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All questions of The d- and f-Block Elements & Coordination Compounds for JEE Exam
A metal, M forms chlorides in its +2 and +4 oxidation states.Which of the following statements about these chlorides is correct?- a)MCl2 is more ionic than MCl4
- b)MCl2 is more easily hydrolysed than MCl4
- c)MCl2 is more volatile than MCl4
- d)MCl2 is more soluble in anhydrous ethanol than MCl4
Correct answer is option 'A'. Can you explain this answer?
A metal, M forms chlorides in its +2 and +4 oxidation states.
Which of the following statements about these chlorides is correct?
a)
MCl2 is more ionic than MCl4
b)
MCl2 is more easily hydrolysed than MCl4
c)
MCl2 is more volatile than MCl4
d)
MCl2 is more soluble in anhydrous ethanol than MCl4
|
Akash Shah answered |
Metal atom in the lower oxidation state forms the ionic bond and in the higher oxidation state the covalent bond. because higher oxidation state means small size and great polarizing power and hence greater the covalent character. Hence MCl2 is more ionic than MCl4.
The species having tetrahedral shape is- a)[PdCl4]2–
- b) [NiCl4]2–
- c)[Pd(CN)4]2–
- d)[Ni(CN)4]2–
Correct answer is option 'B'. Can you explain this answer?
The species having tetrahedral shape is
a)
[PdCl4]2–
b)
[NiCl4]2–
c)
[Pd(CN)4]2–
d)
[Ni(CN)4]2–
|
Mayank Choudhary answered |
The configuration of Ni2+ is 3d8. For the elements of the first transition series, Cl– behaves as a weak field/ high spin ligand. Hence Ni in [NiCl4]2– is sp3 hybridised leading to tetrahedral shape.
Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect ?- a)Ferrous oxide is more basic in nature than the ferric oxide.
- b)Ferrous compounds are relatively more ionic than the corresponding ferric compounds.
- c)Ferrous compounds are less volatile than the corresponding ferric compounds.
- d)Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.
Correct answer is option 'D'. Can you explain this answer?
Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect ?
a)
Ferrous oxide is more basic in nature than the ferric oxide.
b)
Ferrous compounds are relatively more ionic than the corresponding ferric compounds.
c)
Ferrous compounds are less volatile than the corresponding ferric compounds.
d)
Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.
|
Nandini Choudhury answered |
Fe3+ is easily h ydrolysed than Fe2+ due to mor e positive charge.
Iron is rendered passive by treatment with concentrated- a)H2SO4
- b)H3PO4
- c)HCl
- d)HNO3
Correct answer is option 'D'. Can you explain this answer?
Iron is rendered passive by treatment with concentrated
a)
H2SO4
b)
H3PO4
c)
HCl
d)
HNO3
|
Pragati Nair answered |
Conc. HNO3 renders iron passive by forming a thin protective film of Fe3O4 on its surface.
In nitroprusside ion the iron and NO exist as FeII and NO+ rather than FeIII and NO. These forms can be differentiated by- a)estimating the concentration of iron
- b)measuring the concentration of CN–
- c)measuring the solid state magnetic moment
- d)thermally decomposing the compound.
Correct answer is option 'C'. Can you explain this answer?
In nitroprusside ion the iron and NO exist as FeII and NO+ rather than FeIII and NO. These forms can be differentiated by
a)
estimating the concentration of iron
b)
measuring the concentration of CN–
c)
measuring the solid state magnetic moment
d)
thermally decomposing the compound.
|
Deepika Sen answered |
The forms of iron and NO in nitroprusside ion can be differentiated by measuring the concentration of CN (cyanide) rather than by estimating the concentration of iron.
In nitroprusside ion, the FeII (ferrous) form of iron is coordinated to five CN ligands and one NO ligand. On the other hand, the FeIII (ferric) form of iron is coordinated to four CN ligands and one NO ligand.
By measuring the concentration of CN, one can determine the ratio of FeII to FeIII in the nitroprusside ion. A higher concentration of CN would indicate a higher ratio of FeII to FeIII, while a lower concentration of CN would indicate a higher ratio of FeIII to FeII.
In nitroprusside ion, the FeII (ferrous) form of iron is coordinated to five CN ligands and one NO ligand. On the other hand, the FeIII (ferric) form of iron is coordinated to four CN ligands and one NO ligand.
By measuring the concentration of CN, one can determine the ratio of FeII to FeIII in the nitroprusside ion. A higher concentration of CN would indicate a higher ratio of FeII to FeIII, while a lower concentration of CN would indicate a higher ratio of FeIII to FeII.
Reduction of the metal centre in aqueous permanganate ion involves- a)3 electrons in neutral medium
- b)5 electrons in neutral medium
- c)3 electrons in alkaline medium
- d)5 electrons in acidic medium
Correct answer is option 'A,D'. Can you explain this answer?
Reduction of the metal centre in aqueous permanganate ion involves
a)
3 electrons in neutral medium
b)
5 electrons in neutral medium
c)
3 electrons in alkaline medium
d)
5 electrons in acidic medium
|
Subhankar Mukherjee answered |
In acidic medium
Change in oxidation state of Mn = 7 – 2 = 5
Thus electrons lost = 5
In neutral medium
In neutral medium
Change in oxidation state of Mn = 7 – 4 = 3
∴ Electrons lost = 3
∴ Electrons lost = 3
PASSAGE 2Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite (CuSO4.5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance (Cu2S) and malachite (Cu2(OH)2CO3). However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.Q. Partial roasting of chalcopyrite produces- a)Cu2S and FeO
- b)Cu2O and FeO
- c)CuS and Fe2O3
- d)Cu2O and Fe2O3
Correct answer is option 'B'. Can you explain this answer?
PASSAGE 2
Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite (CuSO4.5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance (Cu2S) and malachite (Cu2(OH)2CO3). However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.
Q. Partial roasting of chalcopyrite produces
a)
Cu2S and FeO
b)
Cu2O and FeO
c)
CuS and Fe2O3
d)
Cu2O and Fe2O3
|
Rhea Joshi answered |
The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is:- a)pentaammine nitrito-N-cobalt(II) chloride
- b)pentaammine nitrito-N-cobalt(III) chloride
- c)nitrito-N-pentaamminecobalt(III) chloride
- d)nitrito-N-pentaamminecobalt(II) chloride
Correct answer is option 'B'. Can you explain this answer?
The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is:
a)
pentaammine nitrito-N-cobalt(II) chloride
b)
pentaammine nitrito-N-cobalt(III) chloride
c)
nitrito-N-pentaamminecobalt(III) chloride
d)
nitrito-N-pentaamminecobalt(II) chloride
|
Nandini Nair answered |
[Co(NO)2(NH3)5]Cl2
pentaammine nitrito-N-cobalt (III) chloride
The number of geometric isomers that can exist for square planar complex [Pt (Cl) (py) (NH3) (NH2OH)]+ is(py = pyridine) :- a)4
- b)6
- c)2
- d)3
Correct answer is option 'D'. Can you explain this answer?
