NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

1.

(i). Calculate the Number of Electrons Which Will Together Weigh One Gram.

Ans: Mass of one electron = $$9.10939\times {10}^{-31}kg$$

Number of electrons that weigh $$9.10939\times {10}^{-31}kg$$= 1

Number of electrons that will weigh 1 g = $$1\times {10}^{-3}kg$$

$$\frac{1}{9.10939\times {10}^{-31}kg}(1\times {10}^{-3}kg)=0.1098\times {10}^{-3+31}$$ 

$$=1.098\times {10}^{27}$$

(ii).Calculate the Mass and Charge of One Mole of Electrons.

Ans: Mass of one electron = $$9.10939\times {10}^{-31}kg$$

Mass of one mole of electron = $$(6.022\times {10}^{23})\times (9.10939\times {10}^{-31}kg)$$ = $$5.48\times {10}^{-7}kg$$ 

Charge on one electron = $$1.6022\times {10}^{-19}coulomb$$ 

Charge on one mole of electron = $$(1.6022\times {10}^{-19})(6.022\times {10}^{23})$$ = $$9.65\times {10}^{4}C$$

2.

(i).Calculate the Total Number of Electrons Present in One Mole of Methane.

Ans: Number of electrons present in 1 molecule of methane ($$C{H}_4$$)= $$\{1(6)+4(1)\}=10$$ 

Number of electrons present in 1 mole i.e., $$6.023\times {10}^{23}$$ molecules of methane = $$6.023\times {10}^{23}\times 10=6.023\times {10}^{24}$$ 

(ii). a) Find the Total Number of Neutrons in 7mg of ${}^{14}C$ . (Assume the Mass of a Neutron = $1.675\times {10}^{-27}kg$ )

Ans: Number of atoms of $${}^{14}C$$in 1 mole = $$6.023\times {10}^{23}$$ 

Since 1 atom of $${}^{14}C$$ contains (14-6) i.e., 8 neutrons, the number of neutrons in 14g of $${}^{14}C$$ is $$(6.023\times {10}^{23})\times 8$$ 

Number of neutrons in 7 mg 

$$=\frac{6.022\times {10}^{23}\times 8\times 7}{1400mg}$$ 

$$=2.4092\times {10}^{21}$$ 

(b) Find the Total Mass of Neutrons in 7mg of ${}^{14}C$ . (Assume the Mass of a Neutron = $1.675\times {10}^{-27}kg$ )

Ans: Mass of one neutron = $$1.67493\times {10}^{-27}kg$$ Mass of total neutrons in 7g of $${}^{14}C$$

$$=(2.4092\times {10}^{21})(1.67493\times {10}^{-27}kg)$$ 

$$=4.0352\times {10}^{-6}kg$$ 

(iii). (a) Find the Total Number of Protons in 34mg of $N{H}_3$ at STP. 

Ans: 1 mole of $$N{H}_3$$ = {1(14) +3(1)} g of $$N{H}_3$$ = 17g of $$N{H}_3$$ 

$$=6.023\times {10}^{23}$$ Molecules of $$N{H}_3$$ 

Total number of protons present in 1 molecule of $$N{H}_3$$

 = [1(7) +1(3)] =10

Number of protons in $$6.023\times {10}^{23}$$ molecules of $$N{H}_3$$ 

$$=(6.023\times {10}^{23})(10)=6.023\times {10}^{24}$$ 

 17g of $$N{H}_3$$ contains $$(6.023\times {10}^{24})$$ protons.

Number of protons in 34mg of $$N{H}_3$$ 

$$=\frac{6.022\times {10}^{24}\times 34mg}{1700mg}$$ 

$$=1.2046\times {10}^{22}$$ 

(b) Find the total mass of protons in 34mg of $N{H}_3$ at STP. Will the answer change if the temperature and pressure changed?

Ans: Mass of one proton = $$1.67493\times {10}^{-27}kg$$ 

Total mass of protons in 34mg of $$N{H}_3$$ 

$$=(1.67493\times {10}^{-27}kg)(1.2046\times {10}^{22})$$ 

$$=2.0176\times {10}^{-5}kg$$ 

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

3. How Many Neutrons and Protons are There in the Following Nuclei? $${}_6^{13}C{,}_8^{16}O{,}_{12}^{24}Mg{,}_{26}^{56}Fe{,}_{38}^{88}Sr$$ 

Ans: ${}_6^{13}C$ :

Atomic mass =13

Atomic number = Number of Protons = 6

Number of neutrons = (Atomic Mass)-(Atomic Number)=13-6=7

${}_8^{16}O$ :

