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In the following, two statement are given one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these question from the codes (i), (ii), (iii) and (iv) as given below :

Assertion (A) : Low spin tetrahedral complexes are rarely observed.

Reason (R) : The orbital splitting energies are not sufficiently large to force pairing.

Option: 1

Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).


Option: 2

Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).


Option: 3

Assertion (A) is correct, but Reason (R) is an incorrect statement.


Option: 4

Assertion (A) is incorrect, but Reason (R) is a correct statement.


As we have learnt, 

Tetrahedral splitting is usually small and as a result, the process of pairing requires more energy as opposed to electron occupying the higher energy level in the tetrahedral field.

Hence, spin pairing is not seen and as a result, low spin tetrahedral complexes are seldom seen.

Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).

Hence, the correct answer is Option (1)

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Posted by

vinayak

Two statements are given in the following question one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the options given below :

Assertion (A) : \mathrm{\left[P t(e n)_{2} Cl _{2}\right]^{2+}}complex is less stable than \mathrm{\left[ Pt \left( NH _{3}\right)_{4} C l_{2}\right]^{2+}} complex.

Reason (R) : \mathrm{\left[P t(e n)_{2} Cl _{2}\right]^{2+}}complex shows chelate effect.

Option: 1

Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).


Option: 2

Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).


Option: 3

Assertion (A) is correct, but Reason (R) is incorrect statement.


Option: 4

Assertion (A) is incorrect, but Reason (R) is correct statement.


As we have learnt, 

\mathrm{\left[P t(e n)_{2} Cl _{2}\right]^{2+}}contains bidentate ligand, \mathrm{en} (ethylene diamine) which is capable of showing chelation while \mathrm{\left[P t(NH_3)_{4} Cl _{2}\right]^{2+}} does not show chelation as it contains all monodentate ligands.

Thus, \mathrm{\left[P t(e n)_{2} Cl _{2}\right]^{2+}}is more stable than \mathrm{\left[P t(NH_3)_{4} Cl _{2}\right]^{2+}} and hence, Assertion (A) is incorrect while Reason (R) is correct.

Hence, the correct answer is Option (4)

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Posted by

Divya Prakash Singh

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The crystal field splitting energy for octahedral (\Delta _{0}) and tetrahedral (\Delta _{t}) complexes is related as

Option: 1

\mathrm{\Delta_{t}=\frac{2}{9} \Delta_{0}}


Option: 2

\mathrm{\Delta_{t}=\frac{5}{9} \Delta_{0}}


Option: 3

\mathrm{\Delta_{t}=\frac{4}{9} \Delta_{0}}


Option: 4

\mathrm{\Delta_{t}=2 \Delta_{0}}


The crystal field splitting energy for octahedral (\Delta _{0}) and tetrahedral (\Delta _{t}) complexes is related as 

\mathrm{\Delta _{_{t}}=\frac{4}{9}\; \Delta _{0}}

This is because of the following reasons:

(1) Number of ligands in a tetrahedra complex is 4 whereas that in an octahedral complex is 6. This accounts for a factor of \frac{2}{3} roughly

(2) No ligands directly approach the d- orbitals of the metals in a Tetrahedral field whereas in an Octahedral field, the \mathrm{e_g} orbitals \mathrm{(d_{x^2-y^2}\ and\ d_{z^2})} directly face the incoming ligands which roughly accounts for another factor of  \frac{2}{3}.

Hence, the correct answer is Option (3)

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chirag

Let two fair six-faced dice A and B be thrown simultaneously.  If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ?
Option: 1  E1 and E2 are independent.  
Option: 2   E2 and E3 are independent.  
Option: 3 E1 and E3 are independent.  
Option: 4  E2 and E3 are independent.  
 

  • E1: Die A = 4 → P(E1) = 1/6

  • E2: Die B = 2 → P(E2) = 1/6

  • E3: Sum is odd → P(E3) = 1/2

Now test independence using:
P(A ∩ B) = P(A) × P(B)

  • E1 & E2: (4,2) → 1/36 = 1/6 × 1/6 

  • E2 & E3: B=2, A odd → (1,2), (3,2), (5,2) → 3/36 = 1/6 × 1/2 

  • E1 & E3: A=4, B odd → (4,1), (4,3), (4,5) → 3/36 = 1/6 × 1/2 

  • E2 & E3: Same as above 

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Divya Sharma

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\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}is equal to
Option: 1 \frac{18}{e^{4}}
Option: 2 \frac{27}{e^{2}}
Option: 3 3\log ^{3-2}
Option: 4 3\log ^{3-2}
 

\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}is equal to
Option: 1 \frac{18}{e^{4}}
Option: 2 \frac{27}{e^{2}}
Option: 3 3\log ^{3-2}
Option: 4 3\log ^{3-2}
 

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Posted by

Divya Sharma

Solution of two components containing n1 moles of 1st component and n2 moles of 2nd component is prepared, M1 and M2 are molecular weight of component 1 and 2 respectively. If d is the density of the solution in  g mL-1, C2 is the molarity and x2 is the mole fraction of 2nd component, C2 can be expressed as:
Option: 1 c_{2}=\frac{dx_{2}}{M_{2}+x_{2}(M_{2}-M_{1})}
 