The number of geometric isomers that can exist for square planar complex [Pt (Cl) (py) (NH3) (NH2OH)]+ is
(py = pyridine) :
a)
4
b)
6
c)
2
d)
3
|
Ujwal Chawla answered |
Square planar complexes of type M[ABCD] form three isomers. Their position may be obtained by fixing the position of one ligand and placing at the trans position any one of the remaining three ligands one by one.
The pair (s) of reagents that yield paramagnetic species is/are- a)Na and excess of NH3
- b)K and excess of O2
- c)Cu and dilute HNO3
- d)O2 and 2-ethylanthraquinol
Correct answer is option 'A,B,C'. Can you explain this answer?
The pair (s) of reagents that yield paramagnetic species is/are
a)
Na and excess of NH3
b)
K and excess of O2
c)
Cu and dilute HNO3
d)
O2 and 2-ethylanthraquinol
|
Aryan Sen answered |
Na + NH3 (excess) → Dilute solution of Na in liq.
NH3 → Paramagnetic
K + O2 (excess) → KO2 (O2– is paramagnetic)
Cu + HNO3 (dil.) → Cu(NO3)2 + NO (NO is paramagnetic) 2-Ethylanthraquinol + O2 →
2-Ethylanthraquinone + H2O2
(H2O2 is diamagnetic)
NH3 → Paramagnetic
K + O2 (excess) → KO2 (O2– is paramagnetic)
Cu + HNO3 (dil.) → Cu(NO3)2 + NO (NO is paramagnetic) 2-Ethylanthraquinol + O2 →
2-Ethylanthraquinone + H2O2
(H2O2 is diamagnetic)
Among [Ni(CO)4], [NiCl4]2– , [Co(NH3)4Cl2]Cl, Na3[CoF6], Na2O2 and CsO2, the total number of paramagnetic compounds is- a)2
- b)3
- c)4
- d)5
Correct answer is option 'B'. Can you explain this answer?
Among [Ni(CO)4], [NiCl4]2– , [Co(NH3)4Cl2]Cl, Na3[CoF6], Na2O2 and CsO2, the total number of paramagnetic compounds is
a)
2
b)
3
c)
4
d)
5
|
Aravind Nambiar answered |
So total number of paramagnetic compounds is 3.
Which kind of isomerism is exhibited by octah edral Co(NH3)4Br2Cl?- a)Geometrical and Ionization
- b)Geometrical and Optical
- c)Optical and Ionization
- d)Geometrical only
Correct answer is option 'A'. Can you explain this answer?
Which kind of isomerism is exhibited by octah edral Co(NH3)4Br2Cl?
a)
Geometrical and Ionization
b)
Geometrical and Optical
c)
Optical and Ionization
d)
Geometrical only
|
Anuj Saini answered |
Co(NH3)4Br 2Cl will show both geometr ical a n d ionization isomerism. [Co(NH3)4Br2]Cl and [Co(NH3)4BrCl]Br are ionization isomers and geometrical isomers are
Identify the incorrect statement among the following:- a)4f and 5f orbitals are equally shielded.
- b)d-Block elements show irregular and erratic chemical properties among themselves.
- c)La and Lu have partially filled d-orbitals and no other partially filled orbitals.
- d)The chemistry of various lanthanoids is very similar.
Correct answer is option 'A'. Can you explain this answer?
Identify the incorrect statement among the following:
a)
4f and 5f orbitals are equally shielded.
b)
d-Block elements show irregular and erratic chemical properties among themselves.
c)
La and Lu have partially filled d-orbitals and no other partially filled orbitals.
d)
The chemistry of various lanthanoids is very similar.
|
Om Singh answered |
Incorrect Statement: 4f and 5f orbitals are equally shielded.
Explanation:
- Shielding refers to the reduction in the effective nuclear charge experienced by an electron in an atom due to the presence of other electrons.
- In general, electrons in inner orbitals shield electrons in outer orbitals more effectively than electrons in outer orbitals shield each other.
- The shielding effect decreases as we move from the s orbital to the p orbital, and further decreases from p to d orbitals.
- The 4f and 5f orbitals belong to the f block of the periodic table, and they are located in the innermost part of the atom.
- The 4f orbital is shielded more effectively than the 5f orbital because it is closer to the nucleus and is shielded by the electrons in the 1s, 2s, 2p, 3s, 3p, 3d, and 4s orbitals.
- On the other hand, the 5f orbital is shielded by the electrons in the 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, and 4f orbitals, which are further away from the nucleus.
- Therefore, the 4f and 5f orbitals are not equally shielded, and the statement in option 'A' is incorrect.
Summary:
The incorrect statement among the given options is option 'A' which states that 4f and 5f orbitals are equally shielded. In reality, the 4f orbital is shielded more effectively than the 5f orbital.
Explanation:
- Shielding refers to the reduction in the effective nuclear charge experienced by an electron in an atom due to the presence of other electrons.
- In general, electrons in inner orbitals shield electrons in outer orbitals more effectively than electrons in outer orbitals shield each other.
- The shielding effect decreases as we move from the s orbital to the p orbital, and further decreases from p to d orbitals.
- The 4f and 5f orbitals belong to the f block of the periodic table, and they are located in the innermost part of the atom.
- The 4f orbital is shielded more effectively than the 5f orbital because it is closer to the nucleus and is shielded by the electrons in the 1s, 2s, 2p, 3s, 3p, 3d, and 4s orbitals.
- On the other hand, the 5f orbital is shielded by the electrons in the 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, and 4f orbitals, which are further away from the nucleus.
- Therefore, the 4f and 5f orbitals are not equally shielded, and the statement in option 'A' is incorrect.
Summary:
The incorrect statement among the given options is option 'A' which states that 4f and 5f orbitals are equally shielded. In reality, the 4f orbital is shielded more effectively than the 5f orbital.
PASSAGE 1The coordination number of Ni2+ is 4.NiCl2 + KCN (excess) → A (cyano complex)NiCl2 + Conc . HCl (excess) → B (chloro complex)Q. The hybridization of A and B are- a)dsp2, sp3
- b)sp3, sp3
- c)dsp2, dsp2
- d)sp3d2, d2sp3
Correct answer is option 'A'. Can you explain this answer?
PASSAGE 1
The coordination number of Ni2+ is 4.
NiCl2 + KCN (excess) → A (cyano complex)
NiCl2 + Conc . HCl (excess) → B (chloro complex)
Q. The hybridization of A and B are
a)
dsp2, sp3
b)
sp3, sp3
c)
dsp2, dsp2
d)
sp3d2, d2sp3
|
Dipanjan Majumdar answered |
Discussed above.
The complex ion which has no ‘d’ electron in the central metal atom is- a)[MnO4]-
- b)[Co(NH3)6]3+
- c)[Fe(CN)6]3 –
- d)[Cr(H2O)6]3+
Correct answer is option 'A'. Can you explain this answer?