Atomic mass =16

Atomic number = Number of Protons = 8

Number of neutrons = (Atomic Mass)-(Atomic Number) =16-8=8

${}_{12}^{24}Mg$ :

Atomic mass =24

Atomic number = Number of Protons = 12

Number of neutrons = (Atomic Mass)-(Atomic Number)=24-12 =12

${}_{26}^{56}Fe$ :

Atomic mass = 56

Atomic number = Number of Protons = 26

Number of neutrons = (Atomic Mass)-(Atomic Number) = 56-26=30

${}_{38}^{88}Sr$ :

Atomic mass = 88 

Atomic number = Number of Protons = 38

Number of neutrons = (Atomic Mass)-(Atomic Number)= 88-38=50

4. Write the Complete Symbol for the Atom with the Given Atomic Number (Z) and Atomic mass (A)

a. Z = 17, A = 35

Ans: ${}_{17}^{35}Cl$

b. Z = 92, A = 233

Ans: ${}_{92}^{233}U$

c. Z = 4, A = 9

Ans: ${}_4^{9}Be$ 

5. Yellow light emitted from a sodium lamp has a wavelength ($\lambda$ ) of 580 nm. Calculate the frequency ($\upsilon$) and wave number ( $\stackrel{-}{\upsilon }$ ) of the yellow light. 

Ans:

From the expression,

$$\lambda =\frac{c}{\upsilon }$$ 

We get,

$\upsilon =\frac{c}{\lambda }$ ----- (i)

Where, 

$\upsilon$ = frequency of yellow light

c = velocity of light in vacuum = $3\times {10}^{8}m/s$ 

$\lambda$ = wavelength of yellow light = 580nm = $580\times {10}^{-9}m$

Substituting the values in the expression (i)

$$\upsilon =\frac{3\times {10}^8}{580\times {10}^{-9}}=5.17\times {10}^{14}{S}^{-1}$$ 

Thus, frequency of yellow light emitted from sodium lamp 

= $5.17\times {10}^{14}{S}^{-1}$ 

Wave number of yellow light, $$\stackrel{-}{\upsilon }=\frac{1}{\lambda }=\frac{1}{580\times {10}^{-9}m}=1.72\times {10}^6{m}^{-1}$$   

6.

(i). Find energy of each of the photons which correspond to light of frequency $3\times {10}^{15}Hz$ 

Ans: Energy (E) of a photon is given by the expression,

$$E=h\upsilon$$ 

Where, h = Planck’s constant = $6.626\times {10}^{-34}Js$ 

$\upsilon$ = frequency of light = $3\times {10}^{15}Hz$ 

Substituting the values in the given expression of Energy, $E=(6.626\times {10}^{-34})(3\times {10}^{15})=1.988\times {10}^{-18}J$

(ii). Find energy of each of the photons which have wavelength of $0.50A^o$ 

Ans: Energy (E) of a photon having wavelength ($\lambda$) is given by the expression,

$$E=\frac{hc}{\lambda }$$ 

Where, h = Planck’s constant = $6.626\times {10}^{-34}Js$

c = velocity of light in vacuum = $3\times {10}^{8}m/s$

$\lambda$ = wavelength of yellow light = $0.50A^o$ = $0.5\times {10}^{-10}m$

 $$E=\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{0.5\times {10}^{-10}}J$$ 

$$E=3.98\times {10}^{-15}J$$ 

7. Calculate the wavelength, frequency and wave number of a light wave whose period is $2.0\times {10}^{-10}s$ 

Ans: Frequency ($\upsilon$) of light = $\frac{1}{period}=\frac{1}{2\times {10}^{-10}s}=5\times {10}^9{s}^{-1}$ 

Wavelength of light, $$\lambda =\frac{c}{\upsilon }$$ 

c = velocity of light in vacuum = $3\times {10}^{8}m/s$

Substitution the value in the given expression $\lambda$ ,

 $$\lambda =\frac{3\times {10}^8}{5\times {10}^9}=6.0\times {10}^{-2}m$$ 

Wave number of light, $\stackrel{-}{\upsilon }=\frac{1}{\lambda }=\frac{1}{6.0\times {10}^{-2}}=16.66m$ 

8. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Ans: Energy (E) of a photon is given by the expression,

$$E=h\upsilon$$ 

Energy of ‘n’ photons, 

$$E_n=nh\upsilon$$ 

$$\Rightarrow n=\frac{E_n\lambda }{hc}$$ 

Where, $\lambda$ = wavelength of yellow light = $4000pm=4000\times {10}^{-12}m$ 

c = velocity of light in vacuum = $3\times {10}^{8}m/s$

h = Planck’s constant = $6.626\times {10}^{-34}Js$

Substituting the values in the given expression ‘n’,

$$n=\frac{(1)\times (4000\times {10}^{-12})}{(6.626\times {10}^{-34})(3\times {10}^8)}=2.012\times {10}^{16}$$ 

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are $2.012\times {10}^{16}$ 

9. A photon of wavelength $4\times {10}^{7}m$ strikes on metal surface, the work function of the metal being 2.13eV. 

(i). Calculate the energy of photon (eV)

Ans: Energy of photon, $E=h\upsilon =\frac{hc}{\lambda }$ 

Where, c = velocity of light in vacuum = $3\times {10}^{8}m/s$

h = Planck’s constant = $6.626\times {10}^{-34}Js$

$\lambda$ = wavelength of yellow light = $4\times {10}^{-7}m$

Substituting the values in the given expression of E,

$$E=\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{4\times {10}^{-7}}=4.9695\times {10}^{-19}J$$ 

Hence, the energy of photon is $4.9695\times {10}^{-19}J$

(ii). Calculate the kinetic energy of the emission.

Ans: the kinetic energy of emission is given by,

$$E_k=h\upsilon -h{\upsilon }_0=(E-W)eV$$ 

$$=[\frac{4.9696\times {10}^{-19}}{1.6020\times {10}^{-19}}]-2.13eV=0.9720eV$$ 

Hence, the kinetic energy of emission is 0.97eV

(iv). Calculate the velocity of the photoelectron ($1eV=1.6020\times {10}^{-19}J$ )

Ans: The velocity of a photoelectron (ν) can be calculated by the expression,

$$\frac{1}{2}m{v}^2=h\upsilon -h{\upsilon }_0$$ 

$$\Rightarrow v=\sqrt{\frac{2(h\upsilon -h{\upsilon }_0)}{m}}$$ 

Where, $h\upsilon -h{\upsilon }_0$ is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:

$$v=\sqrt{\frac{2(0.9720\times 1.6020\times {10}^{-19})}{9.10939\times {10}^{-31}kg}J}$$ 

 $$\Rightarrow v=\sqrt{0.3418\times {10}^{12}{m}^2{s}^{-2}}=5.84\times {10}^{5}m{s}^{-1}$$ 

Hence, the velocity of the photoelectron is $5.84\times {10}^{5}m{s}^{-1}$ 

10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the ionization energy of sodium in $kJmo{l}^{-1}$ 

Ans: Energy of sodium, $E=\frac{N_{A}hc}{\lambda }$ 

$$E=\frac{(6.023\times {10}^{23}mo{l}^{-1})(6.626\times {10}^{-34}Js)(3\times {10}^{8}m{s}^{-1})}{242\times {10}^{-9}m}=4.947\times {10}^{5}Jmo{l}^{-1}$$ 

Hence, the ionization energy of sodium is $494kJmo{l}^{-1}$ 

11.  A 25 watt bulb emits monochromatic yellow light of wavelength of $0.57\mu m$ . Calculate the rate of emission of quanta per second.

Ans: power of Bulb, P = 25 Watt = $25J{s}^{-1}$ 

Energy of one photon, $E=h\upsilon =\frac{hc}{\lambda }$

Substituting the values in the given expression of E,

 $$E=\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{0.57\times {10}^{-6}}=34.87\times {10}^{-20}J$$ 

Rate of emission of quanta per second,

$$=\frac{25}{34.87\times {10}^{-20}}=7.169\times {10}^{19}{s}^{-1}$$ 

12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength$6800A^0$. Calculate threshold frequency (${\upsilon }_0$) and the work function ($w_0$) of the metal.

Ans: Threshold wavelength of radian (${\lambda }_0$) = $6800A^0=6800\times {10}^{10}m$ 

Threshold frequency of metal (${\upsilon }_0$) = $\frac{c}{{\lambda }_0}=\frac{3\times {10}^{8}m{s}^{-1}}{6.8\times {10}^{-7}m}=4.41\times {10}^{14}{s}^{-1}$ 

Thus, threshold frequency of the metal is $4.41\times {10}^{14}{s}^{-1}$ 

Hence, the work function of the metal is,

$$w_0=h{v}_0=(6.626\times {10}^{-34}Js)(4.41\times {10}^{14}{s}^{-1})=2.922\times {10}^{-19}J$$ 

13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n =2?

