you explain this answer?
Which one of the following is a diamagnetic ion?
a)
CO2+
b)
Cu2+
c)
Mn2+
d)
Sc3+
|
|
Ameya Pillai answered |
Co2+ (Z = 27) : [Ar]183d7 (3 unpaired electrons)
Cu2+ (Z = 29) : [Ar]183d9 (1 unpaired electrons)
Mn2+ (Z = 25): [Ar]183d5 (5 unpaired electrons)
Sc3+ (Z = 21): [Ar]183d0 (No unpaired electron)
Sc3+ with no unpaired electron will be diamagnetic
Cu2+ (Z = 29) : [Ar]183d9 (1 unpaired electrons)
Mn2+ (Z = 25): [Ar]183d5 (5 unpaired electrons)
Sc3+ (Z = 21): [Ar]183d0 (No unpaired electron)
Sc3+ with no unpaired electron will be diamagnetic
(T) imparts violet colour 
(V) Red gas 
(W) Red ppt.
(W) Red ppt. 
white ppt.(U) 
gas (gives white fumes with HCl)sublimes on heating Identify (T) to (Z).- a)T = KMnO4, U = HCl, V = Cl2, W = HgI2, X =Hg(NH2)NO3, Y = Hg2Cl2, Z = N2
- b)T = K2Cr2O7, U = NH4Cl, V = CrO2Cl2, W = Ag2CrO4, X = [Ag(NH3)2]+, Y = AgCl,Z=NH3
- c)T = K2CrO4, U = KCl, V = CrO2Cl2, W = HgI2, X = Na2CrO4, Y = BaCO3, Z = NH4Cl
- d)T = K2MnO4, U = NaCl, V = CrO3, W = AgNO2, X = (NH4)2CrO4, Y = CaCO3, Z = SO2
Correct answer is option 'B'. Can you explain this answer?
(T) imparts violet colour (V) Red gas (W) Red ppt.
(V) Red gas (W) Red ppt.(W) Red ppt. white ppt.
white ppt.(U) gas (gives white fumes with HCl)
gas (gives white fumes with HCl)sublimes on heating Identify (T) to (Z).
a)
T = KMnO4, U = HCl, V = Cl2, W = HgI2, X =Hg(NH2)NO3, Y = Hg2Cl2, Z = N2
b)
T = K2Cr2O7, U = NH4Cl, V = CrO2Cl2, W = Ag2CrO4, X = [Ag(NH3)2]+, Y = AgCl,Z=NH3
c)
T = K2CrO4, U = KCl, V = CrO2Cl2, W = HgI2, X = Na2CrO4, Y = BaCO3, Z = NH4Cl
d)
T = K2MnO4, U = NaCl, V = CrO3, W = AgNO2, X = (NH4)2CrO4, Y = CaCO3, Z = SO2
|
Sushil Kumar answered |
The correct answer is option B
Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect for this reaction -[AIEEE-04]- a)Evolved I2 is reduced
- b)CuI2 is formed
- c)Na2S2O3 is oxidised
- d)Cu2I2 is formed
Correct answer is option 'B'. Can you explain this answer?
Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect for this reaction -
[AIEEE-04]
a)
Evolved I2 is reduced
b)
CuI2 is formed
c)
Na2S2O3 is oxidised
d)
Cu2I2 is formed
|
|
Swati Verma answered |
The correct answer is option B
It is given that an excess of KI reacts with CuSO4 solution and then Na2S2O3 is added to it. This is hypo test and reactions are as follows:
2CuSO4 + 4KI → 2K2SO4 + 2CuI+I2
2Na2S2O3 + I2→ Na2S4O6 + 2NaI
CuI2 is not formed according to the following reaction.
2CuI2(unstable) → 2CuI+I2
Hence, the statement in option B is incorrect.
It is given that an excess of KI reacts with CuSO4 solution and then Na2S2O3 is added to it. This is hypo test and reactions are as follows:
2CuSO4 + 4KI → 2K2SO4 + 2CuI+I2
2Na2S2O3 + I2→ Na2S4O6 + 2NaI
CuI2 is not formed according to the following reaction.
2CuI2(unstable) → 2CuI+I2
Hence, the statement in option B is incorrect.
Ag+ ion is isoelectronic with- a)Zn2+
- b)Cd2+
- c)Pd2+
- d)Cu2+
Correct answer is option 'B'. Can you explain this answer?
Ag+ ion is isoelectronic with
a)
Zn2+
b)
Cd2+
c)
Pd2+
d)
Cu2+
|
Malavika Shah answered |
Ag+ is isoelectronic with Cd2+
The colourless ions among the lanthanides are- a)La3+
- b)Gd3+
- c)Lu3+
- d)Ce3+
Correct answer is option 'A,B,C,D'. Can you explain this answer?
The colourless ions among the lanthanides are
a)
La3+
b)
Gd3+
c)
Lu3+
d)
Ce3+
|
|
Keerthana Mehta answered |
La3+, Gd3+ and Lu3+ have empty (4f0) half-filled (4f7) and full-filled (4f14) orbitals. Ce3+ (4f1) is colourless because it does not absorb in the visible region.
During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’ These are- a)Sn and Ag
- b)Pb and Zn
- c)Ag and Au
- d)Fe and Ni
Correct answer is option 'C'. Can you explain this answer?
During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’ These are
a)
Sn and Ag
b)
Pb and Zn
c)
Ag and Au
d)
Fe and Ni
|
Hansa Sharma answered |
The correct answer is option C
In Electrolytic refining, the impure metal is made to act as an anode. A strip of the same metal in pure form is used as a cathode.
They are put in a suitable electrolytic bath containing a soluble salt of the same metal.
The more basic metal remains in the solution and the less basic ones go to the anode mud.
Copper is refined using an electrolytic method.
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold, and platinum; recovery of these elements may meet the cost of refining.
In Electrolytic refining, the impure metal is made to act as an anode. A strip of the same metal in pure form is used as a cathode.
They are put in a suitable electrolytic bath containing a soluble salt of the same metal.
The more basic metal remains in the solution and the less basic ones go to the anode mud.
Copper is refined using an electrolytic method.
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold, and platinum; recovery of these elements may meet the cost of refining.
In which of the following oxidation state Cerium achieves the noble gas configuration?- a)+5
- b)+2
- c)+7
- d)+4
Correct answer is option 'D'. Can you explain this answer?
In which of the following oxidation state Cerium achieves the noble gas configuration?
a)
+5
b)
+2
c)
+7
d)
+4
|
|
Mamali . answered |
The reason for this behaviour is a result of the stability of half filled, empty or fulfilled F orbitals that these elements achieve in these Oxidation State.Thats it.
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