The complex ion which has no ‘d’ electron in the central metal atom is
a)
[MnO4]-
b)
[Co(NH3)6]3+
c)
[Fe(CN)6]3 –
d)
[Cr(H2O)6]3+
|
|
Aravind Nambiar answered |
In [MnO4]–, Mn is in +7 oxidation state.
Electronic configuration of Mn (Z = 25) : [Ar] 3d54s2
Electronic configuration of Mn7+ : [Ar] 3d04s0 Central atom in other ions have definite number of d electrons
No. of electrons
Electronic configuration of Mn (Z = 25) : [Ar] 3d54s2
Electronic configuration of Mn7+ : [Ar] 3d04s0 Central atom in other ions have definite number of d electrons
No. of electrons
An aqueous solution of FeSO4, Al2(SO4)3 and chrome alum is heated with excess of Na2O2 and filtered. The materials obtained are :- a)a colourless filtrate and a green residue
- b)a yellow filtrate and a green residue
- c)a yellow filtrate and a brown residue
- d)a green filtrate and a brown residue
Correct answer is option 'C'. Can you explain this answer?
An aqueous solution of FeSO4, Al2(SO4)3 and chrome alum is heated with excess of Na2O2 and filtered. The materials obtained are :
a)
a colourless filtrate and a green residue
b)
a yellow filtrate and a green residue
c)
a yellow filtrate and a brown residue
d)
a green filtrate and a brown residue
|
|
Aravind Nambiar answered |
TIPS/FORMULAE :
Chrome alum is K2SO4.Cr2(SO4)3.24H2O
The filtrate is yellow due to 
ion and residue is brown due to Fe(OH)3.
ion and residue is brown due to Fe(OH)3.Which of the following compounds is metallic and ferromagnetic?- a)VO2
- b)MnO2
- c)TiO2
- d)CrO2
Correct answer is option 'D'. Can you explain this answer?
Which of the following compounds is metallic and ferromagnetic?
a)
VO2
b)
MnO2
c)
TiO2
d)
CrO2
|
Amrita Sarkar answered |
Out of all the four given metallic oxides CrO2 is attracted by magnetic field very strongly. The effect persists even when the magnetic field is removed. Thus CrO2 is metallic and ferromagnetic in nature.
Which pair of compounds is expected to show similar colour in aqueous medium?- a)FeCl2 and CuCl2
- b)VOCl2 and CuCl2
- c)VOCl2 and FeCl2
- d)FeCl2 and MnCl2
Correct answer is option 'B'. Can you explain this answer?
Which pair of compounds is expected to show similar colour in aqueous medium?
a)
FeCl2 and CuCl2
b)
VOCl2 and CuCl2
c)
VOCl2 and FeCl2
d)
FeCl2 and MnCl2
|
Anjali Deshpande answered |
Colour of transition metal ion salt is due to d-d transition of unpaired electrons of d-orbital. Metal ion salt having similar number of unpaired electrons in d-orbitals shows similar colour in aqueous medium.
Statement-1 : The geometrical isomers of the complex [M(NH3)4Cl2] are optically inactive. andStatement-2 : Both geometrical isomers of the complex [M(NH3)4Cl2] possess axis of symmetry. - a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
- b)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
- c)Statement-1 is True, Statement-2 is False
- d)Statement-1 is False, Statement-2 is True
Correct answer is option 'B'. Can you explain this answer?
Statement-1 : The geometrical isomers of the complex [M(NH3)4Cl2] are optically inactive. and
Statement-2 : Both geometrical isomers of the complex [M(NH3)4Cl2] possess axis of symmetry.
a)
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
b)
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c)
Statement-1 is True, Statement-2 is False
d)
Statement-1 is False, Statement-2 is True
|
Rithika Mehta answered |
The geometrical isomers of [M(NH3)4Cl2] can be represented as follows:-
These isomers are optically inactive and they posses axis of symmetry.
Both the statements are thus true. Out of two possible answers i.e. option (a) and (b) option (b) is correct as the statement 2 is not a correct explanation of statement 1.
For a molecule to be optically active it should not possess alternate axis of symmetry.
Both the statements are thus true. Out of two possible answers i.e. option (a) and (b) option (b) is correct as the statement 2 is not a correct explanation of statement 1.
For a molecule to be optically active it should not possess alternate axis of symmetry.
Which compound does not dissolve in hot, dilute HNO3?- a)HgS
- b)PbS
- c)CuS
- d)CdS
Correct answer is option 'A'. Can you explain this answer?
Which compound does not dissolve in hot, dilute HNO3?
a)
HgS
b)
PbS
c)
CuS
d)
CdS
|
|
Anjali Deshpande answered |
HgS does not dissolved in hot dil. HNO3
Statement-1 : Zn2+ is diamagnetic.
Statement-2 : Two electrons are lost from 4s orbital to form Zn2+.- a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
- b)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
- c)Statement-1 is True, Statement-2 is False
- d)Statement-1 is False, Statement-2 is True
Correct answer is option 'B'. Can you explain this answer?
Statement-1 : Zn2+ is diamagnetic.
Statement-2 : Two electrons are lost from 4s orbital to form Zn2+.
Statement-2 : Two electrons are lost from 4s orbital to form Zn2+.
a)
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
b)
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c)
Statement-1 is True, Statement-2 is False
d)
Statement-1 is False, Statement-2 is True
|
|
Tarun Roy answered |
Statement-1 : Zn2 is diamagnetic.
Statement-2 : Two electrons are lost from 4s orbital to form Zn2.
To determine the correctness of the statements and whether Statement-2 is a correct explanation for Statement-1, let's analyze the electronic configuration and magnetic properties of Zn2.
Electronic Configuration of Zinc (Zn) :
The atomic number of zinc (Zn) is 30, which means it has 30 electrons. The electronic configuration of zinc is as follows:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰
Formation of Zn2 :
When zinc loses two electrons, it forms Zn2. The loss of electrons occurs from the highest energy level (outermost shell) first. In the case of zinc, the 4s electrons are lost to form Zn2. The electronic configuration of Zn2 can be represented as:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
Magnetic Properties of Zn2 :
To determine the magnetic properties of an ion or atom, we need to consider the presence of unpaired electrons.
In the case of Zn2, the electronic configuration shows that all the electrons are paired. There are no unpaired electrons in Zn2. According to Hund's rule, if all the electrons are paired, the substance is diamagnetic.
Conclusion :
Based on the analysis, we can conclude the following:
- Statement-1 is True : Zn2 is diamagnetic because it has all its electrons paired.
- Statement-2 is True : Two electrons are lost from the 4s orbital to form Zn2.
- Statement-2 is NOT a correct explanation for Statement-1 : Statement-2 explains the process of electron loss from the 4s orbital to form Zn2 but does not provide an explanation for the diamagnetic nature of Zn2.
Therefore, the correct answer is option B - Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
Statement-2 : Two electrons are lost from 4s orbital to form Zn2.