Ans: The $n_i=4$ to $n_f=2$ transition will rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

$$E=2.18\times {10}^{-18}[\frac{1}{n_i}-\frac{1}{n_f}]$$ 

Substituting the values in the given expression of E,

$$E=2.18\times {10}^{-18}[\frac{1}{{n_i}^2}-\frac{1}{{n^2}_f}]=2.18\times {10}^{-18}[\frac{1}{4^2}-\frac{1}{2^2}]=-4.0875\times {10}^{-19}J$$ 

The negative sign indicates the energy of emission.

Wavelength of light emitted, $$\lambda =\frac{hc}{E}$$ 

Substituting the values in the given expression of $$\lambda$$ :

$$\lambda =\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{4.0875\times {10}^{-19}}$$ 

$$\lambda =4.8631\times {10}^{-7}m$$ 

$$\lambda =486nm$$ 

14. How much energy is required to ionize a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of the H atom (energy required to remove the electron from n =1 orbit).

Ans: The expression of energy is given by,

$$E_n=\frac{-(2.18\times {10}^{-18})Z^2}{n^2}$$ 

Where, Z = atomic number of atom

n = principal quantum number

For ionization from $n_1=5$ to $n_2=\mathrm{\infty }$ 

$$\mathrm{\Delta }E=E_{\mathrm{\infty }}-E_5$$ 

$$\Rightarrow \mathrm{\Delta }E=[\left(\frac{(-2.18\times {10}^{-18}J)(1^2)}{{\mathrm{\infty }}^2}\right)-\left(\frac{(-2.18\times {10}^{-18}J)(1^2)}{5^2}\right)]=2.18\times {10}^{-18}J$$ 

Hence, the energy required for ionization from $$n_1=5$$ to $$n_2=\mathrm{\infty }$$ is $$8.72\times {10}^{20}J$$ 

The energy required for ionization from $$n_1=1$$ to $$n_2=\mathrm{\infty }$$

$$\mathrm{\Delta }E=E_{\mathrm{\infty }}-E_1$$ 

$$\Rightarrow \mathrm{\Delta }E=[\left(\frac{(-2.18\times {10}^{-18}J)(1^2)}{{\mathrm{\infty }}^2}\right)-\left(\frac{(-2.18\times {10}^{-18}J)(1^2)}{1^2}\right)]=2.18\times {10}^{-18}J$$ 

Hence, less energy is required to ionize an electron in the $5^{th}$  orbital of a hydrogen atom as compared to that in the ground state.

15. What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

Ans: When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum. The number of spectral lines produced when an electron in the $n^{th}$ level drops down to the ground state is given by. $\frac{n(n-1)}{2}$ 

Number of spectral lines for n=6, $\frac{6(6-1)}{2}=15$ 

16.

(i). The energy associated with the first orbit in the hydrogen atom is $-2.18\times {10}^{18}J$&$ato{m}^{-1}$. What is the energy associated with the fifth orbit?

Ans: Energy associated with the fifth orbit of hydrogen atom is calculated as:

$$E_5=\frac{-(2.18\times {10}^{-18})}{{(5)}^2}=\frac{-2.18\times {10}^{-18}}{25}$$ 

$$E_5=-8.72\times {10}^{-20}J$$

(ii). Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Ans: Radius of Bohr’s $n^{th}$ orbit for hydrogen atom is given by,

 $$r_n=(0.0529nm)n^2$$ 

For n = 5,

$$r_5=(0.0529nm)\times 5^2$$ 

 $$r_5=1.3225nm$$ 

17. Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans: For the Balmer series, $n_i=2$ . Thus, the expression of wave number ($\stackrel{-}{\upsilon }$) is given by

$$\stackrel{-}{\upsilon }=(\frac{1}{{(2)}^2}-\frac{1}{{n^2}_f})(1.097\times {10}^7{m}^{-1})$$ 

Wave number ($\stackrel{-}{\upsilon }$) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ($\stackrel{-}{\upsilon }$) has to be the smallest. 

