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This is a charge balancing question.  Fe exists in two forms mainly Ferrous (Fe2+) and Ferric (Fe3+). Let there are 100 oxygen atoms and x number of Fe3+ atoms and 96-x number of Fe2+ atoms. Balance charge The percentage of  Fe3+ ions is,     Option (1) is correct.
Engineering
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   If \lambda _{0} and \lambda be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

  • Option 1)

    \sqrt{\frac{2h}{m}\left ( \lambda _{0}-\lambda \right )}

  • Option 2)

    \sqrt{\frac{2hc}{m}\left ( \lambda _{0}-\lambda \right )}

  • Option 3)

    \sqrt{\frac{2hc}{m}\left ( \frac{\lambda _{0}-\lambda }{\lambda \lambda _{0}} \right )}

  • Option 4)

    \sqrt{\frac{2h}{m}\left ( \frac{1}{\lambda _{0}}-\frac{1}{\lambda } \right )}

 

As discussed in the concept

Photoelectric Effect -

\frac{1}{2}mu^{2}= hv-hv_{0}

- wherein

where

m is the mass of the electron

u is the velocity associated with the ejected electron.

h is plank’s constant.

v is frequency of photon,

v0 is threshold frequency of metal.

 

 v^{2} = \frac{2hc}{m}\left ( \frac{1}{\lambda } - \frac{1}{\lambda _{0}} \right )

v =\sqrt{ \frac{2hc}{m}\left ( \frac{1}{\lambda } - \frac{1}{\lambda _{0}} \right )}

 

Hence the correct option is 3

 


Option 1)

\sqrt{\frac{2h}{m}\left ( \lambda _{0}-\lambda \right )}

Option 2)

\sqrt{\frac{2hc}{m}\left ( \lambda _{0}-\lambda \right )}

Option 3)

\sqrt{\frac{2hc}{m}\left ( \frac{\lambda _{0}-\lambda }{\lambda \lambda _{0}} \right )}

Option 4)

\sqrt{\frac{2h}{m}\left ( \frac{1}{\lambda _{0}}-\frac{1}{\lambda } \right )}

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Engineering
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 Chloro compound of Vanadium has only spin magnetic moment of 1.73 BM.  This Vanadium chloride has the formula :

(at. no. of V=23)

  • Option 1)

    VCl2

  • Option 2)

    VCl4

  • Option 3)

    VCl3

  • Option 4)

    VCl5

 

As discussed in

Magnetic Quantum Number (m) -

It  gives information about the spatial orientation of the orbital with respect to standard set of co-ordinate axis.

-

 

 The value of magnetic moment is 1.73 BM

1.73 = \sqrt{n\left ( n+2 \right )} where n is the number of unpaired electrons

3 = n \left ( n+2 \right )

After calculation n = 1

V_{23} => 1S^{2},2S^{2},2p^{6},3S^{2},3p^{6},4S^{2},3d^{3}

To obtain one unpaired electron V should be tetrapositive ion and the formula of its chlorid should be VCl_{4}

Option 2 is correct

 

 

 

 


Option 1)

VCl2

Incorrect option 

Option 2)

VCl4

Correct option

Option 3)

VCl3

Incorrect option

Option 4)

VCl5

Incorrect option

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Engineering
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The energy absorbed in the transition of the electron from n=2 to 1 in H-atom will be 

  • Option 1)

    10.2 meV

  • Option 2)

    3.4 meV

  • Option 3)

    13.6 meV

  • Option 4)

    12.75 meV

 
As we learnt in  Lyman Series spectrum -   Where This lies in Ultraviolet region -     where for Lyman series Z=1 (hydrogen) Rydberg constant, R = Energy absorbed = =        = 10.2 meV Option 1) 10.2 meV Option 2) 3.4 meV Option 3) 13.6 meV Option 4) 12.75 meV
Engineering
94 Views   |  

Element with electronic configuration 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}3p^{6}3d^{10},4s^{2}4p^{6}4d^{10}, 5s^{2}5p^{3} belongs to which group of the periodic table?

  • Option 1)

    3rd

  • Option 2)

    15th

  • Option 3)

    7th

  • Option 4)

    2nd

 
As we learnt in  Aufbau Principle - In the ground state of the atoms, the orbitals are filled in order of their increasing energies. - wherein    Given configuration: Element of the above configuration belongs in p block and for p block elements, group no = valence electrons...
Engineering
161 Views   |  

The number of quanta of radiations of V=4.67\times10^{13} HZ that must be absorbed in order to melt 5g of ice is (the energy required to melt 1g of ice is 333 J)

  • Option 1)

    5.38\times10^{22}

  • Option 2)

    30.9\times10^{21}

  • Option 3)

    4.67\times10^{13}

  • Option 4)

