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10 months ago

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Payal Gupta

Contributor-Level 10

5. (A) (i) Out of all the elements of the first transition series copper has the highest second ionisation enthalpy.

Electronic configuration of Copper is: 3d104s1

After the Loss of first electron from the 4s copper acquires 3d10 configuration which is stable. Therefore, removal of second electron from the field 3-D orbital is very difficult and requires high amount of energy.

(ii) Among the elements of first transition series zinc has the highest third ionisation enthalpy. Electronic configuration of zinc is: 3d104s2

After the loss of two electrons from 4s orbital, Z and +2 Ion acquires 3d10 fully filled configuration which is highly sta

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10 months ago

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Payal Gupta

Contributor-Level 10

4. (a) As per fajan's rule, smaller the size of ion, greater is its tendency to make covalent bonds as the positively charged cation strongly attract the negatively charged electron cloud.

On moving from La to Lu in the lanthanoid series, the atomic size decreases so the covalent character increases. Hence, La2O3 is ionic and Lu2O3 is covalent.

(b) The stability of Oxo salts is directly proportional to the size of atom, as we move from La to Lu, the size of the atom decreases and hence the stability of oxo salts also decreases.

(c) As we move along the lanthanide series, the atomic size decreases. As a result, the charge/size ratio incr

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New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

4. As per fajan's rule, smaller the size of ion, greater is its tendency to make covalent bonds as the positively charged cation strongly attract the negatively charged electron cloud.

On moving from La to Lu in the lanthanoid series, the atomic size decreases so the covalent character increases. Hence, La2O3 is ionic and Lu2O3 is covalent.

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

3. (A)  =  MnO2

(B)  = K2MnO4

(C)  =  KMnO4 

(D)  =  KIO3 

2MnO2 +  4KOH  + O2 → 2K2MnO4 +  2H2O

  (A)                                              (B)

3MnO42-  +  4H+  → 2MnO4-  +  MnO2 +  2H2O

  (C)

2MnO42-  +  2H2O  +  KI→ 2MnO2 +  2SO4-  +  KIO3

  (A)                

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Payal Gupta

Contributor-Level 10

2.  (A)  =  FeCr2O4

(B)  =  Na2CrO4

(C)  =  Na2Cr2O7

(D)  = K2Cr2O7 

4FeCrO4 +  8NaCO3 +  7O2 → 8Na2CrO4 +  2Fe2O3 +  8CO2

  (A)                                                     (B)

2NaCrO4 +  2H +  → Na2Cr2O7 +  2Na +  + H2O

Na2Cr2O7 +  KCl→ K2Cr2O7 +  2NaCl 

  (C)            &

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Payal Gupta

Contributor-Level 10

1. CuCO3 CuO  +  CO2

  (D) 

Ca (OH)2 +  CO2 → CaCO3 + H2O

  (E) 

CaCO3 +  CO2 + H2O → Ca (HCO3)2

                                        clear sol. 

CuO   +  CuS 3Cu  +  SO2 

  (A)  

 Cu + 4HNO3 → Cu (NO3)2 + 2NO2 +  2H2O

         &nb

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New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

77. Option (iii) Assertion is correct statement, but reason is wrong statement is the answer since intermediate conductivity in semiconductor is due to the small energy gap between valence band and conduction band and hence the assertion is correct, but the reason is wrong.

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

76. Option (ii) Assertion and reason both are correct statements, but reason is not correct explanation for assertion is the answer since in fcc structure the number of atoms present = 4 per unit cell which provides a maximum packing efficiency of 74%. 

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

75. Option (iii) Assertion is the correct statement, but the reason is wrong is the answer since besides the body centre there is one octahedral void at the centre of each 12 edges which is shared between four adjacent unit cells. Thus, Octahedral voids present at the body centre of the cube = 1 

12 octahedral voids are located at each edge and shared between four-unit cells = 12 * 1 2  = 3

Total number of octahedral voids = 4. 

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

74. Option (ii) Assertion and reason both are correct statements, but reason is not correct explanation for assertion is the answer since in graphite carbon atoms are arranged in different layers and each atom is covalently bonded to three of its neighbouring atoms in the same layer due to which the fourth valence electron of each atom, present between the different layers is free to move and this makes graphite a good conductor of electricity.

As different layers can slide over each other which makes it soft. while in case of diamond C atom is bonded covalently with their adjacent atoms throughout the crystal. Thus, all the four v

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