Solved Exercise, Chem-11, Ch-05

EXPERIMENTAL TECHNIQUES IN CHEMISTRY

Q.01: Select the most suitable answer for the given one.

(i) The nature of positive rays depends on:

(a) The nature of electrodes

(b) The nature of the discharge tube

(c) The nature of the residual gas

(d) All of the above

Ans: (c) The nature of the residual gas

Explanation:

Explanation: This is because the positive rays are produced by the ionization of gas molecules present inside gas discharge tube. Thus, they are positively charged nuclei of the gas which have one or more electrons lost in the process of ionization. Therefore, e/m value of positive rays depends upon the nature of the gas. The lighter the gas is, the greater is the e/m value and vice versa.

(ii) The velocity of photon is:

(a) Independent of wave length

(b) Depends on its wave length

(c) Depends on it source

(d) Equal to square of its amplitude

Ans: (a) Independent of wavelength

Explanation:

Explanation: The velocity of photon does not depend upon any of the factors like wavelength, frequency, energy of the photon or the nature of source emitting it. Photons of different wavelength travel in vacuum or space time fabric with the same speed because it is a constant, according to Einstein’s theory of relativity. This may also be explained as photons are non-material particles which have no effect of gravity, so they travel in space-time fabric with maximum speed, that is, the velocity of light.

(iii) The wave number of the light emitted by a certain source is 2×106 m–1. The wave length of this light will be:

(a) 500 nm                     

(b) 500 m

(c) 200 nm                     

(d) 5⨯107m

Ans: (a) 500 nm

Explanation:

Explanation: This is because wave number is the number of waves per meter length, so both are inversely proportional to each other. By applying the following formula: Wave number = 1/Wavelength, we can calculate wavelength of the wave if its wave number is given. Thus, the answer is Wavelength = 1/2×106 m-1 =500 nm

(iv) Rutherford’s model of the atom failed because:

(a) The atom did not have a nucleus and electrons

(b) It did not account for the attraction between protons and neutrons.

(c) It did not account for the stability of the atom.

(d) There is usually no space between the nucleus and the electrons.

Ans: (c) It did not account for the stability of the atom.

Explanation:

Explanation: The reason is that if electrons are constantly moving around the nucleus as suggested by Rutherford in his model, they should utilize their energy, and finally fall into the nucleus due to the attraction of the nucleus, collapsing the whole system. But, actually, it does not happen. The electrons neither radiate energy continuously nor they fall into nucleus. It means atom is stable enough. But, Rutherford’s model failed to explain this stability of atom.

(v) Bohr model of atom is contradicted by:

(a) Plank’s quantum theory

(b) Dual nature of matter

(c) Heisenberg’s uncertainty principal

(d) All of the above

Ans: (c) Heisenberg’s uncertainty principle

Explanation:

Explanation: Bohr, in his model, says that electrons revolve around nucleus in certain circular fix orbits. So, the position of electron and its velocity both can be calculated accurately. He rather gave formulas to calculate these values. Heisenberg, on the other hand, says in his uncertainty principle that the position and momentum of an electron in an atom cannot be determined simultaneously with accuracy. In other words, you cannot say with certainty, where the electron is at a particular moment of time and what its velocity or momentum is at that moment. This is clear contradiction to Bohr’s model.

(vi) Splitting of spectral lines when atoms are subjected to strong electric field is called:

(a) Zeeman effect           

(b) Stark effect

(c) Photoelectric effect    

(d) Compton effect

Ans: (b) Stark effect

Explanation:

Explanation: Stark proved that when spectral lines emitting from excited hydrogen atom or some other element, are passed through an electric field, they further split into several closely spaced lines. This splitting of spectral lines in the presence of an electric field is called ‘Stark Effect’.

(vii) In the ground state of an atom, the electron is present:

(a) In the nucleus

(b) In the second shell

(c) Nearest to the nucleus

(d) Farthest from the nucleus

Ans: (c) Nearest to the nucleus

Explanation:

Explanation: In the ground state, when electrons are not excited to higher energy levels, they tend to go close to the nucleus due to the attraction of the protons in nucleus. But, their energy and motion prevents them from falling into nucleus. So, they are closest to the nucleus but not in the nucleus.

(viii) Quantum number values for 2p orbital are:                                  

(a) n = 2, ℓ = 1             

(b) n = 1, ℓ = 2

(c) n = 1, ℓ = 0             

(d) n = 2, ℓ = 0

Ans: (a) n=2, ℓ=1

Explanation:

Explanation: 2p orbital means p orbital or sub-shell present in 2nd or L-shell around the atom. So, the value of principal quantum number is n=2, while p is sub-shell, so its value is given by its azimuthal quantum number which is ℓ=1.

