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Kiara KapoorEduRev Mathematics |
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Kiara Kapoor
EduRev Mathematics
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Top Scoring Tests by Kiara
Test: VSEPR Theory & shapes of molecules | 20/20 |
Test: Molecular Orbital Theory | 20/20 |
Test: Nucleophilic Substitution Reaction | 20/20 |
Test: Periodic Properties- 1 | 44/46 |
Test: Hybridisation & Shapes of Molecules | 18/20 |
Test: Conductance - 1 | 36/40 |
Test: Grignard Reagent | 17/20 |
Test: Chemical Bonding- 2 | 51/60 |
Test: General Characteristics of the Compounds of the Alkali Metals | 25/30 |
Test: Bragg’s Law | 16/20 |
Discussed Questions
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Biswa Nath upvoted • Jun 25, 2023 |
The complex [Fe(phen)2(NCS)2] (phen = 1,10–phenanthroline) shows spin cross–over behavior.CFSE and µeff at 250 and 150 K, respectively are:
- a)0.4 ∆o, 4.90 BM and 2.4 ∆o, 0.00 BM
- b)2.4 ∆o, 2.90 BM and 0.4 ∆o, 1.77 BM
- c)2.4 ∆o, 0.00 BM and 0.4 ∆o, 4.90 BM
- d)1.2 ∆o, 4.90 BM and 2.4 ∆o, 0.00 BM
Correct answer is option 'A'. Can you explain this answer?
The complex [Fe(phen)2(NCS)2] (phen = 1,10–phenanthroline) shows spin cross–over behavior.CFSE and µeff at 250 and 150 K, respectively are:
a)
0.4 ∆o, 4.90 BM and 2.4 ∆o, 0.00 BM
b)
2.4 ∆o, 2.90 BM and 0.4 ∆o, 1.77 BM
c)
2.4 ∆o, 0.00 BM and 0.4 ∆o, 4.90 BM
d)
1.2 ∆o, 4.90 BM and 2.4 ∆o, 0.00 BM
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Yash Roy answered |
Spin Crossover Behavior in [Fe(phen)2(NCS)2]
CFSE and eff at 250 and 150 K, respectively are given for the complex [Fe(phen)2(NCS)2]. Let's understand the spin crossover behavior of this complex and how it affects the CFSE and effective magnetic moment (eff) values.
Spin Crossover Behavior
Spin crossover is a phenomenon where a coordination compound undergoes a rev... more
CFSE and eff at 250 and 150 K, respectively are given for the complex [Fe(phen)2(NCS)2]. Let's understand the spin crossover behavior of this complex and how it affects the CFSE and effective magnetic moment (eff) values.
Spin Crossover Behavior
Spin crossover is a phenomenon where a coordination compound undergoes a rev... more
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Biswa Nath upvoted • Jun 25, 2023 |
The complex [Fe(phen)2(NCS)2] (phen = 1,10–phenanthroline) shows spin cross–over behavior.CFSE and µeff at 250 and 150 K, respectively are:
- a)0.4 ∆o, 4.90 BM and 2.4 ∆o, 0.00 BM
- b)2.4 ∆o, 2.90 BM and 0.4 ∆o, 1.77 BM
- c)2.4 ∆o, 0.00 BM and 0.4 ∆o, 4.90 BM
- d)1.2 ∆o, 4.90 BM and 2.4 ∆o, 0.00 BM
Correct answer is option 'A'. Can you explain this answer?
The complex [Fe(phen)2(NCS)2] (phen = 1,10–phenanthroline) shows spin cross–over behavior.CFSE and µeff at 250 and 150 K, respectively are:
a)
0.4 ∆o, 4.90 BM and 2.4 ∆o, 0.00 BM
b)
2.4 ∆o, 2.90 BM and 0.4 ∆o, 1.77 BM
c)
2.4 ∆o, 0.00 BM and 0.4 ∆o, 4.90 BM
d)
1.2 ∆o, 4.90 BM and 2.4 ∆o, 0.00 BM
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Himanshu Trivedi answered |
At 250K temp it will behave as high spin complex, at 150K it will behave like as low spin complex so option A would be right. Good Luck
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Biswa Nath upvoted • Jun 23, 2023 |
The infra–red starching frequency vCO of P-S follows the order:
- a)P > R > S > Q
- b)S > P > R > Q
- c)Q > S > P > R
- d)R > Q > P > S
Correct answer is option 'D'. Can you explain this answer?
The infra–red starching frequency vCO of P-S follows the order:
a)
P > R > S > Q
b)
S > P > R > Q
c)
Q > S > P > R
d)
R > Q > P > S
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Suman answered |
If we look at the M.O of CO we find HOMO is the non bonding orbital which has some antibonding character also so when CO adduct with BH3 it acts only as sigma bond donner .For this purpose CO give electron from non bonding orbital which has some antibonding character also so bond strangth increases for BH3 adduct ..and adduct with Mn(+) and V(-) it works as sigma donner aswell as pi accepter and a... more
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Biswa Nath upvoted • Feb 04, 2023 |
Which of the following is most basic?
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Which of the following is most basic?
a)
b)
c)
d)
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Subha Som answered |
How many signals are obtained for the following compound in 1H NMR spectroscopy?
Correct answer is '5'. Can you explain this answer?
How many signals are obtained for the following compound in 1H NMR spectroscopy?
