NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

Question 2.1 (i) Calculate the number of electrons that will together weigh one gram.

Answer :

As the mass of one electron we know is $9.11\times {10}^{-31}kg$ .

Therefore,

$1g={10}^{-3}kg=\left(\frac{1}{9.11\times {10}^{-31}}\right)\times {10}^{-3}electrons$

$\Rightarrow 1.098\times {10}^{27}electrons.$ .

Question 2.1 (ii) Calculate the mass and charge of one mole of electrons.

Answer :

As the mass of one electron is equal to $9.11\times {10}^{-31}kg$

Therefore, Mass of 1 mole or $6.022\times {10}^{23}$ electrons $=(9.11\times {10}^{-31})\times (6.022\times {10}^{23})=5.48\times {10}^{-7}kg$ .

Charge on one electron is $1.602\times {10}^{-19}coulomb$ .

Therefore, the charge on 1 mole of electrons will be:

$\Rightarrow (1.602\times {10}^{-19})\times (6.022\times {10}^{23})=9.65\times {10}^{4}coulombs$ .

Question 2.2 (i) Calculate the total number of electrons present in one mole of methane.

Answer :

1 molecule of methane $C{H}_4$ contains $6+4=10$ electrons.

Therefore, 1 mole of methane will contain:

$6.022\times {10}^{23}\times 10=6.022\times {10}^{24}$ electrons.

Question 2.2 (ii) Find (a) the total number of neutrons in 7 mg of ¹⁴ C. (Assume that mass of a neutron $=1.675\times {10}^{-27}kg$ ).

Answer :

As 1 atom of ¹⁴ C contains $=14-6=8neutrons$ .

and the number of atoms in ¹⁴ C in 1 mole is $6.022\times {10}^{23}$ atoms.

Therefore, the number of neutrons in 14g of ¹⁴ C in 1 mole $=8\times 6.022\times {10}^{23}neutrons$ .

The number of neutrons in $7mg$ :

$=\left(\frac{8\times 6.022\times {10}^{23}\times 7}{14000}\right)$

$=2.4088\times {10}^{21}neutrons$ .

Question 2.2 (ii) Find (b) the total mass of neutrons in 7 mg of ¹⁴ C. (Assume that mass of a neutron = 1.675 × 10 kg).

Answer :

As the mass of one neutron is $1.674\times {10}^{-27}kg$ .

Then the mass of total neutrons in 7grams of ¹⁴ C:

$=(2.4088\times {10}^{21})(1.675\times {10}^{-27}kg)=4.035\times {10}^{-6}kg$ .

Question 2.2 (iii) Find (a) the total number of protons in 34 mg of NH ₃ at STP.
Will the answer change if the temperature and pressure are changed ?

Answer :

1 mole of ammonia $N{H}_3=$ $17gN{H}_3=6.022\times {10}^{23}moleculesofN{H}_3$ .

and 1 atom of $N{H}_3$ contains $=7+3=10protons$ .

Therefore, the number of protons in 1 mole of $N{H}_3=6.022\times {10}^{24}protons$ .

Number of protons in 3mg of $N{H}_3$ : $=\frac{(6.022\times {10}^{24}\times 34)}{17}\times 1000=1.2044\times {10}^{22}protons$

No, there will be no effect of temperature and pressure.

Question 2.2 (iii) Find (b) the total mass of protons in 34 mg of NH ₃ at STP.

Will the answer change if the temperature and pressure are changed?

Answer :

As the mass of one proton is $1.6726\times {10}^{-27}kg$

Therefore, the mass of $1.2044\times {10}^{22}protons$ will be:

$=(1.6726\times {10}^{-27})\times (1.2044\times {10}^{22})kg=2.0145\times {10}^{-5}kg.$

No, there will be no effect of temperature and pressure.

Question 2.3(i) How many neutrons and protons are there in the following nuclei?

${}_6^{13}\text{C}$

Answer :

Given the nucleus of carbon: ${}_6^{13}\text{C}$

Atomic number (Z) = 6

Mass number (A) = 13

Number of protons (Z) = 6

Number of neutrons (A-Z) = 13-6 = 7

Question 2.3(ii) How many neutrons and protons are there in the following nuclei?