(ix) Orbitals having same energy are called:

(a) Hybrid orbitals          

(b) Valence orbitals

(c) Degenerate orbitals    

(d) d-orbitals

Ans: (c) Degenerate orbitals

Explanation:

When excited atoms emitting spectral lines are placed in an electric or magnetic field, their sub-shells or orbitals are arranged in particular directions in response to these fields. Now, the electrons and their orbitals can be distinguished from each other because each acquires particular unique energy depending upon its orientation. When electric and magnetic fields are removed, the electrons and the orbitals go into their initial indistinguishable state where no electron or orbital can be identified from other. So, in this state, they are termed as degenerate because they are same in every respect. The term degenerate comes from the Latin word ‘degenerate’ which means ‘to fall away from one’s kind’.

(x) When 6d orbital is complete, the entering electron goes into:

(a) 7f          

(b) 7s          

(c) 7p         

(d) 7d

Ans: (c) 7p

Explanation:

Electrons fills various orbitals around nucleus in increasing energy order or increasing order of n+l value. If n+l values of two orbitals are the same, the electron will go into that orbital which have lower ‘n’ value. This is called n+l rule which is also depicted as ‘aufbau principle’. n+l value for 6d orbital is 6+2=8. The n+l value for 7p is also 7+1=8. Both have same n+l values, so according to n+l rule or aufbau principle, the electrons will fill 6d orbital first and then 7p orbital.

Q.02: Fill in the blanks:

(i) b-particles are nothing but ___________ moving with a very high speed.

(ii) The charge on one mole of electrons is___________coulombs.

(iii) The mass of hydrogen atom is_____________ grams.

(iv) The mass of one mole of electrons is_______________.

(v) Energy is ___________ when electron jumps from higher to a lower orbit.

(vi) The ionization energy of hydrogen atom can be calculated from______model of atom.

(vii) For d-subshell, the azimuthal quantum number has value of ___________.

(viii) The number of electrons in a given subshell is given by formula__________.

ix)The electronic configuration of H+ is ___________.

Q.03: Indicate true or false as the case may be.

(i) A neutron is slightly lighter particle than a proton.

(ii) A photon is the massless bundle of energy but has momentum.

(iii)  The unit of Rydberg constant is the reciprocal of unit of length.

(iv)  The actual isotopic mass is a whole number.

(v)  Heisenberg’s uncertainty principle is applicable to macroscopic bodies.

(vi)  The nodal plane in an orbital is the plane of zero electron density.

(vii)  The number of orbitals present in a sub-level is given by the formula (2l+1).

(viii) The magnetic quantum number was introduced to explain Zeeman and Stark effect.

(ix) Spin quantum number tells us the direction of spin of the nucleus

Q.04: Keeping in mind the discharge tube experiment, answer the following questions:

(a) Why is it necessary to decrease the pressure in the discharge tube to get the cathode rays?

Ans:  It is necessary to decrease the pressure in the discharge tube to get cathode rays because at ordinary pressure, gases don’t conduct electricity, even if very high voltage of about 5000 V is applied. This is because, at ordinary pressure, gas molecules hinder the movement of electrons. But, when pressure is decreased, this hindrance is reduced, and the electrons move easily from cathode to anode, causing ionization of the gas. 

(b) Whichever gas is used in the discharge tube, the nature of the cathode rays remains the same. Why?

Ans: Whichever gas is used in the discharge tube, the nature of the cathode rays remains the same because cathode rays consist of electrons, which are produced by the ionization of the gas inside the tube. Since, electrons in every gas are same because it is fundamental particle of matter, so cathode rays obtained from every gas have same nature. 

(c) Why e/m value of the cathode rays is just equal to that of electron?

Ans: The e/m value of cathode rays is just equal to that of electrons because cathode rays are, actually, the streams of electrons, produced by the ionization of the gas inside the gas discharge tube.

(d) How the bending of the cathode rays in the electric and magnetic fields shows that they are negatively charged?

Ans: When cathode rays are passed through an electric field, they bend towards anode. This shows that these rays are negatively charged.

When cathode rays are passed between the poles of a magnet, they neither bend towards anode nor towards cathode. They curve inwards to the line, joining the two poles of the magnet. This also shows that they are negatively charged. 

(e) Why the positive rays are also called canal rays?