To determine the correctness of the statements and whether Statement-2 is a correct explanation for Statement-1, let's analyze the electronic configuration and magnetic properties of Zn2.
Electronic Configuration of Zinc (Zn) :
The atomic number of zinc (Zn) is 30, which means it has 30 electrons. The electronic configuration of zinc is as follows:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰
Formation of Zn2 :
When zinc loses two electrons, it forms Zn2. The loss of electrons occurs from the highest energy level (outermost shell) first. In the case of zinc, the 4s electrons are lost to form Zn2. The electronic configuration of Zn2 can be represented as:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
Magnetic Properties of Zn2 :
To determine the magnetic properties of an ion or atom, we need to consider the presence of unpaired electrons.
In the case of Zn2, the electronic configuration shows that all the electrons are paired. There are no unpaired electrons in Zn2. According to Hund's rule, if all the electrons are paired, the substance is diamagnetic.
Conclusion :
Based on the analysis, we can conclude the following:
- Statement-1 is True : Zn2 is diamagnetic because it has all its electrons paired.
- Statement-2 is True : Two electrons are lost from the 4s orbital to form Zn2.
- Statement-2 is NOT a correct explanation for Statement-1 : Statement-2 explains the process of electron loss from the 4s orbital to form Zn2 but does not provide an explanation for the diamagnetic nature of Zn2.
Therefore, the correct answer is option B - Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
Sodium thiosulphate is used in photography because of its- a)reducing behaviour
- b)oxidising behaviour
- c)complex forming behaviour
- d)reaction with light
Correct answer is option 'C'. Can you explain this answer?
Sodium thiosulphate is used in photography because of its
a)
reducing behaviour
b)
oxidising behaviour
c)
complex forming behaviour
d)
reaction with light
|
Sinjini Datta answered |
Hypo solution (Na2S2O3) is used in photography to remove the unaffected AgBr in the form of soluble complex.
Coordination compounds have great importance in biological systems. In this context which of the following statements is incorrect ?- a)Cyanocobalamin is B12 and contains cobalt
- b)Haemoglobin is the red pigment of blood and contains irons
- c)Chlorophylls are green pigments in plants and contain calcium
- d)Carboxypeptidase - A is an exzyme and contains zinc.
Correct answer is option 'C'. Can you explain this answer?
Coordination compounds have great importance in biological systems. In this context which of the following statements is incorrect ?
a)
Cyanocobalamin is B12 and contains cobalt
b)
Haemoglobin is the red pigment of blood and contains irons
c)
Chlorophylls are green pigments in plants and contain calcium
d)
Carboxypeptidase - A is an exzyme and contains zinc.
|
Hiral Choudhary answered |
Chlorophylls are green pigments in plants and contain calcium
Chlorophylls are essential pigments in plants responsible for capturing light energy during photosynthesis. However, it is important to note that chlorophylls do not contain calcium. The central atom in the structure of chlorophyll is magnesium, not calcium. This is a common misconception, but the correct information is that chlorophyll contains magnesium at its core, not calcium.
Explanation:
- Chlorophyll is a complex molecule that consists of a porphyrin ring with a magnesium ion at its center. This magnesium ion plays a crucial role in the absorption of light energy in plants.
- The presence of magnesium in chlorophyll allows plants to convert sunlight into chemical energy through the process of photosynthesis.
- Calcium, on the other hand, is not a component of chlorophyll and does not play a direct role in the light-capturing function of this pigment.
- Therefore, the statement that chlorophylls contain calcium is incorrect. The correct information is that chlorophylls contain magnesium.
In conclusion, while coordination compounds such as cyanocobalamin, hemoglobin, and carboxypeptidase contain cobalt, iron, and zinc respectively, chlorophylls do not contain calcium. It is important to have a clear understanding of the composition of biological molecules to appreciate their functions in living organisms.
Chlorophylls are essential pigments in plants responsible for capturing light energy during photosynthesis. However, it is important to note that chlorophylls do not contain calcium. The central atom in the structure of chlorophyll is magnesium, not calcium. This is a common misconception, but the correct information is that chlorophyll contains magnesium at its core, not calcium.
Explanation:
- Chlorophyll is a complex molecule that consists of a porphyrin ring with a magnesium ion at its center. This magnesium ion plays a crucial role in the absorption of light energy in plants.
- The presence of magnesium in chlorophyll allows plants to convert sunlight into chemical energy through the process of photosynthesis.
- Calcium, on the other hand, is not a component of chlorophyll and does not play a direct role in the light-capturing function of this pigment.
- Therefore, the statement that chlorophylls contain calcium is incorrect. The correct information is that chlorophylls contain magnesium.
In conclusion, while coordination compounds such as cyanocobalamin, hemoglobin, and carboxypeptidase contain cobalt, iron, and zinc respectively, chlorophylls do not contain calcium. It is important to have a clear understanding of the composition of biological molecules to appreciate their functions in living organisms.
The octah edral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is:- a)L4 <L3 < L2 < L1
- b)L1 < L3 < L2 < L4
- c)L3 < L2 < L4 < L1
- d)L1 < L2 < L4 < L3
Correct answer is option 'B'. Can you explain this answer?
The octah edral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is:
a)
L4 <L3 < L2 < L1
b)
L1 < L3 < L2 < L4
c)
L3 < L2 < L4 < L1
d)
L1 < L2 < L4 < L3
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|
Nabanita Basu answered |
For a given metal ion, weak field ligands create a complex with smaller Δ, which will absorbs light of longer l and thus lower frequency. Conservely, stronger field ligands create a larger Δ, absorb light of shorter l and thus higher v i.e. higher energy.
So order of ligand strength is
L1 < L3 < L2 < L4
The complex showing a spin-only magnetic moment of 2.82 B.M. is :- a)Ni(CO)4
- b)[NiCl4]2–
- c)Ni(PPh3)4
- d)[Ni(CN)4]2–
Correct answer is option 'B'. Can you explain this answer?
The complex showing a spin-only magnetic moment of 2.82 B.M. is :
a)
Ni(CO)4
b)
[NiCl4]2–
c)
Ni(PPh3)4
d)
[Ni(CN)4]2–
|
Nitya Yadav answered |
The correct answer is b) [NiCl4]2.
To determine the spin-only magnetic moment of a complex, we can use the formula:
μ = √(n(n+2)) where n is the number of unpaired electrons.
For a), Ni(CO)4, we need to determine the number of unpaired electrons in the Ni atom. The electron configuration of Ni is 1s2 2s2 2p6 3s2 3p6 4s2 3d8. Since Ni has 8 valence electrons in the 3d shell, it can potentially have 2 unpaired electrons. However, in Ni(CO)4, each CO ligand donates a pair of electrons to the Ni atom, resulting in the formation of a d10 configuration with all electrons paired. Therefore, the spin-only magnetic moment of Ni(CO)4 is 0 B.M.