For ($\stackrel{-}{\upsilon }$) to be minimum, $n_f$ should be minimum. For the Balmer series, a transition from $n_i$  = 2 to $n_f$= 3 is allowed. Hence, taking $n_f$ = 3, we get: 

$$\stackrel{-}{\upsilon }=(\frac{1}{{(2)}^2}-\frac{1}{{(3)}^2})(1.097\times {10}^7{m}^{-1})$$ 

$$\stackrel{-}{\upsilon }=(\frac{1}{4}-\frac{1}{9})(1.097\times {10}^7{m}^{-1})$$ 

 $$\stackrel{-}{\upsilon }=1.5236\times {10}^6{m}^{-1}$$ 

18. What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is $-2.18\times {10}^{-11}ergs$ 

Ans: Energy (E) of the $n^{th}$ Bohr orbit of an atom is given by,

 $$E_n=\frac{-(2.18\times {10}^{-18})Z^2}{n^2}$$ 

Where, Z = atomic number of the atom

Ground state energy=  $-2.18\times {10}^{-11}ergs=-2.18\times {10}^{-11}\times {10}^{-7}J=-2.18\times {10}^{-18}J$

Energy required to shift the electron from n = 1 to n = 5 is given as:

$$\mathrm{\Delta }E=E_5-E_1$$ 

 $$=\frac{-(2.18\times {10}^{-18}){(1)}^2}{{(5)}^2}-(-2.18\times {10}^{-18})$$ 

$$=(2.18\times {10}^{-18})\left(\frac{24}{25}\right)=2.0928\times {10}^{-18}$$ 

Weight of emitted light = $\frac{hc}{E}$ 

 $$=\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{(2.0928\times {10}^{-18})}=9.498\times {10}^{-8}m$$ 

19. The electron energy in hydrogen atom is given by $E_n=\frac{(-2.18\times {10}^{-18})}{n^2}J$ Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Ans: $E_n=\frac{(-2.18\times {10}^{-18})}{n^2}J$

Given, 

$$\mathrm{\Delta }E=E_{\mathrm{\infty }}-E_2$$ 

$$\mathrm{\Delta }E=\left(\frac{(-2.18\times {10}^{-18})}{{\mathrm{\infty }}^2}\right)-\left(\frac{(-2.18\times {10}^{-18})}{2^2}\right)=0.545\times {10}^{-18}J$$ 

Energy required for ionization from n = 2 is given by, $$0.545\times {10}^{-18}J$$

 $$\mathrm{\Delta }E=5.45\times {10}^{-19}J$$ 

 $$\lambda =\frac{hc}{\mathrm{\Delta }E}$$ , here $\lambda$ is the longest wavelength causing the transition.

 $$\lambda =\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{5.45\times {10}^{-19}}=3.647\times {10}^{-7}m$$ 

 $$\lambda =3647A^0$$ 

20. Calculate the wavelength of an electron moving with a velocity of $2.05\times {10}^{7}m{s}^{-1}$ 

Ans: According to de Broglie’s equation, $\lambda =\frac{h}{mv}$ 

Where,$\lambda$ = wavelength of moving particle

m = mass of particle, v = velocity of particle, h = Planck’s constant

Substituting the values in the expression of $\lambda$:

$$\lambda =\frac{6.626\times {10}^{-34}Js}{(9.10939\times {10}^{-31}kg)(2.05\times {10}^{7}m{s}^{-1})}=3.548\times {10}^{-11}m$$ 

Hence, the wavelength of an electron moving with a velocity of $2.05\times {10}^{7}m{s}^{-1}$ is $3.548\times {10}^{-11}m$ 

21. The mass of an electron is $9.1\times {10}^{-31}kg$ . If it is K.E. is $3.0\times {10}^{-25}J$ . Calculate its wavelength.

Ans: According to de Broglie’s equation, $\lambda =\frac{h}{mv}$

Given, the K.E. of electrons is$3.0\times {10}^{-25}J$.

Since, $K.E=\frac{1}{2}m{v}^2$ 

Velocity (v) = $\sqrt{\frac{2K.E}{m}}$ 

$$\Rightarrow v=\sqrt{\frac{2(3.0\times {10}^{-25}J)}{9.10939\times {10}^{-31}kg}=}\sqrt{6.5866\times {10}^4}$$ 

V= $811.579m{s}^{-1}$

 Substituting the value in the expression of $\lambda$ :

$$\lambda =\frac{6.626\times {10}^{-34}Js}{(9.10939\times {10}^{-31}kg)(811.579m{s}^{-1})}$$ 

$$\lambda =8.9625\times {10}^{-7}m$$ 

22. Which of the following are isoelectronic species i.e., those having the same number of electrons? $N{a}^{+},K^{+},M{g}^{+2},C{a}^{+2},S^{-2},Ar$ 

Ans: Isoelectronic species have the same number of electrons.

Number of electrons in sodium (Na) = 11

Number of electrons in ($N{a}^{+}$) = 10

A positive charge denotes the loss of an electron.