Ans: When E. Goldstein repeated Crooks’ discharge tube experiment in 1886, using perforated cathode, he observed another type of rays that were travelling from anode to cathode. These rays, after passing through perforated cathode, produced glow on the glass wall opposite to anode. Since, these rays passed through holes or canals present in the cathode, these were named as canal rays.

(f) The e/m value of positive rays for different gases are different but those for cathode rays the e/m values are the same. Justify it.

Ans: Positive rays consist of positive ions, produced by the ionization of gas molecules, inside the gas discharge tube. Since, different gases have different masses, so their e/m values are also different.

Cathode rays, on the other hand, consist of electrons which are produced by the ionization of gas inside the gas discharge tube. Since, electrons in every gas are the same, so e/m value of cathode rays remains same, whichever gas is used in the discharge tube.

(g) The e/m value for positive rays obtained from hydrogen gas is 1836 times less than that of cathode rays. Justify it

Ans: e/m value of positive rays obtained from hydrogen gas is 1836 times less than that of cathode rays  because cathode rays consists of electrons while positive rays of hydrogen are positively charged hydrogen nuclei or protons which are produced by the ionization of hydrogen gas within gas discharged tube. Since proton is 1836 times heavier than electron while the amount of charge on both is equal, so the charge to mass ratio or e/m value for positive rays of hydrogen is 1836 times smaller than that of cathode rays

Q.05: (a) Explain Millikan’s oil drop experiment to determine the charge of an electron.

Ans:

(b) What is J.J Thomson’s experiment for determining e/m value of electron?

Ans:

(c) Evaluate mass of electron from the above two experiments.

Ans:The charge on the electron  e  1.602 × 10-19 coulomb   e/m  1.7588 × 1011 coulombs kg-1

Putting the value of e,

Q.06:

(a)Explain Millikan’s oil drop experiment to determine the charge of an electron.

Ans:

(b) What is J.J Thomson’s experiment for determining e/m value of electron?

(c) Evaluate mass of electron from the above two experiments.

The charge on the electron  e  1.602 × 10-19 coulomb   e/m  1.7588 × 1011 coulombs kg-1

Putting the value of e,

(a)Discuss Chadwick’s experiment for the discovery of neutron. Compare the properties of electron, proton and neutron.

Ans: Discovery of Neutron: Neutrons were discovered by James Chadwick in 1932, and he was awarded Nobel prize in 1935. He obtained a-particles from Polonium and directed them at Beryllium target. Some highly penetrating rays were produced which were called neutrons because the charge detector showed them to be neutral.

(b)Rutherford’s atomic model is based on the scattering of a-particles from a thin gold foil. Discuss it and explain the conclusions

Ans:

Q.07

(c) Give the postulates of Bohr’s atomic model. Which postulate tells us that orbits are stationary and energy is quantized?

Ans: Postulates of Bohr’s Atomic Model: The main postulates of Bohr’s atomic model are:

  1. Electrons revolve around the nucleus in fixed circular orbits, having a definite amount of energy and quantum number.
  2. Electron revolving in a particular orbit does not radiate energy. It radiates energy only when it jumps from one orbit to another orbit.
  3. When electron jumps from lower to higher orbit, it absorbs energy. When it jumps from higher to lower orbit, it releases energy. The energy absorbed or released is given by Plank’s equation:

∆E  E2 – E1  h𝛎

Where E2 is the higher orbit and E1 is the lower orbit and ∆E is the energy difference between the two orbits.

The electron can revolve only in those orbits, having a fixed angular momentum (mvr). The angular momentum of an orbit depends upon its quantum number, and it is an integral multiple of the factor . i.e.,  (where, n 1, 2, 3 …).

The permitted values of angular momentum are, therefore, , , , …………. The electron is bound to remain in these orbits and not in between them. So, angular momentum is quantized.

(b) Derive the equation for the radius of nth orbit of hydrogen atom using Bohr’s model.

(c)How does the above equation tell you that (i) radius is directly proportional to the square of the number of orbit. (ii) Radius is inversely proportional to the number of protons in the nucleus.

Q.08:

Derive the formula for calculating the energy of an electron in nth orbit using Bohr’s model. Keeping in view this formula explain the following:

  • The potential energy of the bounded electron is negative.

Ans:

According to Bohr, the equation for potential energy of an electron in an atom is:

This equation shows that the potential energy of a bounded electron is always negative. This is because when an electron is present at infinity i.e., outside the atom, it is not attracted by the nucleus so its P.E. is considered as zero. When it enters the orbit of the atom, it is attracted by the nucleus. The potential energy of electron decreases due to attraction and becomes less than zero. The quantity less than zero is negative. So, the P.E. of the bounded electron is always negative.