${}_8^{16}\text{O}$

Answer :

Given the nucleus of oxygen: ${}_8^{16}\text{O}$

Atomic number (Z) = 8

Mass number (A) = 16

Number of protons (Z) = 8

Number of neutrons (A-Z) = 16-8 = 8

2.3 How many neutrons and protons are there in the following nuclei?
Edit Q

Question 2.3(iii) How many neutrons and protons are there in the following nuclei?

${}_{12}^{24}\text{Mg}$

Answer :

Given the nucleus of Magesium: ${}_{12}^{24}\text{Mg}$

Atomic number (Z) = 12

Mass number (A) = 24

Number of protons (Z) = 12

Number of neutrons (A-Z) = 24-12 = 12

Question 2.3(iv) How many neutrons and protons are there in the following nuclei?

${}_{26}^{56}\text{Fe}$

Answer :

Given the nucleus of Iron: ${}_{26}^{56}\text{Fe}$

Atomic number (Z) = 26

Mass number (A) = 56

Number of protons (Z) = 26

Number of neutrons (A-Z) = 56-26 = 30

Question 2.3(v) How many neutrons and protons are there in the following nuclei?

${}_{38}^{88}\text{Sr}$

Answer :

Given the nucleus of Strontium: ${}_{38}^{88}\text{Sr}$

Atomic number (Z) = 38

Mass number (A) = 88

Number of protons (Z) = 38

Number of neutrons (A-Z) = 88-38 = 50

Question 2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(i) Z = 17 , A = 35.

Answer :

For the given atomic number Z=17 and mass number A=35;

Atom is ${}^{35}C{l}_{17}$ .

Question 2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(ii) Z = 92 , A = 233.

Answer :

For the given atomic number Z=92 and mass number A=233;

Atom is ${}^{233}{U}_{92}$ .

Question 2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(iii) Z = 4 , A = 9.

Answer :

For the given atomic number Z=4 and mass number A=9;

Atom is ${}^{9}B{e}_4$ .

Question 2.5 Yellow light emitted from a sodium lamp has a wavelength ( $\lambda$ ) of 580 nm. Calculate the frequency (ν) and wavenumber ( $\nu$ ) of the yellow light.

Answer :

Given the wavelength of the yellow light emitted from a sodium, lamp is $\lambda =580nm=580\times {10}^{-9}m$ .

And the frequency will be:

$\nu =\frac{c}{\lambda }=\frac{3.0\times {10}^{8}m/s}{580\times {10}^{-9}m}=5.17\times {10}^{14}{s}^{-1}$

Therefore the wavenumber,

$\vartheta =\frac{1}{\lambda }=\frac{1}{580\times {10}^{-9}m}=1.72\times {10}^6{m}^{-1}$

Question 2.6 Find energy of each of the photons which
(i) correspond to light of frequency $3\times {10}^{15}\text{Hz}$ .

Answer :

If a photon has a frequency of $\vartheta =3\times {10}^{15}Hz$ .

Then, the energy of each of the photons will be:

$E=h\vartheta =(6.626\times {10}^{-34}J.s)\times (3\times {10}^{15}{s}^{-1})$

$=1.988\times {10}^{-18}J$

Question 2.6 Find energy of each of the photons which

(ii) have wavelength of 0.50 Å.

Answer :

For the wavelength $\lambda =0.50\times {10}^{-10}m$ .

The energy of each of the photons will be:

$E=h\vartheta =\frac{hc}{\lambda }=\frac{(6.626\times {10}^{-34}J.s)\times (3\times {10}^{15}{s}^{-1})}{0.50\times {10}^{-10}m}=3.98\times {10}^{-15}J$

Question 2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is $2.0\times {10}^{-10}\text{s}$ .

Answer :

Given frequency, wavelength, and the wave number of a light wave:

$Frequency(\nu )=\frac{1}{Period}=\frac{1}{2.0\times {10}^{-10}s}=5\times {10}^9{s}^{-1}$

$Wavelength(\lambda )=\frac{c}{\nu }=\frac{3.0\times {10}^{8}m{s}^{-1}}{5\times {10}^9{s}^{-1}}=6.0\times {10}^{2}m$

$Wavenumber\left(\vartheta \right)=\frac{1}{\lambda }=\frac{1}{6.0\times {10}^2}m=16.66m^{-1}$

Question 2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?