Option: 2 c_{2}=\frac{dx_{2}}{M_{2}+x_{2}(M_{2}-M_{1})}
 
Option: 3 c_{2}=\frac{dx_{2}}{M_{2}+x_{2}(M_{2}-M_{1})}
 
Option: 4 c_{2}=\frac{dx_{2}}{M_{2}+x_{2}(M_{2}-M_{1})}
 
Option: 5 c_{2}=\frac{dx_{1}}{M_{2}+x_{2}(M_{2}-M_{1})}
Option: 6 c_{2}=\frac{dx_{1}}{M_{2}+x_{2}(M_{2}-M_{1})}
Option: 7 c_{2}=\frac{dx_{1}}{M_{2}+x_{2}(M_{2}-M_{1})}
Option: 8 c_{2}=\frac{dx_{1}}{M_{2}+x_{2}(M_{2}-M_{1})}
Option: 9 c_{2}=\frac{1000dx_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}  
Option: 10 c_{2}=\frac{1000dx_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}  
Option: 11 c_{2}=\frac{1000dx_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}  
Option: 12 c_{2}=\frac{1000dx_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}  
Option: 13 c_{2}=\frac{1000x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}
Option: 14 c_{2}=\frac{1000x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}
Option: 15 c_{2}=\frac{1000x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}
Option: 16 c_{2}=\frac{1000x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}

Molarity \(C_2\) is moles of solute per liter of solution.

Using mole fraction \(x_2\), molecular weights \(M_1, M_2\), and density \(d\), we derive: \[ C_2=\frac{1000 \cdot d \cdot x_2}{M_1+x_2\left(M_2-M_1\right)} \]

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Posted by

Divya Sharma

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Assertion (A): Ethane polymerized in the presence of Ziegler Natta catalyst at high temperature and pressure is used to make bucket and dustbin.
Reason (R): High-density polymers are closely packed and are chemically inert. Choose the correct answer:
 
Option: 1 (A) and (R) both are wrong
Option: 2 (A) is correct but (R) is wrong
Option: 3 Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Option: 4 Both (A) and (R) are correct and (R) is the correct explanation of (A)
 

Both (A) and (R) are correct and (R) is the correct explanation of (A).:
Ethene polymerizes using Ziegler-Natta catalyst to form high-density polyethylene (HDPE), which is rigid, chemically inert, and closely packed, which makes it ideal for making buckets and dustbins.

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Posted by

Divya Sharma

Among the statements (a) - (d), the incorrect ones are: (a) Octahedral Co(III) complexes with strong field ligands have very high magnetic moments. (b) When \Delta _{0}<P, the d-electron configuration of Co(III) is an octahedral is t_{eg}^{4}e_{g}^{2}. (c) Wavelength of light absorbed by \left [ Co(en)_{3} \right ]^{3+} is lower than that of \left [ CoF_{6} \right ]^{3+} (d) If the \Delta _{0}  for an octahedral complex of Co(III) is 18,000 cm-1, the \Delta _{t} for tetrahedral complex with the same ligand will be 16,000cm-1
Option: 1 b and c only
Option: 2 a and b only
Option: 3 a and d only
Option: 4 c and d only
 

  • (a) Incorrect – Co(III) with strong field ligands forms low-spin complexes → few or no unpaired electrons → low magnetic moment, not high.

  • (b) Incorrect – Co(II) = 3d?. If Δ? < P → high spin → correct configuration is t_{2 g}^5 e_{g^{\prime}}^2 \operatorname{not} t_{2 g}^4 e_{g^{\prime}}^2

  • (c) Correct – [Co(en)?]3+ has strong field ligand → absorbs shorter wavelength light than [CoF?]3-

  • (d) Incorrect – Tetrahedral field splitting Δ? ≈ \(\frac{4}{9} \Delta_0 \rightarrow \frac{4}{9} \times 18,000=8,000 cm^{-1}\), not \(16,000 cm^{-1}\).

  • So, (a), (b), and (d) are incorrect → Correct option is: 3 (though it says a and d only, it should be a, b, and d)..

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Posted by

Divya Sharma

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Among the statements (a) - (d), the correct ones are: (a) Decomposition of hydrogen peroxide gives  dioxygen. (b) Like hydrogen peroxide, compounds, such as KClO_{3}, Pb(NO_{3})_{2} and NaNO_{3}  when heated, also liberate dioxygen. (c) 2-Ethylanthraquinol is useful fo the industrial prepartion of hydrogen peroxide. (d) Hydrogen peroxide is used for the manufacture of sodium perborate.
Option: 1 a, b, c and d
Option: 2 a , b and c only
Option: 3 a, c and d only
Option: 4  a  and c only
 

(a) Decomposition of hydrogen peroxide gives dioxygen.
— Correct. Hydrogen peroxide decomposes as:
2 H?O? → 2 H?O + O?

(b) Compounds like KClO?, Pb(NO?)?, and NaNO? liberate dioxygen on heating.
— Correct. These compounds release oxygen gas upon thermal decomposition.

(c) 2-Ethylanthraquinol is useful for the industrial preparation of hydrogen peroxide.
— Correct. It is used in the anthraquinone process, which is a commercial method for producing H?O?.

(d) Hydrogen peroxide is used in the manufacture of sodium perborate.
— Correct. H?O? reacts with borax and sodium hydroxide to produce sodium perborate.

Correct Option: 1 — a, b, c and d

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Posted by

Divya Sharma

An Ellingham diagram provides information about :
Option: 1 The conditions of pH and potential under which a species is  thermodynamically stable
Option: 2 The temperature dependence of the standard Gibbs energies of  formation of some metal oxides.
Option: 3 The pressure dependence of the standard electrode potentials of the extraction of metals.
Option: 4 The kinetics of the reduction process.

The temperature dependence of the standard Gibbs energies of formation of some metal oxides.

An Ellingham diagram plots ΔG° (standard Gibbs free energy change) vs temperature for the formation of metal oxides. It helps predict the feasibility of reducing metal oxides at different temperatures.

 

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Posted by

Divya Sharma

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