(b)Total energy of the bounded electron is also negative.

Ans:

(c)Energy of an electron is inversely proportional to n2, but energy of higher orbits are always greater than those of the lower orbits.

Ans:

(d) The energy difference between adjacent levels goes on decreasing sharply.

Ans:

(e) The energy difference goes on decreasing sharply from lower to higher orbits. This can be justified by calculating the energies of electrons present in different orbits of hydrogen atom by using Bohr’s equation, and knowing the difference between them. Bohr’s equation for the energy of electron in H-atom is:  En   2.178 x 10-18 ( ) J    or    En    kJ mol-1

Q.09:

(a) Derive the following equations for hydrogen atom, which are related to the

(i)  Energy difference between two levels, n1 and n2.

 (ii) Frequency of photon emitted when an electron jumps from n2to n1.

 (iii)  Wave number of the photon when the electron jumps from n2 to n1.

Ans:

(b)Justify that Bohr’s equation for the wave number can explain the spectral lines of Lyman, Balmer and Paschen series.

Ans:

Q.10:

(a) What is spectrum? Differentiate between continuous spectrum and line spectrum.

Ans:

Spectrum: “A visual display or the dispersion of the components of white light, when it is passed through a prism is called spectrum”.  

Spectrum is of two types:

(1)Continuous Spectrum: When the colours are diffused into each other. e.g., rainbow.

(2)Atomic or Line Spectrum: When the spectral lines (bright or dark) are separated by empty spaces. e.g., hydrogen spectrum.

(b)Compare line emission and line absorption spectra.

Ans:

line emission:

  1. The spectrum which consists of bright lines separated by dark spaces

It is obtained when an element or a compound is volatalized on a flame, or its vapours are subjected to high temperature or electric discharge.

line absorption:

  1. The spectrum which consists of dark lines separated by bright spaces.

It is obtained when white light is passed through the gaseous sample of an element.

(c)What is the origin of line spectrum?

Ans:

Origin of Line Spectrum: Line spectrum originates in two ways. Firstly, when the electrons absorb particular wavelength of light and jump from lower to higher orbits. The absorbed energy appears as dark lines separated by bright spaces (Absorption spectrum). Secondly, when they jump from higher to lower orbits and release particular wavelength of light. This released energy appears as bright lines separated by dark spaces (Emission spectrum).

Q.11:

(a) Hydrogen atom and He+ are mono-electronic system, but the size of He+ is much smaller than H+ , why?

Ans:

According to Bohr, the radius of an orbit is:                               

Putting the values of ‘n’ and ‘Z’, the radii of H and He+ come out to be:

These relations show that the size of He+ ion is about half to that of H-atom. i.e.,  

Thisis because He+ ion has higher atomic number, greater nuclear charge and stronger attraction for its valence electron than H-atom, so its valence shell is more contracted.

(b)Do you think that the size of Li+2 is even smaller than He+? Justify with calculations.

Ans:

According to Bohr, the radius of an orbit is:                              

Putting the values of ‘n’ and ‘Z’, the radii of He+ and come out to be:

These relations show that the size of Li2+ ion is even smaller than the size of He+ ion. i.e.,

Thisis because Li2+ ion has higher atomic number, greater nuclear charge and stronger attraction for its valence electron than He+ ion, so its valence shell is more contracted.

Q.12:

(a) What are X-rays? What is their origin? How was the idea of atomic number derived from the discovery of X-rays?

Ans:

(b)How does the Bohr’s model justify the Moseley’s equation

Ans:

Q.13:

Point out the defects of Bohr’s model. How these defects are partially covered by dual nature of electron and Heisenberg’s uncertainty principle?

Ans:

Defects in Bohr’s Model:

(1)Bohr’s model is applicable only to mono-electronic systems like H, He+, Li2+, Be3+ etc., not to poly-electronic systems.

(2)It does not explain the fine or multiple structure of hydrogen spectrum.

(3)It suggests that atom is flat but actually it is three dimensional.

It does not explain Zeeman effect and Stark effect. i.e., the splitting of spectral lines in magnetic and electric fields respectively.

Q.14:

(a) Briefly discuss the wave mechanical model of atom. How has it given the idea of orbital? Compare orbit and orbital.

Ans:

Orbit:

  1. The circular paths around the nucleus, in which the electrons revolve, are called orbits.
  2. Orbits are flat.
  3. The maximum number of electrons in an orbit is given by 2n2.

Examples: K, L, M, N, etc.

Orbital :

  1. The volume of space around the nucleus in which there in 95% chance of finding the electron is called orbital.
  2. Orbitals are three dimensional.
  3. An orbital can have maximum of two electrons.