For b), [NiCl4]2, we need to determine the number of unpaired electrons in the Ni atom. The electron configuration of Ni is 1s2 2s2 2p6 3s2 3p6 4s2 3d8. In [NiCl4]2, each Cl ligand donates one electron to the Ni atom, resulting in the formation of a d8 configuration with two unpaired electrons. Plugging n = 2 into the formula, we get:
μ = √(2(2+2)) = √(2(4)) = √8 = 2.82 B.M.
Therefore, the complex showing a spin-only magnetic moment of 2.82 B.M. is [NiCl4]2.
To determine the spin-only magnetic moment of a complex, we can use the formula:
μ = √(n(n+2)) where n is the number of unpaired electrons.
For a), Ni(CO)4, we need to determine the number of unpaired electrons in the Ni atom. The electron configuration of Ni is 1s2 2s2 2p6 3s2 3p6 4s2 3d8. Since Ni has 8 valence electrons in the 3d shell, it can potentially have 2 unpaired electrons. However, in Ni(CO)4, each CO ligand donates a pair of electrons to the Ni atom, resulting in the formation of a d10 configuration with all electrons paired. Therefore, the spin-only magnetic moment of Ni(CO)4 is 0 B.M.
For b), [NiCl4]2, we need to determine the number of unpaired electrons in the Ni atom. The electron configuration of Ni is 1s2 2s2 2p6 3s2 3p6 4s2 3d8. In [NiCl4]2, each Cl ligand donates one electron to the Ni atom, resulting in the formation of a d8 configuration with two unpaired electrons. Plugging n = 2 into the formula, we get:
μ = √(2(2+2)) = √(2(4)) = √8 = 2.82 B.M.
Therefore, the complex showing a spin-only magnetic moment of 2.82 B.M. is [NiCl4]2.
Which of the following has a square planar geometry?(At. nos.: Fe = 26, Co = 27, Ni = 28, Pt = 78)- a)[PtCl4]2–
- b)[CoCl4]2–
- c)[FeCl4]2–
- d)[NiCl4]2–
Correct answer is option 'A'. Can you explain this answer?
Which of the following has a square planar geometry?
(At. nos.: Fe = 26, Co = 27, Ni = 28, Pt = 78)
a)
[PtCl4]2–
b)
[CoCl4]2–
c)
[FeCl4]2–
d)
[NiCl4]2–
|
Mehul Chavan answered |
Complexes with dsp2 hybridisation are square planar.
So [PtCl4]2– is square planar in shape.
So [PtCl4]2– is square planar in shape.
The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is- a)0
- b)2.84
- c)4.90
- d)5.92
Correct answer is option 'A'. Can you explain this answer?
The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is
a)
0
b)
2.84
c)
4.90
d)
5.92
|
Avantika Choudhary answered |
Understanding the Magnetic Moment in Cr(CO)6
The magnetic moment of a complex is determined by the presence of unpaired electrons. In the case of Cr(CO)6, we need to analyze its electronic configuration and the effect of the ligands.
Electronic Configuration of Chromium
- Chromium (Cr) has an atomic number of 24.
- Its ground state electronic configuration is [Ar] 3d5 4s1.
- In Cr(CO)6, chromium is in a +3 oxidation state, which results in an electronic configuration of 3d4 (after losing 3 electrons).
CO as a Ligand
- Carbon monoxide (CO) is a strong-field ligand and causes pairing of electrons.
- In an octahedral field created by six CO ligands, the 3d orbitals split into two groups: lower-energy t2g and higher-energy eg orbitals.
Electron Configuration in Cr(CO)6
- The 3d4 configuration under the influence of CO will lead to all four electrons being paired in the t2g orbitals.
- Thus, there are no unpaired electrons in Cr(CO)6.
Calculating the Magnetic Moment
- The spin-only magnetic moment formula is given by the equation: μ = √(n(n+2)), where n is the number of unpaired electrons.
- Since Cr(CO)6 has 0 unpaired electrons (n=0), substituting into the formula gives: μ = √(0(0+2)) = 0.
Conclusion
- Therefore, the spin-only magnetic moment of Cr(CO)6 is 0 Bohr magnetons, confirming that the correct answer is option 'A'.
The magnetic moment of a complex is determined by the presence of unpaired electrons. In the case of Cr(CO)6, we need to analyze its electronic configuration and the effect of the ligands.
Electronic Configuration of Chromium
- Chromium (Cr) has an atomic number of 24.
- Its ground state electronic configuration is [Ar] 3d5 4s1.
- In Cr(CO)6, chromium is in a +3 oxidation state, which results in an electronic configuration of 3d4 (after losing 3 electrons).
CO as a Ligand
- Carbon monoxide (CO) is a strong-field ligand and causes pairing of electrons.
- In an octahedral field created by six CO ligands, the 3d orbitals split into two groups: lower-energy t2g and higher-energy eg orbitals.
Electron Configuration in Cr(CO)6
- The 3d4 configuration under the influence of CO will lead to all four electrons being paired in the t2g orbitals.
- Thus, there are no unpaired electrons in Cr(CO)6.
Calculating the Magnetic Moment
- The spin-only magnetic moment formula is given by the equation: μ = √(n(n+2)), where n is the number of unpaired electrons.
- Since Cr(CO)6 has 0 unpaired electrons (n=0), substituting into the formula gives: μ = √(0(0+2)) = 0.
Conclusion
- Therefore, the spin-only magnetic moment of Cr(CO)6 is 0 Bohr magnetons, confirming that the correct answer is option 'A'.
The pair having the same magnetic moment is:[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]- a)[Mn(H2O)6]2+ and [Cr(H2O)6]2+
- b)[CoCl4]2– and [Fe(H2O)6]2+
- c)[Cr(H2O)6]2+ and [CoCl4]2–
- d)[Cr(H2O)6]2+ and [Fe(H2O)6]2+
Correct answer is option 'D'. Can you explain this answer?
The pair having the same magnetic moment is:
[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
a)
[Mn(H2O)6]2+ and [Cr(H2O)6]2+
b)
[CoCl4]2– and [Fe(H2O)6]2+
c)
[Cr(H2O)6]2+ and [CoCl4]2–
d)
[Cr(H2O)6]2+ and [Fe(H2O)6]2+
|
|
Nabanita Basu answered |
Since (a) and (b), each has 4 unpaired electrons, they will have same magnetic moment.
Which of the following complex species is not expected to exhibit optical isomerism ?- a)[Co(en)3]3+
- b)[Co(en)2 Cl2]+
- c)[Co(NH3)3 Cl3]
- d)[Co(en) (NH3)2 Cl2]+
Correct answer is option 'C'. Can you explain this answer?
Which of the following complex species is not expected to exhibit optical isomerism ?
a)
[Co(en)3]3+
b)
[Co(en)2 Cl2]+
c)
[Co(NH3)3 Cl3]
d)
[Co(en) (NH3)2 Cl2]+
|
|
Ujwal Chawla answered |
Octahedral coordination entities of the type Ma3b3 exhibit Geometrical isomerism. The compound exists both as facial and meridional isomers.