Answer :

Given the wavelength of light $\lambda =4000pm=4000\times {10}^{-12}m=4\times {10}^{-9}m$ .

and Energy is 1J of energy:

$E=Nh\nu =Nh\frac{c}{\lambda }$

Therefore, the number of photons of light with a wavelength of 4000 pm that provides 1J of energy is:

$N=E\times \frac{\lambda }{hc}=(1J)\times \left(\frac{4\times {10}^{-9}m}{(6.626\times {10}^{-34}J.s)\times (3.0\times {10}^{8}m{s}^{-1})}\right)$

$=2.012\times {10}^{16}photons.$

Question 2.9(i) A photon of wavelength $4\times {10}^{-7}m$ strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV)

Answer :

The photon is having a wavelength of $4\times {10}^{-7}m$ strikes on a metal surface, where the work function of the metal being is $2.13eV$ .

So, Energy of the photon:

$E=h\nu =\frac{hc}{\lambda }=\frac{(6.626\times {10}^{-34}J.s)\times (3.0\times {10}^{8}m{s}^{-1})}{(4\times {10}^{-7}m)}$

$=4.97\times {10}^{-19}J$

$=\frac{4.97\times {10}^{-19}J}{1.602\times {10}^{-19}}eV$

$=3.10eV$

Question 2.9(ii) A photon of wavelength 4 × 10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (ii) the kinetic energy of the emission

Answer :

The photon is having a wavelength of $4\times {10}^{-7}m$ strikes on a metal surface, where the work function of the metal being is $2.13eV$ .

The kinetic energy of the emission will be:

$K.E.=h\nu -h{\nu }_o=3.10-2.13eV=0.97eV$

Question 2.9 A photon of wavelength 4 × 10 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (iii) the velocity of the photoelectron ( $1\text{eV}=1.6020\times {10}^{-19}\text{J}$ ) .

Answer :

The photon is having a wavelength of $4\times {10}^{-7}m$ strikes on a metal surface, where the work function of the metal being is $2.13eV$ .

From the previous part, we have the Kinetic Energy (K.E.):

$K.E.=\frac{1}{2}m{v}^2=0.97eV=0.97\times 1.602\times {10}^{-19}J$

$\Rightarrow \frac{1}{2}(9.11\times {10}^{-31}kg)\times v^2=0.97\times 1.602\times {10}^{-19}J$

$\because (massofanelectron=9.11\times {10}^{-31}kg)$

$\Rightarrow v^2=0.341\times {10}^{12}=34.1\times {10}^{10}m/s$

$\Rightarrow v=5.84\times {10}^{5}m/s$

Question 2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol ^–1 .

Answer :

Given the wavelength of the electromagnetic radiation is $242nm$ which is just sufficient to ionize the sodium atom.

So, the ionization energy required will be:

$E=Nh\nu =N\frac{hc}{\lambda }$

$=(6.022\times {10}^{23}mo{l}^{-1})\left(\frac{6.022\times {10}^{23}J.s\times 3.0\times {10}^{8}m{s}^{-1}}{242\times {10}^{-9}m}\right)$

$=4.945\times {10}^{5}Jmo{l}^{-1}=494.5kJmo{l}^{-1}$

Question 2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.

Answer :

Given that the light is monochromatic yellow of wavelength $\lambda =0.57\mu m$ .

Hence the energy emitted by the bulb will be:

$E=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m/s}{0.57\times {10}^{-6}m}=3.48\times {10}^{-19}J$

Therefore, the number of photons emitted per second:

$\frac{25J{s}^{-1}}{3.48\times {10}^{-19}J}=7.18\times {10}^{19}$

Question 2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν ₀ ) and work function (W ₀ ) of the metal.

Answer :

Given the wavelength of radiation is $6800A$ .

$Energygiven=Workfunction+Kineticenergy.$

But the electrons are emitted with zero velocity from a metal surface when it is exposed to radiation. That means the kinetic energy will be zero .

So, the Threshold frequency ${\nu }_o$ will be:

${\nu }_o=\frac{c}{{\lambda }_o}=\frac{3.0\times {10}^{8}m/s}{6800\times {10}^{-10}m}=4.14\times {10}^{14}{s}^{-1}$

and the Work function will be:

$W_o=h{\nu }_o=6.626\times {10}^{-34}Js\times 4.14\times {10}^{14}{s}^{-1}=2.92\times {10}^{-19}J$

Question 2.13 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?

