Test: Valence Bond Theory, Effective Atomic Number And Magnetic Properties


25 Questions MCQ Test Chemistry Class 12 | Test: Valence Bond Theory, Effective Atomic Number And Magnetic Properties


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This mock test of Test: Valence Bond Theory, Effective Atomic Number And Magnetic Properties for Class 12 helps you for every Class 12 entrance exam. This contains 25 Multiple Choice Questions for Class 12 Test: Valence Bond Theory, Effective Atomic Number And Magnetic Properties (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Valence Bond Theory, Effective Atomic Number And Magnetic Properties quiz give you a good mix of easy questions and tough questions. Class 12 students definitely take this Test: Valence Bond Theory, Effective Atomic Number And Magnetic Properties exercise for a better result in the exam. You can find other Test: Valence Bond Theory, Effective Atomic Number And Magnetic Properties extra questions, long questions & short questions for Class 12 on EduRev as well by searching above.
QUESTION: 1

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. 

Which one of the following is wrongly matched?

Solution:

Due to strong ligand, this is inner orbital complex and the hybridisation of Fe is d2sp3.

QUESTION: 2

In which of the following the central atom has sp3d2-hybridisation?

Solution:

Since, F- ion is a weak ligand hence, pairing of electrons does not occur and outer orbital complex is formed in [CoF6]3-.

QUESTION: 3

Which is the diamagnetic?

Solution:

[Ni(CN)4]2- in this, Ni has dsp2 hybridisation with square planar geometry and it is diamagnetic.

QUESTION: 4

A magnetic moment of 1.73 BM will be shown by one among following.

Solution:

A magnetic moment of 1.73 BM corresponds to one unpaired electron. Among the given, (a) and (b) are diamagnetic and (c) is paramagnetic with 3 unpaired electrons (3d7 configuration).

QUESTION: 5

Which one of the following complex species does not involve inner orbital hybridisation?

Solution:

F- ion is weak ligand hence, pairing of electrons does not occur and outer orbital complex is formed in [CoF6]3-.

QUESTION: 6

Which of the following complex has zero magnetic moment (spin only)?

Solution:

The potassium ferrocyanide, K4 [Fe(CN)6] due to strong ligand pairing of electrons occur and it has no unpaired electrons.

QUESTION: 7

The number of unpaired electrons calculated in [Co(NH3)6]3+ and [CoF6]3- are

Solution:

 In [Co(NH3)6]3+ due to pairing cobalt configuration and it is diamagnetic.
In [CoF6]3- , F- being weak ligand pairing does not occur and Co3+ ion has configuration . It has 4 unpaired electrons.

QUESTION: 8

The tetrachloro complexes of Ni(ll) and Pd(ll) respectively are (atomic number of Ni and Pd are 28 and 46 respectively).

Solution:

ln [NiCI4]2- , Ni has sp3 hybridisation with 2 unpaired electrons, i.e. paramagnetic.
In [PdCI4]2- Pd has dsp2 hybridisation and it has no unpaired electrons, i.e. diamagnetic.

QUESTION: 9

Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Atomic number, Zn = 30, Cr= 24, Co = 27, Ni = 28)

Solution:

Zn and Ni always give outer orbital complexes with coordination number 6.
[Co(NH3)6]3+ is diamagnetic complex.
[Cr(NH3)6]3+ is param agnetic with 3d3 configuration.

QUESTION: 10

The pair of compounds having the same hybridisation for the central atom is

Solution:

[Cu(NH3)4]2+ and [Ni(NH3)4]2+ the central atoms have dsp2 hybridisation.

*Multiple options can be correct
QUESTION: 11

One or More than One Options Correct Type

Direction (Q. Nos. 11-15) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q. 

The formation of complex which involves d-orbitals of outer shell are called

Solution:

The complex which involves d-orbitals of outer shell are called high spin complex as well as outer orbital complex.

*Multiple options can be correct
QUESTION: 12

sp3-hybridisation is found in

Solution:

[ZnCI4]2-, [CuCI4]2- and [Ni(CO)4] have sp3 hybridisation and all three contains weak field ligands.

*Multiple options can be correct
QUESTION: 13

Regarding the complex [Co(H2O)6]2- , correct statement(s) is/are

Solution:

Since, F- ion is weak ligand it forms outer orbital complex arid it has 3d7 configuration with 3 unpaired electrons.

*Multiple options can be correct
QUESTION: 14

For which the EAN value is equal to the atomic number of a noble gas?