Similarly, 

Number of electrons in $K^{+}$  = 18 

Number of electrons in $M{g}^{+2}$  = 10 

Number of electrons in $C{a}^{+2}$  = 18 

A negative charge denotes the gain of an electron by a species. 

Number of electrons in sulphur (S) = 16 

$\therefore$ Number of electrons in $S^{-2}$  = 18 

Number of electrons in argon (Ar) = 18 

Hence, the following are isoelectronic species: 

(1) $N{a}^{+}$and $M{g}^{+2}$ (10 electrons each) 

(2) $K^{+}$,$C{a}^{+2}$, $S^{-2}$and Ar (18 electrons each).

23. 

(i). Write the Electronic Configurations of the Following ions:

(a) $H^{-}$ ion

Ans: The electronic configuration of H atom is $1s^1$ 

A negative charge on the species indicates the gain of an electron by it.

$\therefore$ Electronic configuration of $H^{-}$ is $1s^2$ 

(b) $N{a}^{+}$ion

Ans: The electronic configuration of Na atom is $1s^{2}2s^{2}2p^{6}3s^1$ 

A positive charge on the species indicates the loss of an electron by it.

$\therefore$Electronic configuration of $N{a}^{+}$is $1s^{2}2s^{2}2p^{6}3s^{0}or1s^{2}2s^{2}2p^6$ 

(c) $O^{-2}$ ion

Ans: The electronic configuration of O atom is $1s^{2}2s^{2}2p^4$ 

A negative charge on the species indicates the gain of two electrons by it.

$\therefore$Electronic configuration of $O^{-2}$is $1s^{2}2s^{2}2p^6$

(d) $F^{-}$ ion

Ans: The electronic configuration of F atom is $1s^{2}2s^{2}2p^5$ 

A negative charge on the species indicates the gain of an electron by it.

$\therefore$Electronic configuration of $F^{-}$is $1s^{2}2s^{2}2p^6$

(ii). What are the Atomic Numbers of Elements Whose Outermost Electrons are Represented by 

(a) $3s^1$ 

Ans: Completing the electron configuration of the element as $1s^{2}2s^{2}2p^{6}3s^1$

$\therefore$ Number of electrons present in the atom of the element 

= 2 + 2 + 6 + 1 = 11

$\therefore$Atomic number of the element = 11

(b) $2p^3$ 

Ans: Completing the electron configuration of the element as $1s^{2}2s^{2}2p^3$

$\therefore$ Number of electrons present in the atom of the element 

= 2+2+3=7

$\therefore$Atomic number of the element = 7

(c) $3p^5$ 

Ans: Completing the electron configuration of the element as $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$

$\therefore$ Number of electrons present in the atom of the element 

= 2+2+6+2+5=17

$\therefore$Atomic number of the element = 17

(iii).Which Atoms are Indicated by the Following Configurations?

(a) $[He]2s^1$ 

Ans: The electronic configuration of the element is $[He]2s^1$= $1s^{2}2s^1$ 

$\therefore$ Atomic number of the element = 3 

Hence, the element with the electronic configuration $[He]2s^1$is lithium (Li).

(b) $[Ne]3s^{2}3p^3$

Ans:  The electronic configuration of the element is $[Ne]3s^{2}3p^3$= $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$ . 

$\therefore$ Atomic number of the element = 15 

Hence, the element with the electronic configuration $[Ne]3s^{2}3p^3$is phosphorus (P).

(c) $[Ar]4s^{2}3d^1$ 

Ans: The electronic configuration of the element is $[Ar]4s^{2}3d^1$=$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^1$  . 

$\therefore$Atomic number of the element = 21 

Hence, the element with the electronic configuration $[Ar]4s^{2}3d^1$is scandium (Sc).

24. What is the Lowest Value of n That Allows G Orbitals To Exist?

Ans: For g-orbitals, l = 4. 

As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1). 

$\therefore$ For l = 4, minimum value of n = 5.

25. An Electron Is in One of the 3D Orbitals. Give the Possible Values of n, l and m, for This Electron.

Ans: For the 3d orbital: 

Principal quantum number (n) = 3 

Azimuthal quantum number (l) = 2 

Magnetic quantum number ($m_l$) = –2, –1, 0, 1, 2

26. 

(i). An Atom of an Element Contains 29 Electrons and 35 Neutrons. Deduce the Number of Protons.

Ans: For an atom to be neutral, the number of protons is equal to the number of electrons. 