Examples: s, px, py, pz, etc.

(b)What are quantum numbers? Discuss their significance.

Ans:

(c)When azimuthal quantum number has a value 3, then there are seven values of magnetic quantum number. Give reasons

Ans:

Q.15:

(a) Discuss rules for the distribution of electrons in energy subshells and in orbitals.

Ans:

(b) What is (n + l ) rule. Arrange the orbitals according to this rule. Do you think that this rule is applicable to degenerate orbitals?

Ans: ‘n + Rule: n+ ℓ rule says that: “The sub-shells are arranged in increasing order of ‘n + value, and if any two sub-shells have same ‘n + value, then that sub-shell is placed first whose ‘n’ value is smaller.”

For example, the arrangement of sub-shells according to this rule is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s. So, 4s is filled before 3d because its ‘n + ℓ’ value is lower.   

For 4s, ‘n + ℓ’  4 + 0  4, and for 3d, ‘n + ℓ’  3 + 2  5

(c)Distribute electrons in orbitals of 57La, 29Cu, 79Au, 24Cr, 531, 86Rn.

Ans:

Q.16: Draw the shapes of s, p and d-orbitals. Justify these by keeping in view the azimuthal and magnetic quantum numbers.

Ans: Shapes of s & p-Orbitals:

Q.17:   A photon of light with energy 10-19 J is emitted by a source of light

 (a)Convert this energy into the wavelength, frequency and wave number of the photon in terms of meters, hertz and m-1, respectively.

(Ans:1.51xl014s-1; 1.98×10-6m; 5xl05m-1)

Ans:    

 (b)Convert this energy of the photon into ergs and calculate the wavelength in cm, frequency in Hz and wave number in cm-1.

 [h = 6.626x 10-34 Js or 6.625x 10-27 ergs, c = 3×108 ms-1 or 3x 10+10 cms-1]

Ans:    

Q.18: The formula for calculating the energy of an electron in hydrogen atom given by Bohr’s model.

En =

Calculate the energy of the electron in first orbit of hydrogen atom. The values of various parameters are same as provided in.

Ans:    

Q.19:    Bohr’s equation for the radius of nth orbit of electron in hydrogen atom is.

(a)When the electron moves from n= 1 to n =2, how does the radius of the orbit increases.

Ans:    

 (b)What is the distance travelled by the electron when it goes from n=2 to n=3 and n=9 to n=10?

While doing calculation take care of units of energy parameter.

[ ]

Ans:    

Q.20:   Answer the following questions, by performing the calculations.

 (a)Calculate the energy of first five orbits of hydrogen atom and determine the energy differences between them.

Ans:    

 (b)Justify that energy difference between second and third orbits is approximately five times smaller than that between first and second orbits.

Ans:    

 (c)Calculate the energy of electron in He+ in first five orbits and justify that the energy differences are different from those of hydrogen atom.

Ans:    

 (d)Do you think that groups of the spectral lines of He+ are at different places than those for hydrogen atom? Give reasons.

Ans:    

Q.21:   Calculate the value of principal quantum number if an electron in hydrogen atom revolves in an orbit of energy -0.242×10-18J.

Ans:

Q.22:    Bohr’s formula for the energy levels of hydrogen atom for any system say H, He+, Li2+, etc. is

         Or      

For hydrogen: Z=1 and for He+, Z=2

 (a)Draw an energy level diagram for hydrogen atom and He+.

Ans:    

 (b)Thinking that K= 2.18×10-18J, calculate the energy needed to remove the electron from hydrogen atom and from He+.

Ans:    

 (c)How do you justify that the energies calculated in (b) are the ionization energies of H and He+?

Ans:    

 (d)Use Avogadro’s number to convert ionization energy values in kJmole-1 for H and He+.

Your values with experimental values?

Ans:    

 (e)The experimental values of ionization energy of H and He+ are 1331 KJ mol-1 and 5250 kJ molrespectively. How do you compare

Ans:    

Q.23:    Calculate the wave number of the photon when the electron jumps from

i)          n=5 to n=2      (ii) n=5 to n=1

In which series of spectral lines and spectral regions these photons will appear.

Ans:    

Q.24:  

 (a)      What is de-Broglie’s Wave length of an electron in meters travelling at half a speed of light?

Ans:    

 (b)      Convert the mass of electron into grams and velocity of light into cms-1 and then calculate the wave length of an electron in cm.

Ans:    

 (c)       Convert the wav length of electron from meters to :

  • nm
  • Ao
  • Pm

  Ans:  

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