Solution:

K2[Hgl4] In this EAN = 80 - 2 + 8 = 86 (Equal to noble gas, radon).
[Pd(NH3)CI2] In this EAN = 46 - 2 + 8 = 52 (Not equal to noble gas, xenon)
[Cdl4]2- In this EAN = 48 - 2 + 8 = 54 (Equal to noble gas, xenon)
Co2(CO)8 In this EAN = Electrons from Co-atom (27) + electrons from 4 CO molecules (8) + one electron from Co— Co bond (1) = 36 (Equal to noble gas krypton)

*Multiple options can be correct
QUESTION: 15

[Fe(H2O)6]2+ and [Fe(CN)6]4- differ in

Solution:

[Fe(H2O)6]2+ and [Fe(CN)6]4- have different hybridisation, magnetic moment and colour also.

QUESTION: 16

Comprehension Type

Direction (Q. Nos. 16 and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Passage

When crystals of CuSO4 . 4NH3 are dissolved in water, there is hardly any evidence for the presence of Cu2+ ions or ammonia molecules. A new ion [Cu(NH3)4]2+ is furnished in which ammonia molecules are directly linked with the metal ion. Similarly, aqueous solution of Fe(CN)2 .4KCNdoes not give tests of Fe2+ and CN- ions but give test for new ion [Fe(CN)6]4- called ferrocyanide ion.

Q. 

The hybridisation and geometry and magnetic property of [Cu(NH3)4]2+ ion are

Solution:

Cu(Z = 29) copper configuration (3d104s1)

Hence, [Cu(NH3)4]2+ has square planar geometry and 1 unpaired electron in Ad orbital which is transferred from 3d orbital with magnetic moment 1.73 BM.

QUESTION: 17

When crystals of CuSO4 . 4NH3 are dissolved in water, there is hardly any evidence for the presence of Cu2+ ions or ammonia molecules. A new ion [Cu(NH3)4]2+ is furnished in which ammonia molecules are directly linked with the metal ion. Similarly, aqueous solution of Fe(CN)2 .4KCNdoes not give tests of Fe2+ and CN- ions but give test for new ion [Fe(CN)6]4- called ferrocyanide ion.

Q. 

The hybridisation, geometry and magnetic property of [Fe(CN)6]4- are

Solution:

It is inner orbital complex, diamagnetic.

QUESTION: 18

Matching List Type

Direction (Q. Nos. 18 and 19) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d) out of which one is correct.

Q. 

Match the Column I with Column II and mark the correct option from the codes given bleow.

Solution:

(i) → (p.s), (ii) → (p,s), (iii) → (q,t), (iv) → (q.r)

QUESTION: 19

Match the Column I with Column il and mark the correct option from the codes given bleow.

Solution:

A → (iii) (b) B → (i) (c) C → (iv) (d) D → (ii)

*Answer can only contain numeric values
QUESTION: 20

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. 

How many of the following species obey Sidgwick EAN rule?

K3[Fe(CN)6], [Ru(CO)5], [Cr(NH3)6] 3+, [Co(NH3)6]3+, [Ni(NH3)6]2+, [Fe(CO)5],[W(CO)6]


Solution:


*Answer can only contain numeric values
QUESTION: 21

How many of the following are diamagnetic?

[Zn(OH)4]2-,[Ni(NH3)6]2+, K4[Fe(CN)6], K3[Fe(CN)6], [Cu(NH3)4]2+,[PdBr4]2-, [Ni(CO)4], [CoF6] 3-


Solution:

Diamagnetic are
[Zn(OH)4]2-, K4[Fe(CN)6], [PdBr4]2-,[Ni(CO)4]

*Answer can only contain numeric values
QUESTION: 22

The number of unpaired electrons in the complex [Mn(acac)3] is

(Atomic number of Mn= 25)


Solution:

In this, Mn has +3 oxidation state. Mn3+ configuration = [Ar]3d4. This forms outer complex. Hence, unpaired electrons are 4.

*Answer can only contain numeric values
QUESTION: 23

The magnetic moment of [Ru(H2O)6]2+ corresponds to the presence of ...... unpaired electrons.


Solution:

Ru2+ =[Kr] 4d6. This forms outer complex. Hence, unpaired electrons are 4.

*Answer can only contain numeric values
QUESTION: 24

The number of unpaired electrons in [Fe(dipy)3]2+ are


Solution:

In [Fe(dipy)3]2+ complex ion, Fe has 3d6 configuration. Due to pairing, this has no unpaired electrons.

QUESTION: 25

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 : Both [Ni(CN)4]2- and [NiCI4]2- have same shape and same magnetic behaviour.

Statement II : Both are square planar and diamagnetic

Solution:

Statement I [Ni(CN)4]2- is square planar and diamagnetic whereas [NiCI4]2- is tetrahedral and paramagnetic .
Statement II [Ni(CN)4]2- is square planar while [NiCI4]2- is tetrahedral.