$\therefore$ Number of protons in the atom of the given element = 29

(II). An Atom of an Element Contains 29 Electrons and 35 Neutrons. Deduce the Electronic Configuration of the Given Element.

Ans: The electronic configuration of the atom is $1s^{2}2s^{2}3s^{2}3p^{6}4s^{2}3d^{10}$ 

The name of the element is Copper ${}_{29}^{35}Cu$ 

27. Give the Number of Electrons in the Species  ${H_2}^{+}$ , $H_2$ and ${O_2}^{+}$ 

Ans: ${H_2}^{+}$:

Number of electrons present in hydrogen molecule ($H_2$) = 1 + 1 = 2

$\therefore$Number of electrons in ${H_2}^{+}$= 2 – 1 = 1

$H_2$:

Number of electrons in$H_2$ = 1 + 1 = 2

${O_2}^{+}$:

Number of electrons present in oxygen molecule ($O_2$) = 8 + 8 = 16 

$\therefore$ Number of electrons in ${O_2}^{+}$= 16-1 = 15

28.

(i). An atomic orbital has n = 3. What are the possible values of l and $m_l$ ?

Ans: n = 3 (Given) 

For a given value of n, l can have values from 0 to (n – 1). 

$\therefore$For n = 3 

l = 0, 1, 2 

For a given value of l, ml can have (2l + 1) values. 

$\therefore$For l = 0, m = 0 

l = 1, m = –1, 0, 1 

l = 2, m = –2, –1, 0, 1, 2 

$\therefore$For n = 3 

l = 0, 1, 2 

$m_0$ = 0 

$m_1$  = –1, 0, 1 

$m_2$  = –2, –1, 0, 1, 2

(ii). List the quantum numbers ($m_l$ and l) of electrons for 3d orbital.

Ans: For 3d orbital, l = 2. 

For a given value of l, $m_l$ can have (2l + 1) values i.e., 5 values. 

 $\therefore$For l = 2 

$m_2$ = –2, –1, 0, 1, 2

(iii). Which of the following orbitals are possible? 

1p, 2s, 2p and 3f

Ans: Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist. For p-orbital, l = 1. 

For a given value of n, l can have values from zero to (n – 1). 

Therefore for l is equal to 1, the minimum value of n is 2. 

Similarly, 

For f-orbital, l = 4. 

For l = 4, the minimum value of n is 5. Hence, 1p and 3f do not exist.

29. Using s, p, d notations, describes the orbital with the following quantum numbers.

a)  n = 1, l = 0;

Ans: n = 1, l = 0 (Given) the orbital is 1s.

b)  n = 3; l =1

Ans: For n = 3 and l = 1 the orbital is 3p.

c)  n = 4; l = 2;

Ans: For n = 4 and l = 2 the orbital is 4d.

d)  n = 4; l =3.

Ans: For n = 4 and l = 3 the orbital is 4f.

30. Explain, Giving Reasons, Which of the Following Sets of Quantum Numbers are Not

Ans:

a)  The given set of quantum numbers is not possible because the value of the   principal quantum number (n) cannot be zero.

b)  The given set of quantum numbers is possible.

c)  The given set of quantum numbers is not possible. For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.

d)  The given set of quantum numbers is possible.

e)  The given set of quantum numbers is not possible. For n = 3, 

l = 0 to (3 – 1) 

l = 0 to 2 i.e., 0, 1, 2

f)  The given set of quantum numbers is possible.

31. How Many Electrons in an Atom May Have the Following Quantum Numbers?

a)  n = 4, and $m_s=-\frac{1}{2}$ 

Ans: 

Total number of electrons in an atom for a value of n = $2n^2$ 

$\therefore$ For n = 4,

Total number of electrons = $2{(4)}^2=32$

The given element has a fully filled orbital as $1s^{2}2s^{2}3s^{2}3p^{6}4s^{2}3d^{10}$ 

Hence, all the electrons are paired. 

$\therefore$ Number of electrons (having n = 4 and $m_s=-\frac{1}{2}$ ) = 16 

b)  n = 3, l = 0

Ans: n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

32. Show That the Circumference of the Bohr Orbit for the Hydrogen Atom is an Integral Multiple of the De Broglie Wavelength Associated With the Electron Revolving Around the Orbit.