Answer :

When an electron in a hydrogen atom undergoes a transition from an energy level with $n=4$ to an energy level $n=2$ , there will be an emission of energy whose wavelength can be found by:

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

$=109677(\frac{1}{2^2}-\frac{1}{4^2})c{m}^{-1}=20564.4c{m}^{-1}$

and wavelength will be equal to:

$\lambda =\frac{1}{\vartheta }=\frac{1}{20564.4}=486\times {10}^{-7}cm=486\times {10}^{-9}m=486nm$

Question 2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom ( energy required to remove the electron from n =1 orbit).

Answer :

The energy which is required to ionize an H atom if the electron occupies n=5 orbit is:

$E_n=\frac{-21.8\times {10}^{-19}}{n^2}Jato{m}^{-1}$

For ionization from 5th orbit, $n_1=5and{n}_2=\mathrm{\infty }$

Therefore, $\mathrm{△}E=E_2-E_1=-21.8\times {10}^{-19}\times (\frac{1}{n_2^2}-\frac{1}{n_1^2})=21.8\times {10}^{-19}\times (\frac{1}{n_1^2}-\frac{1}{n_2^2})$ $=21.8\times {10}^{-19}\times ({\frac{1}{5}}^2-\frac{1}{\mathrm{\infty }})=8.72\times {10}^{-20}J$

For ionization from 1st orbit, $n_1=1and{n}_2=\mathrm{\infty }$

Therefore,

$\mathrm{△}{E}^{\prime }=21.8\times {10}^{-19}\times ({\frac{1}{1}}^2-\frac{1}{\mathrm{\infty }})=21.8\times {10}^{-19}J$

$\frac{\mathrm{△}{E}^{\prime }}{\mathrm{△}E}=\frac{21.8\times {10}^{-19}}{8.72\times {10}^{-20}}=25$

Hence, 25 times less energy is required to ionize an electron in the 5th orbital of the hydrogen atom as compared to that in the ground state.

Question 2.15 What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Answer :

The number of lines produced when an electron from $n^{th}$ shell drops to the ground state:

$=\frac{n(n-1)}{2}$ .

According to the question, the maximum number of emission lines when the excited electron of an H atom in $n=6$ drops to the ground state will be:

$\frac{6(6-1)}{2}=15$

These are produced because of the following transitions:

$6\to 5$ $6\to 4$ $6\to 3$ $6\to 2$ $6\to 1$

$5\to 4$ $5\to 3$ $5\to 2$ $5\to 1$

$4\to 3$ $4\to 2$ $4\to 1$

$3\to 2$ $3\to 1$

$2\to 1$

Question 2.16 (i) The energy associated with the first orbit in the hydrogen atom is $-2.18\times {10}^{-18}{\text{J atom}}^{-1}$ . What is the energy associated with the fifth orbit?

Answer :

The energy associated with the first orbit in the hydrogen atom is $-2.18\times {10}^{-18}{\text{J atom}}^{-1}(Given)$

The energy of an electron in $n^{th}$ shell is given by:

$E=\frac{-2.18\times {10}^{-18}}{n^2}Jato{m}^{-1}$

So, the energy associated with the fifth orbit would be:

$E=\frac{-2.18\times {10}^{-18}}{5^2}Jato{m}^{-1}=\frac{-2.18\times {10}^{-18}}{25}Jato{m}^{-1}$

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

Question 2.16 (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Answer :

The radius of Bohr's $n^{th}$ orbit for the hydrogen atom is given by,

$r_n=(0.0529nm)n^2$

So, for $n=5$ we have

$r_5=(0.0529nm)(5{)}^2$

$r_5=1.3225nm$

Question 2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer :

Balmer formula:

$\vartheta =\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

As we can note from the formula that the wavenumber is inversely proportional to the wavelength.

Hence, for the longest wavelength transition in the Balmer series of atomic hydrogen wavenumber has to be the smallest or $n_2$ should be minimum i.e., $n_2=3$ .