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Biswa Nath answered • Feb 04, 2023 |
4 types of protons are present. so it will give 4 signal
The change in the hybrid state of BeCl2 in the solid state from vapour state is:- a)sp2 to sp3
- b)sp to sp3
- c)sp to sp2
- d)No change
Correct answer is option 'B'. Can you explain this answer?
The change in the hybrid state of BeCl2 in the solid state from vapour state is:
a)
sp2 to sp3
b)
sp to sp3
c)
sp to sp2
d)
No change
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Biswa Nath answered • Feb 04, 2023 |
When it is in vapour state the coordination number is 2 i.e sp hybridization.
In solid phase coordination number is 4 i.e sp³
In solid phase coordination number is 4 i.e sp³
Alkaline earth metals show +1 Oxidation state and their atomic volume ______ down the group.
- a)is irregular
- b)decrease
- c)increase
- d)do not change
Correct answer is option 'C'. Can you explain this answer?
Alkaline earth metals show +1 Oxidation state and their atomic volume ______ down the group.
a)
is irregular
b)
decrease
c)
increase
d)
do not change
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Biswa Nath answered • Feb 04, 2023 |
B should be the correct answer.
How can we prepare RR’ R” OH by the action of excess of a suitable Grignard reagent on which of the following reactants?
- a)ketone or an ester
- b)ester or an alcohol
- c)aldehyde or a ketone
- d)nitrile or an aldehyde
Correct answer is option 'A'. Can you explain this answer?
How can we prepare RR’ R” OH by the action of excess of a suitable Grignard reagent on which of the following reactants?
a)
ketone or an ester
b)
ester or an alcohol
c)
aldehyde or a ketone
d)
nitrile or an aldehyde
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Biswa Nath answered • Jan 25, 2023 |
The answer will be D not A
A sigma bond may be formed by the overlap of 2 atomic orbitals of atoms A and B. If the bond is formed along the x-axis, which of the following overlaps is acceptable ?- a)s orbital of A and pz orbital of B
- b)px orbital of A and py orbital of B
- c)pz orbital of A and px orbital of B
- d)px orbital of A and s orbital of B
Correct answer is option 'D'. Can you explain this answer?
A sigma bond may be formed by the overlap of 2 atomic orbitals of atoms A and B. If the bond is formed along the x-axis, which of the following overlaps is acceptable ?
a)
s orbital of A and pz orbital of B
b)
px orbital of A and py orbital of B
c)
pz orbital of A and px orbital of B
d)
px orbital of A and s orbital of B
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Biswa Nath answered • Jan 21, 2023 |
Sigma bond is formed because of the Head to Head overlap.
Now Sigma bond is formed along X axis...
So only if Px orbital and S orbital overlapped to each other then a bond will formed
Now Sigma bond is formed along X axis...
So only if Px orbital and S orbital overlapped to each other then a bond will formed
In the formation of N2+ from N2, the electron is removed from- a)σ orbital
- b)π orbital
- c)σ* orbital
- d)π* orbital
Correct answer is option 'A'. Can you explain this answer?
In the formation of N2+ from N2, the electron is removed from
a)
σ orbital
b)
π orbital
c)
σ* orbital
d)
π* orbital
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Biswa Nath answered • Jan 21, 2023 |
Because Sigma is the HOMO for N2...
So electron will removed from Sigma
So electron will removed from Sigma
In the cyanide ion the negative charge is on
- a)N
- b)C
- c)Both C and N
- d)Resonate between C and N
Correct answer is option 'A'. Can you explain this answer?
In the cyanide ion the negative charge is on
a)
N
b)
C
c)
Both C and N
d)
Resonate between C and N
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Biswa Nath answered • Jan 21, 2023 |
For cyanide -ve charge is on Carbon
and for Isocyanide on Nitrogen... there have resonance.
and for Isocyanide on Nitrogen... there have resonance.
In which of the following molecule are all the bonds not equal ?- a)NF3
- b)CIF3
- c)BF3
- d)AlF3
Correct answer is option 'B'. Can you explain this answer?
In which of the following molecule are all the bonds not equal ?
a)
NF3
b)
CIF3
c)
BF3
d)
AlF3
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Biswa Nath answered • Jan 21, 2023 |
In case of ClF3, there have two axial bond and one equatorial bond.
Amongst LiCl , RDCl , BeCl2 and MgCl2 the compounds with the greatest and the least ionic character, respectively are:- a)LiCl and RbCl
- b)RbCl and BeCl2
- c)RbCl and MgCl2
- d)MgCl2 and BeCl2
Correct answer is option 'B'. Can you explain this answer?
Amongst LiCl , RDCl , BeCl2 and MgCl2 the compounds with the greatest and the least ionic character, respectively are:
a)
LiCl and RbCl
b)
RbCl and BeCl2
c)
RbCl and MgCl2
d)
MgCl2 and BeCl2
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Biswa Nath answered • Jan 20, 2023 |
The size of Rb+ is bigger than others i.e low polarising power. Hence ionic.
similarly Be2+ is smaller i.e High polarising power.
covalent
similarly Be2+ is smaller i.e High polarising power.
covalent
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Biswa Nath asked • Aug 26, 2022 |
Interconvertion between Weiss indices and Miller indices?
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Mosim Khan answered |
Bhai edu rev premium lena worth it h ya nahi h
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