# Q&A - Ask Doubts and Get Answers

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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
89 Views   |

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...
Engineering
135 Views   |

The maximum KE of the photoelectron is found to be $6.63\times10^{-19}J$, when the metal is irradiated with a radiation of frequency $2\times10^{15} H_{2}$ . The threshold frequency of the metal is about

• Option 1)

$1\times10^{15}s^{-1}$

• Option 2)

$2\times10^{15}s^{-1}$

• Option 3)

$3\times10^{15}s^{-1}$

• Option 4)

$1.5\times10^{15}s^{-1}$

As learnt in Photoelectric Effect - - 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.     maximum KE  threshold frequency   we have   Option 1) This solution is correct. Option 2) This solution is incorrect. Option 3) This solution is...
Engineering
128 Views   |

The incorrect order of energy of following orbitals is

• Option 1)

3s < 3p < 4s < 4d

• Option 2)

5p < 5d < 5f < 6p

• Option 3)

7s < 5f < 6d < 7p

• Option 4)

4d < 5p < 6s < 4f < 5d

As we learned in concept Aufbau Principle - In the ground state of the atoms, the orbitals are filled in order of their increasing energies. - wherein    The order of energy of orbitals is determined by n+l where n= principal quantum number and l = azimuthal quantum number So, 6p = 6+1 = 7       5f = 5+3 = 8       5d = 5+2 = 7        5p = 5+1 = 6 Clearly, to state that 5p<5d<5f<6p is...
Engineering
77 Views   |

The energy of one mole of photons of radiation whose frequency is $1\times 10^{14}Hz$

• Option 1)

$39.85\times 10^{3}\:J/mol$

• Option 2)

$39.90 \times10^{3}\:KJ/mol$

• Option 3)

$398.5 \times10^{3}\:J/mol$

• Option 4)

$3.985 \times10^{6}\:J/mol$

As we learned in concept The energy (E) of a quantum of radiation - Where h is plank’s constant and  is frequency -    E=hv Energy with one mole of photons = = = Option 1) This option is correct. Option 2) This option is incorrect. Option 3) This option is incorrect. Option 4) This option is incorrect.
Engineering
110 Views   |

The elements X, Y, Z and J have the indicated electron configurations starting with the innermost shell. The most metallic element is:

• Option 1)

X = 2,8,3

• Option 2)

Y=2,8,8

• Option 3)

Z= 2, 8, 8, 1

• Option 4)

J = 2, 8, 8, 7

As we learned in concept Metals - –  Form metallic bonds. –  78% of all known elements. –  good conductor of heat & electricity. –  Malleable and ductile. -    When we look at electronic config of these elements, we find that Y is a noble gas and J is a halogen (non metal). Between X to Z, Zis an alkali metal an also it is lower down the group than X and metallic character increases down the...
Engineering
107 Views   |

The element whose electronic configuration is $1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}$ is

• Option 1)

Metalloid

• Option 2)

Metal

• Option 3)

Noble Gas

• Option 4)

Non-metal

As we discussed in concept S- Block elements - The elements of group 1 (alkali metal) and elements of group 2 (alkaline metal) which have ns1 and ns2 configuration belong to s-block elements. -    The element with configuration is in group 2 and period 3. It is called magnesium. Elements with this configuration attain noble gas config by losing 2 electrons in 3s arbital. Therefore it is a...
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