Ans: Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

$mvr=n\frac{h}{2\pi }$ ----------- (i)

Where, n = 1, 2, 3 …

According to de Broglie’s equation:

 $\lambda =\frac{h}{mv}$ Or $mv=\frac{h}{\lambda }$ ------- (ii)

Substituting the value of ‘mv’ from expression (ii) in expression (i):

$\frac{hr}{\lambda }=n\frac{h}{2\pi }$ ----------------- (iii)

Since $2\pi r$  represents the circumference of the Bohr orbit (r), it is proved by equation (iii) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

33. What Transition in the Hydrogen Spectrum Would have the Same Wavelength as the Balmer Transition n = 4 to n = 2 of $H{e}^{+}$ spectrum?

Ans: for $H{e}^{+}$ion, the wave number $\stackrel{-}{v}$ associated with the Balmer transition, n=4 to n =2 given by:

 $$\stackrel{-}{v}=\frac{1}{\lambda }=R{Z}^2(\frac{1}{{n_1}^2}-\frac{1}{{n^2}_2})$$ 

Where, $n_1=2and{n}_2=4$ , Z = atomic number of Helium

  $$\stackrel{-}{v}=\frac{1}{\lambda }=R{2}^2(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{4}$$ 

 $$\Rightarrow \lambda =\frac{4}{3R}$$ 

According to the question, the desired transition for hydrogen will have the same wavelength as that of $H{e}^{+}$.

$$R{(1)}^2(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})=\frac{3R}{4}$$ 

 $$(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})=\frac{3}{4}$$ -------- (1)

By hit and trial method, the equality given by equation (1) is true only when $n_1=1$ and $n_2=2$ 

The transition for $n_2=2$to $n_1=1$ in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of $H{e}^{+}$spectrum.

34. Calculate the energy required for the process $H{e^{+}}_{(g)}\to H{e_{(g)}}^{2+}+e^{-}$ 

The ionization energy for the H atom in the ground state is $2.18\times {10}^{-18}Jato{m}^{-1}$ .

Ans: Energy associated with hydrogen-like species is given by,

$$E_n=\frac{-(2.18\times {10}^{-18})Z^2}{n^2}$$

For ground state of hydrogen atom,

 $$\mathrm{\Delta }E=E_{\mathrm{\infty }}-E_1$$ 

$$\Rightarrow \mathrm{\Delta }E=0-[-2.18\times {10}^{-18}\left(\frac{{(1)}^2}{{(1)}^2}\right)]J$$ 

$$\therefore \mathrm{\Delta }E=2.18\times {10}^{-28}J$$ 

For the given process, $H{e^{+}}_{(g)}\to H{e_{(g)}}^{2+}+e^{-}$

An electron is removed from n = 1 to n = ∞.

$$\mathrm{\Delta }E=E_{\mathrm{\infty }}-E_1$$

$$\Rightarrow \mathrm{\Delta }E=0-[-2.18\times {10}^{-18}\left(\frac{{(2)}^2}{{(1)}^2}\right)]J$$ 

 $$\therefore \mathrm{\Delta }E=8.72\times {10}^{-18}J$$ 

The energy required for the process $8.72\times {10}^{-18}J$ 

35. If the Diameter of a Carbon Atom Is 0.15 nm, Calculate the Number of Carbon Atoms Which Can Be Placed Side by Side in a Straight Line Across a Length of Scale of Length 20 cm Long.

Ans: 1 m = 100 cm

1cm = ${10}^{-2}m$ 

Length of the scale = 20 cm = $20\times {10}^{-2}m$ 

Diameter of a carbon atom = 0.15 nm = $0.15\times {10}^{-9}m$ 

One carbon atom occupies $0.15\times {10}^{-9}m$

Number of carbon atoms that can be placed in a straight line = $$\frac{20\times {10}^{-2}m}{0.15\times {10}^{-9}m}=1.33\times {10}^9$$ 

36. $2\times {10}^8$ atoms of carbon are arranged side by side. Calculate the radius of the carbon atom if the length of this arrangement is 2.4 cm.

Ans: Length of the given arrangement = 2.4 cm

Number of carbon atoms present = $2\times {10}^8$

$\therefore$ Diameter of carbon atom = $=\frac{2.4\times {10}^{-2}}{2\times {10}^8}=1.2\times {10}^{-16}m$ 

Radius of carbon atom = $\frac{diameter}{2}=\frac{1.2\times {10}^{-10}m}{2}=6\times {10}^{-11}m$ 

37. The diameter of the zinc atom is $2.6A^0$ . 

a)  Calculate radius of zinc atom in pm

Ans: Radius of zinc atom = $\frac{diameter}{2}=\frac{2.6A^0}{2}=1.3\times {10}^{-11}m=13\times {10}^{-12}m=13pm$