Consider the following complex ions, P, Q and R.P = [FeF6]3–, Q = [V(H2O)6]2+ and R = [Fe(H2O)6]2+The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is- a)R < Q < P
- b)Q < R < P
- c)R < P < Q
- d)Q < P < R
Correct answer is option 'B'. Can you explain this answer?
Consider the following complex ions, P, Q and R.
P = [FeF6]3–, Q = [V(H2O)6]2+ and R = [Fe(H2O)6]2+
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is
a)
R < Q < P
b)
Q < R < P
c)
R < P < Q
d)
Q < P < R
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|
Anjali Deshpande answered |
The electronic configuration of central metal ion in complex ions P, Q and R are
Higher the no. of unpaired electron(s), higher will be magnetic moment.
Thus the correct order of spin only magnetic moment is Q < R < P
Thus the correct order of spin only magnetic moment is Q < R < P
Among the following ions which one has the highest paramagnetism?- a)[Cr(H2O)6]3+
- b)[Fe(H2O)6]2+
- c)[Cu(H2O)6]2+
- d)[Zn(H2O)6]2+
Correct answer is option 'B'. Can you explain this answer?
Among the following ions which one has the highest paramagnetism?
a)
[Cr(H2O)6]3+
b)
[Fe(H2O)6]2+
c)
[Cu(H2O)6]2+
d)
[Zn(H2O)6]2+
|
Sagar Mukherjee answered |
Highest parmagnetic character will be shown by the ion having maximum number of unpaired electrons in their d-subshells.
25Cr3+ has 3 unpaired electrons ; 26Fe2+ has 4 unpaired electrons
29Cu2+ has 1 unpaired electrons ; 30Zn2+ has no unpaired electrons
So (a), (b) & (c) show paramagnetism. Out of which (b) has the highest paramagnetism.
25Cr3+ has 3 unpaired electrons ; 26Fe2+ has 4 unpaired electrons
29Cu2+ has 1 unpaired electrons ; 30Zn2+ has no unpaired electrons
So (a), (b) & (c) show paramagnetism. Out of which (b) has the highest paramagnetism.
Which one of the following complexes shows optical isomerism?(en = ethylenediamine)- a)trans [Co(en)2Cl2]Cl
- b)[Co(NH3)4Cl2]Cl
- c)[Co(NH3)3Cl3]
- d)cis[Co(en)2Cl2]Cl
Correct answer is option 'C'. Can you explain this answer?
Which one of the following complexes shows optical isomerism?
(en = ethylenediamine)
a)
trans [Co(en)2Cl2]Cl
b)
[Co(NH3)4Cl2]Cl
c)
[Co(NH3)3Cl3]
d)
cis[Co(en)2Cl2]Cl
|
Aryan Dasgupta answered |
Optical isomerism occurs when a molecule is non-superimposable with its mirror image hence the complex cis-[Co(en)2Cl2]Cl is optically active.
NiCl2 {P(C2H5)2(C6H5)}2 exhibits temperature depend-ent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively- a)tetrahedral and tetrahedral
- b)square planar and square planar
- c)tetrahedral and square planar
- d)square planar and tetrahedral
Correct answer is option 'C'. Can you explain this answer?
NiCl2 {P(C2H5)2(C6H5)}2 exhibits temperature depend-ent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively
a)
tetrahedral and tetrahedral
b)
square planar and square planar
c)
tetrahedral and square planar
d)
square planar and tetrahedral
|
Sanchita Patel answered |
In both states (paramagnetic and diamagnetic) of the given complex, Ni exists as Ni2+ whose electronic configuration is [Ar] 3d84s0.
In the above paramagnetic state the geometry of the complex is sp3 giving tetrahedral geometry.
The diamagnetic state is achieved by pairing of electrons in 3d orbital.
Thus the geometry of the complex will be dsp2 giving square planar geometry.
PASSAGE 1The coordination number of Ni2+ is 4.NiCl2 + KCN (excess) → A (cyano complex)NiCl2 + Conc . HCl (excess) → B (chloro complex)Q. Predict the magnetic nature of A and B- a)Both are diamagnetic
- b)A is diamagnetic and B is paramagnetic with one unpaired electron
- c)A is diamagnetic and B is paramagnetic with two unpaired electrons
- d)Both are paramagnetic
Correct answer is option 'C'. Can you explain this answer?
PASSAGE 1
The coordination number of Ni2+ is 4.
NiCl2 + KCN (excess) → A (cyano complex)
NiCl2 + Conc . HCl (excess) → B (chloro complex)
Q. Predict the magnetic nature of A and B
a)
Both are diamagnetic
b)
A is diamagnetic and B is paramagnetic with one unpaired electron
c)
A is diamagnetic and B is paramagnetic with two unpaired electrons
d)
Both are paramagnetic
|
|
Rithika Mehta answered |
However, CN– is a strong field ligand so it forces the 3d electrons to pair up and hence the effective configuration in this case will be Ni2+ in presence of CN–
Thus [Ni (CN)4]2– exhibits dsp2 hybridization and square planar shape. Since here number of unpaired electrons is zero the complex will be diamagnetic.
In case of [NiCl4]2–, Cl– is a weak field ligand, so the effective configuration of Ni2+ in this complex will be as follows : Ni2+ in presence of Cl–
In case of [NiCl4]2–, Cl– is a weak field ligand, so the effective configuration of Ni2+ in this complex will be as follows : Ni2+ in presence of Cl–
So here Ni2+ is sp3 hybridised and thus tetrahedral in shape. Since the complex has two unpaired electrons, it will be paramagnetic.
As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is :- a)Tetraaquadiaminecobalt (III) chloride
- b)Tetraaquadiamminecobalt (III) chloride
- c)Diaminetetraaquacoblat (II) chloride
- d)Diamminetetraaquacobalt (III) chloride
Correct answer is option 'D'. Can you explain this answer?
As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is :
a)
Tetraaquadiaminecobalt (III) chloride
b)
Tetraaquadiamminecobalt (III) chloride
c)
Diaminetetraaquacoblat (II) chloride
d)
Diamminetetraaquacobalt (III) chloride
|
Amrutha Chaudhary answered |
[Co(H2O)4(NH3)2]Cl3
= Diamminetetraaquacobalt (III) chloride.
= Diamminetetraaquacobalt (III) chloride.
Arrange Ce+3, La+3, Pm+3 and Yb+3 in increasing order of their ionic radii.- a)Yb+3 < Pm+3 < Ce+3 < La+3
- b)Ce+3 < Yb+3 < Pm+3 < La+3
- c)Yb+3 < Pm+3 < La+3 < Ce+3
- d)Pm+3 < La+3 < Ce+3 < Yb+3.
Correct answer is option 'A'. Can you explain this answer?