    6.62\times10^{34}

 
As we learnt in  The energy (E) of a quantum of radiation - Where h is plank’s constant and  is frequency -    Energy required=   Option 1) Correct Option 2) Incorrect Option 3) Incorrect Option 4) Incorrect
Engineering
622 Views   |  

Zero group was introduced by

  • Option 1)

    Lothar Meyer

  • Option 2)

    Lockyer

  • Option 3)

    Mendeleev

  • Option 4)

    Ramsay

 
  Noble Gas - Each period ends with a noble gas of outermost electronic configuration  ns2 np6  except  He. - wherein The electronic configuration of  He  is 1s2    The group cantaining moles or inert gases named the zero group by scattish chemist william Ramsay. It was called so an act of the zero valency eschiluted by its elements. Option 1) Lothar Meyer Incorrect Option...
Engineering
144 Views   |  

Which pair of elements belongs to same group

  • Option 1)

    Elements with atomic No. 17 and 28

  • Option 2)

    Elements with atomic No. 20 and 40

  • Option 3)

    Elements with atomic No. 17 and 53

  • Option 4)

    Elements with atomic No. 11 and 33

 
  Modern periodic law - The physical and chemical properties of elements are periodic functions of their atomic number. -    Atomic number 17: Chlorine Atomic number 53: Iodine Clearly they both belongs to the same group called halogen Option 1) Elements with atomic No. 17 and 28 Incorrect Option 2) Elements with atomic No. 20 and 40 Incorrect Option 3) Elements with atomic No. 17 and...
Engineering
106 Views   |  

Which of the following, the highest oxidation state is achieved by:

  • Option 1)

    (n-1)d^{5}ns^{2}

  • Option 2)

    (n-1)d^{5}ns^{1}

  • Option 3)

    (n-1)d^{3}ns^{2}

  • Option 4)

    1 (n-1)d^{8}ns^{2}

 
  Electronic configuration of d block elements - These elements have other electronic configuration of    -    Let see the oxidation state achieved by each one. Option 1) Correct Option 2) Incorrect Option 3) Incorrect Option 4) Incorrect
Engineering
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Which of the following options does not represent ground state electronic configuration of an atom?

  • Option 1)

    1s^{2}2s^{2}2p^{6}3s^{3}3p^{6}3d^{8}4s^{2}

  • Option 2)

    1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{1}

  • Option 3)

    1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^{1}

  • Option 4)

    1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{9}4s^{2}

 
  Aufbau Principle - In the ground state of the atoms, the orbitals are filled in order of their increasing energies. - wherein   Electrons fill the orbitals in the order of increasing energy. By t hat rule 3d orbit should fill up before 4s. Option 1) Incorrect Option 2) Incorrect Option 3) Incorrect Option 4) Correct
Engineering
124 Views   |  

Which of the following is not correctly matched?

  • Option 1)

    Lyman;\:\:n_{1}=1,\:n_{2}=2,3\cdot \cdot \cdot \cdot UV\:region

  • Option 2)

    Balmer;\:\:n_{1}=2,\:n_{2}=2,3\cdot \cdot \cdot \cdot visible\:region

  • Option 3)

    Paschen;\:\:n_{1}=3,\:n_{2}=4,5\cdot \cdot \cdot \cdot 1R\:region

  • Option 4)

    Brackett;\:\:n_{1}=4,\:n_{2}=5,6\cdot \cdot \cdot \cdot 1R\:region

 
As we learned in concept no. Balmer Series Spectrum - Where It lies in visible region -    Balmer series,  Option 1) Incorrect Option 2) Correct Option 3) Incorrect Option 4) Incorrect
Engineering
357 Views   |  

Which of the following is Dobereiner triad?

  • Option 1)

    Cl, Br, I

  • Option 2)

    Zn, Cr, Na

  • Option 3)

    Ne, Ar, K

  • Option 4)

    B, C, Si

 
As learnt in Mendeleev's periodic table - Mendeleev arranged the 63 discovered elements in the periodic table in to 7 horizontal rows known as periods and vertical column known as groups numbered 1 to 8. -    According to this principle, the atomic weight of the middle element in the triad is nearly the same as average of the atomic weights of other two elements. Cl (35.5)  Br (80)  I...
Engineering
94 Views   |  

Which is not arranged in the correct sequence?

  • Option 1)

    MO, M_{2}O_{3}, MO_{2}, M_{2}O_{5}   decreasing basic strength

  • Option 2)

    d^{5},d^{3},d^{1},d^{4} increasing magnetic moment

  • Option 3)

    Sc, V, Cr, Mn - Increasing number of oxidation states

  • Option 4)

    CO^{2+}, fe^{3+}, Cr^{3+},Sc^{3+} - Increasing stability

 
As learnt in Magnetic Quantum Number (m) - It  gives information about the spatial orientation of the orbital with respect to standard set of co-ordinate axis. -   The magnetic moment is maximum when the maximum number of orbitals are occupied by single electrons. Clearly, that is not the case in option 2 as d5 has maximum magnetic moment.                                    d5 Option 1)  ...
Engineering
87 Views   |  

Which electronic configurations represent to a transition element?