Arrange Ce+3, La+3, Pm+3 and Yb+3 in increasing order of their ionic radii.
a)
Yb+3 < Pm+3 < Ce+3 < La+3
b)
Ce+3 < Yb+3 < Pm+3 < La+3
c)
Yb+3 < Pm+3 < La+3 < Ce+3
d)
Pm+3 < La+3 < Ce+3 < Yb+3.
|
|
Aryan Sen answered |
In lanthanides there is a regular decrease in the atomic radii as well as ionic radii of trivalent ions as the atomic number increases from Ce to Lu. This decrease in size of atoms and ions is known as lanthanide contraction.
Although the atomic radii do show some irregualrities but ionic radii decreases from La to Lu. Thus the correct order is.
Although the atomic radii do show some irregualrities but ionic radii decreases from La to Lu. Thus the correct order is.
Yb+3 < Pm +3 < Ce +3 < La +3
86.8pm 97pm 1 02pm 103pm
86.8pm 97pm 1 02pm 103pm
The chemical processes in the production of steel from haematite ore involve- a)reduction
- b)oxidation
- c)reduction followed by oxidation
- d)oxidation followed by reduction
Correct answer is option 'C'. Can you explain this answer?
The chemical processes in the production of steel from haematite ore involve
a)
reduction
b)
oxidation
c)
reduction followed by oxidation
d)
oxidation followed by reduction
|
Sanaya Menon answered |
Haematite ore (Fe2O3) is first reduced to cast iron which is then oxidised for removing carbon (impurity) as CO2.
The magnetic moment (spin only) of [NiCl4]2– is :- a)1.82 BM
- b)5.46 BM
- c)2.82 BM
- d)1.41 BM
Correct answer is option 'C'. Can you explain this answer?
The magnetic moment (spin only) of [NiCl4]2– is :
a)
1.82 BM
b)
5.46 BM
c)
2.82 BM
d)
1.41 BM
|
Shail Saha answered |
Understanding the Complex: [NiCl4]2–
The magnetic moment of a complex ion is determined by the presence of unpaired electrons. In the case of [NiCl4]2–, we need to analyze the electronic configuration of nickel (Ni) and its interaction with chloride ligands.
Electron Configuration of Nickel
- Nickel has an atomic number of 28.
- The electron configuration is: [Ar] 3d8 4s2.
- In the +2 oxidation state (as in [NiCl4]2–), nickel loses 2 electrons, leading to the configuration: 3d8.
Effect of Chloride Ligand
- Chloride (Cl–) is a weak field ligand and does not cause significant pairing of electrons.
- Therefore, in a tetrahedral complex like [NiCl4]2–, the 3d electrons remain unpaired.
Count of Unpaired Electrons
- For 3d8 configuration in a tetrahedral field:
- The arrangement of electrons will have 2 unpaired electrons.
Calculating the Magnetic Moment
- The formula for magnetic moment (μ) is given by:
μ = √(n(n + 2)), where n is the number of unpaired electrons.
- For [NiCl4]2–, n = 2 (unpaired electrons).
Substituting the Values
- μ = √(2(2 + 2)) = √(2 * 4) = √8 = 2.82 BM.
Thus, the magnetic moment of [NiCl4]2– is 2.82 Bohr Magnetons (BM), confirming that option 'C' is correct.
The magnetic moment of a complex ion is determined by the presence of unpaired electrons. In the case of [NiCl4]2–, we need to analyze the electronic configuration of nickel (Ni) and its interaction with chloride ligands.
Electron Configuration of Nickel
- Nickel has an atomic number of 28.
- The electron configuration is: [Ar] 3d8 4s2.
- In the +2 oxidation state (as in [NiCl4]2–), nickel loses 2 electrons, leading to the configuration: 3d8.
Effect of Chloride Ligand
- Chloride (Cl–) is a weak field ligand and does not cause significant pairing of electrons.
- Therefore, in a tetrahedral complex like [NiCl4]2–, the 3d electrons remain unpaired.
Count of Unpaired Electrons
- For 3d8 configuration in a tetrahedral field:
- The arrangement of electrons will have 2 unpaired electrons.
Calculating the Magnetic Moment
- The formula for magnetic moment (μ) is given by:
μ = √(n(n + 2)), where n is the number of unpaired electrons.
- For [NiCl4]2–, n = 2 (unpaired electrons).
Substituting the Values
- μ = √(2(2 + 2)) = √(2 * 4) = √8 = 2.82 BM.
Thus, the magnetic moment of [NiCl4]2– is 2.82 Bohr Magnetons (BM), confirming that option 'C' is correct.
The most stable ion is- a)[Fe(OH)3]3-
- b)[Fe(Cl)6]3-
- c)[Fe(CN)6]3-
- d)[Fe(H2O)6]3+.
Correct answer is option 'C'. Can you explain this answer?
The most stable ion is
a)
[Fe(OH)3]3-
b)
[Fe(Cl)6]3-
c)
[Fe(CN)6]3-
d)
[Fe(H2O)6]3+.
|
|
Nandini Nair answered |
The cyano and hydroxo complexes are far more stable than those formed by halide ion. This is due to the fact that CN– and OH– are strong lewis bases (nucleophiles).
Further [Fe(OH)5]3– is not formed. hence most stable ion is [Fe(CN)6]3-
Further [Fe(OH)5]3– is not formed. hence most stable ion is [Fe(CN)6]3-
The oxidation state Cr in [Cr (NH3)4 Cl2]+ is- a)0
- b)+ 1
- c)+ 2
- d)+ 3
Correct answer is option 'D'. Can you explain this answer?
The oxidation state Cr in [Cr (NH3)4 Cl2]+ is
a)
0
b)
+ 1
c)
+ 2
d)
+ 3
|
Gayatri Patel answered |
Oxidation state of Cr in [Cr (NH3)4 Cl2]+.
Let it be x, 1 × x + 4 × 0 + 2 × (–1) = 1 Therefore x = 3.
Let it be x, 1 × x + 4 × 0 + 2 × (–1) = 1 Therefore x = 3.
The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is- a)2/5
- b)3/5
- c)4/5
- d)1
Correct answer is option 'A'. Can you explain this answer?
The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is
a)
2/5
b)
3/5
c)
4/5
d)
1
|
|
Charvi Chopra answered |
To determine the number of moles of KMnO4 needed to react with one mole of sulphite ion in acidic solution, we need to first write the balanced chemical equation for the reaction between KMnO4 and sulphite ion.
The balanced equation is as follows:
5SO3^2- + 2MnO4^- + 6H+ -> 5SO4^2- + 2Mn^2+ + 3H2O
From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of sulphite ion. Therefore, the ratio of moles of KMnO4 to sulphite ion is 2:5.
Now, we can use this ratio to determine the number of moles of KMnO4 needed to react with one mole of sulphite ion.
Let's assume x moles of KMnO4 are needed to react with 1 mole of sulphite ion.
According to the ratio, we can set up the following equation:
2 moles of KMnO4 / 5 moles of sulphite ion = x moles of KMnO4 / 1 mole of sulphite ion
Cross-multiplying and solving for x, we get:
2 * 1 = 5 * x
2 = 5x
x = 2/5
Therefore, the number of moles of KMnO4 needed to react with one mole of sulphite ion is 2/5.
Hence, the correct answer is option A) 2/5.
The balanced equation is as follows:
5SO3^2- + 2MnO4^- + 6H+ -> 5SO4^2- + 2Mn^2+ + 3H2O
From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of sulphite ion. Therefore, the ratio of moles of KMnO4 to sulphite ion is 2:5.
Now, we can use this ratio to determine the number of moles of KMnO4 needed to react with one mole of sulphite ion.
Let's assume x moles of KMnO4 are needed to react with 1 mole of sulphite ion.
According to the ratio, we can set up the following equation:
2 moles of KMnO4 / 5 moles of sulphite ion = x moles of KMnO4 / 1 mole of sulphite ion
Cross-multiplying and solving for x, we get:
2 * 1 = 5 * x
2 = 5x
x = 2/5
Therefore, the number of moles of KMnO4 needed to react with one mole of sulphite ion is 2/5.
Hence, the correct answer is option A) 2/5.
The geometry of Ni(CO)4 and Ni(PPh3)2Cl2 are- a)both square planar
- b)tetrahedral and square planar, respectively
- c)both tetrahedral
- d)square planar and tetrahedral, respectively
Correct answer is option 'C'. Can you explain this answer?
The geometry of Ni(CO)4 and Ni(PPh3)2Cl2 are
a)
both square planar
b)
tetrahedral and square planar, respectively
c)
both tetrahedral
d)
square planar and tetrahedral, respectively
|
|
Sahana Ahuja answered |
NOTE : In metal carbonyl the metal is in zero oxidation state.
In Ni(CO)4 , O.N. of Ni = 0
For Ni (Z = 28)
In Ni(CO)4 , O.N. of Ni = 0
For Ni (Z = 28)
In presence of CO two 4s electrons pair up, thus
In Ni(PPh3)2Cl2, O.N. of Ni = +2 For Ni2+
PPh3 and Cl– can’t pair up d-electrons, leading to sp3 hybridization leading to tetrahedral geometry.
Statement-1 : [Fe(H2O)5NO]SO4 is paramagnetic. andStatement-2 : The Fe in [Fe(H2O)5NO]SO4 has three unpaired electrons.- a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
- b)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
- c)Statement-1 is True, Statement-2 is False
- d)Statement-1 is False, Statement-2 is True
Correct answer is option 'A'. Can you explain this answer?
Statement-1 : [Fe(H2O)5NO]SO4 is paramagnetic. and
Statement-2 : The Fe in [Fe(H2O)5NO]SO4 has three unpaired electrons.
a)
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
b)
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c)
Statement-1 is True, Statement-2 is False
d)
Statement-1 is False, Statement-2 is True
|
|
Akash Shah answered |
In [Fe(H2O)5NO] SO4, Let the oxidation state of Fe be x. Then for [Fe(H2O)5 NO]2–,
x + 1 = + 2 or x = + 2 – 1 = + 1
Hence in this complex the oxidation state of Fe is + 1 Electronic configuration of Fe+ can be represented as Fe+ = 1s22s22p63s23p63d7. This unexpected configuration is due to presence of strong ligand field.
Due to which 1 electron from 4s1 gets shifted to 3d– orbitals.
The 3d7 electrons in five 3d– orbitals can be shown as
x + 1 = + 2 or x = + 2 – 1 = + 1
Hence in this complex the oxidation state of Fe is + 1 Electronic configuration of Fe+ can be represented as Fe+ = 1s22s22p63s23p63d7. This unexpected configuration is due to presence of strong ligand field.
Due to which 1 electron from 4s1 gets shifted to 3d– orbitals.
The 3d7 electrons in five 3d– orbitals can be shown as
In it we find 3 unpaired electrons.
Because of the presence of unpaired electrons the complex is paramagnetic i.e. statement 1 is true.
Because of the presence of unpaired electrons the complex is paramagnetic i.e. statement 1 is true.
As is clear from above there are three unpaired electrons in this complex i.e. statement 2 is true.
Since paramagnetic behaviour is due to presence of unpaired electrons in it so statement 2 is correct explanation of statement 1.
In view of the above facts the correct answer is option (a)
Since paramagnetic behaviour is due to presence of unpaired electrons in it so statement 2 is correct explanation of statement 1.
In view of the above facts the correct answer is option (a)
The IUPAC name of [Ni (NH3)4] [NiCl4] is- a)Tetrachloronickel (II) - tetraamminenickel (II)
- b)Tetraamminenickel (II) - tetrachloronickel (II)
- c)Tetraamminenickel (II) - tetrachloronickelate (II)
- d)Tetrachloronickel (II) - tetrachloronickelate (0) Ans. (C)
Correct answer is option 'C'. Can you explain this answer?
The IUPAC name of [Ni (NH3)4] [NiCl4] is
a)
Tetrachloronickel (II) - tetraamminenickel (II)
b)
Tetraamminenickel (II) - tetrachloronickel (II)
c)
Tetraamminenickel (II) - tetrachloronickelate (II)
d)
Tetrachloronickel (II) - tetrachloronickelate (0) Ans. (C)
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Amrutha Chaudhary answered |
The correct IUPAC name of the given compound is tetramminenickel (II) - tetrachloronickelate (II) thus (c) is the correct answer.
Which one of the following complexes is an outer orbital complex ?(Atomic nos. : Mn = 25; Fe = 26; Co = 27, Ni = 28)- a)[Co(NH3)6]3+
- b)[Mn(CN)6]4–
- c)[Fe(CN)6]4–
- d)[Ni(NH3)6]2+
Correct answer is option 'D'. Can you explain this answer?
Which one of the following complexes is an outer orbital complex ?
(Atomic nos. : Mn = 25; Fe = 26; Co = 27, Ni = 28)
a)
[Co(NH3)6]3+
b)
[Mn(CN)6]4–
c)
[Fe(CN)6]4–
d)
[Ni(NH3)6]2+
|
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Aryan Dasgupta answered |
Hybridisation
Hence [ Ni(NH3)6]2+ is outer orbital complex.
The color of KMnO4 is due to :- a)L → M charge transfer transition
- b)α - α* transition
- c)M → L charge transfer transition
- d)d – d transition
Correct answer is option 'A'. Can you explain this answer?
The color of KMnO4 is due to :
a)
L → M charge transfer transition
b)
α - α* transition
c)
M → L charge transfer transition
d)
d – d transition
|
Avantika Joshi answered |
L → M charge transfer spectra. KMnO4 is colored because it absorbs light in the visible range of electromagnetic radiation. The permanganate ion is the source of color, as a ligand to metal, (L → M) charge transfer takes place between oxygen's p orbitals and the empty d-orbitals on the metal. This charge transfer takes place when a photon of light is absorbed, which leads to the purple color of the compound.
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