  • Option 1)

    1s^{2}, 2s^{2} 2p^{6}, 3s^{2}3p^{6}3d^{10}, 4s^{2}4p^{6}

  • Option 2)

    1s^{2}, 2s^{2} 2p^{6}, 3s^{2}3p^{6}3d^{10}, 4s^{2}4p^{1}

  • Option 3)

    1s^{2}, 2s^{2} 2p^{6}, 3s^{2}3p^{6}3d^{2}, 4s^{2}

  • Option 4)

    1s^{2}, 2s^{2} 2p^{6}, 3s^{2}3p^{6}, 4s^{2}

 
As learnt in Electronic configuration of d block elements - These elements have other electronic configuration of    -    Transition elements have the following configuration: 1s2 2s2 . . . . . (n - 1)d 1 - 10  ns2 Clearly, only 3 fits with 1s2 2s2 2p6 3s2 3p6 3d2 4s2 Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is correct Option 4) This...
Engineering
101 Views   |  

The wavelength of a ball of mass 1kg moving with a velocity of 10ms^{-1} ?

  • Option 1)

    6.626\times 10^{-34} m

  • Option 2)

    6.626\times 10^{-35} m

  • Option 3)

    6.626\times 10^{-30} m

  • Option 4)

    6.626\times 10^{-33} m

 
As learnt in De-broglie wavelength - - wherein where m is the mass of the particle v its velocity  p its momentum    De Broglie's wavelength is given by:                                 Option 1) This option is incorrect Option 2) This option is correct Option 3) This option is incorrect Option 4) This option is incorrect
Engineering
99 Views   |  

The wave number of the spectral line corresponding to the transition from n_{1}=2\:to\:n_{2}=4 is

  • Option 1)

    20565 per m

  • Option 2)

    205.65 per cm

  • Option 3)

    20565 per cm

  • Option 4)

    205.65 per m

 
As learnt in Line Spectrum of Hydrogen like atoms -   - wherein Where R is called Rhydberg constant, R = 1.097 X 107 , Z is atomic number n1= 1,2 ,3…. n2= n1+1, n1+2 ……                       cm-1       cm  Option 1) 20565 per m This option is incorrect Option 2) 205.65 per cm This option is incorrect Option 3) 20565 per cm This option is correct Option 4) 205.65 per m This option is incorrect
Engineering
85 Views   |  

The ratio of the energy of the electron in the ground state of hydrogen to the electron in the first excited state of Be^{3+} is 

  • Option 1)

    1:4

  • Option 2)

    1:8

  • Option 3)

    1:16

  • Option 4)

    16:1

 
As learnt in Total energy of elctron in nth orbit - Where z is atomic number -    We know that total energy of electron in nth orbit,  E1 (ground state of hydrogen) =  E2 (first excited state of Be 3+) =          (n = 2, z = 4)                                                   = -13.6 x 4 eV Option 1) 1:4 This option is correct Option 2) 1:8 This option is incorrect Option 3) 1:16 This...
Engineering
312 Views   |  

The photoelectric effect was not observed for

  • Option 1)

    Potassium

  • Option 2)

    Rubidium

  • Option 3)

    Lithium 

  • Option 4)

    Caesium

 
As learnt in Photoelectric Effect - The electrons are ejected from the metal surface as soon as the beam of light of a particular frequency strikes the surface. -    Lithium has a very high ionization potential on account of its very small size thus making photoelectric effect difficult. Option 1) Potassium The option is incorrect. Option 2) Rubidium The option is incorrect. Option...
Engineering
85 Views   |  

The number of photons of light with the wavelength of 4000 pm that provides 1 J of energy is

  • Option 1)

    2\times10^{16}

  • Option 2)

    1\times10^{18}

  • Option 3)

    3\times10^{15}

  • Option 4)

    3\times10^{16}

 
As learnt in The energy (E) of a quantum of radiation - Where h is plank’s constant and  is frequency -    Thus,  Option 1) This solution is correct. Option 2) This solution is incorrect. Option 3) This solution is incorrect. Option 4) This solution is incorrect.
Engineering
124 Views   |  

The maximum number of emission lines when the excited electron of an H atom in n=5 drops to the ground state?

  • Option 1)

    15

  • Option 2)

    10

  • Option 3)

    20

  • Option 4)

    5

 
As learnt in Paschen , Bracket and Pfund Series spectrums - Infrared Region -    Each different transition of the electron results in a different emission line. So, as far the figure 5   The figure contains all possible transitions.  possible emission lines (max.) = 10 Option 1) 15 This solution is incorrect. Option 2) 10 This solution is correct. Option 3) 20 This solution is...
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