For the Balmer series, $n_1=2$

Thus, the expression of wavenumber is given by,

$\vartheta =(1.097\times {10}^7{m}^{-1})[\frac{1}{2^2}-\frac{1}{3^2}]$

$\vartheta =(1.097\times {10}^7{m}^{-1})[\frac{1}{4}-\frac{1}{9}]$

$\vartheta =(1.097\times {10}^7{m}^{-1})\left[\frac{9-4}{36}\right]$

$\vartheta =(1.097\times {10}^7{m}^{-1})\left[\frac{5}{36}\right]=1.5236\times {10}^6{m}^{-1}$

Question 2.18 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 $-2.18\times {10}^{-11}\text{ergs}$ .

Answer :

The ground state energy:

$E_1=-2.18\times {10}^{-11}ergs$

$=-2.18\times {10}^{-11}\times {10}^{-7}J$

$=-2.18\times {10}^{-18}J$

The energy required to shift the electron from the first Bohr orbit to the fifth Bohr orbit is:

$\mathrm{△}E=E_5-E_1$

And the expression for the energy of an electron is given by:

$E_n=-\frac{2n^{2}m{e}^4{Z}^2}{n^2{h}^2}$

where m is mass of an electron, Z is the atomic mass of an atom, e is a charge of an electron, and h is the Planck's constant.

Now, substituting the values in the equation, we get

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

$=(2.18\times {10}^{-18})[1-\frac{1}{25}]$

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

Hence, the wavelength of the emitted light will be:

$\lambda =\frac{hc}{E}=\frac{(6.626\times {10}^{-34}Js)(3\times {10}^{8}m/s)}{(2.0928\times {10}^{-18}J)}$

$=9.498\times {10}^{-8}m$

Question 2.19 The electron energy in hydrogen atom is given by $E_n=(-2.18\times {10}^{-18})/n^2\text{J}$ . 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?

Answer :

The expression for the energy of an electron in hydrogen is:

$E_n=-\frac{2n^{2}m{e}^4{Z}^2}{n^2{h}^2}$

Where m is mass of electrons, Z is the atomic mass of an atom, e is the charge of an electron, and h is the Planck's constant.

and electron energy in the hydrogen atom is given by,

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

The electron energy in $n=2$ orbit is:

$E_n=-\frac{(2.18\times {10}^{-18})}{2^2}J=0.5465\times {10}^{-18}J$

Therefore, the energy required for the ionization from $n=2$ is $5.45\times {10}^{-19}J$

Now, the longest wavelength of light that can be used to cause this transition will be:

$E=\frac{hc}{\lambda }$

$\lambda =\frac{hc}{E}=\frac{(6.626\times {10}^{-34}Js)(3\times {10}^8)m/s}{5.45\times {10}^{-19}J}$

$=3.674\times {10}^{-7}m=3.674\times {10}^{-5}cm$

Question 2.20 Calculate the wavelength of an electron moving with a velocity of $2.05\times {10}^7{\text{ms}}^{-1}$ .

Answer :

The wavelength of an electron is given by the de Broglie's equation:

$\lambda =\frac{h}{mv}$

Where,

$\lambda$ is the wavelength of moving particle,

m is the mass of the particle, i.e., $9.11\times {10}^{-31}kg$

v is the velocity of the particle,i.e., $2.05\times {10}^7{\text{ms}}^{-1}$ ( Given)

and h is the Planck's constant value, i.e., $(6.626\times {10}^{-34}Js)$

Now, substituting the values in the equation, we get

$\lambda =\frac{(6.626\times {10}^{-34}Js)}{(9.11\times {10}^{-31}kg)(2.05\times {10}^{7}m/s)}=3.548\times {10}^{-11}m$

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

Question 2.21 The mass of an electron is $9.1\times {10}^{-31}\text{kg}$ If its K.E. is $3.0\times {10}^{-25}\text{J}$ , calculate its wavelength.

Answer :

The wavelength of an electron can be found by de Broglie's equation:

$\lambda =\frac{h}{mv}$

Given the K.E. of electron $3.0\times {10}^{-25}\text{J}$ which is equal to $\frac{1}{2}m{v}^2$ .

Hence we get,

$\frac{1}{2}m{v}^2=3.0\times {10}^{-25}J$

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(3.0\times {10}^{-25}J)}{9.1\times {10}^{-31}kg}}=811.579m/s$

Hence the wavelength is given by,

$\lambda =\frac{6.626\times {10}^{-34}Js}{(9.1\times {10}^{-31}kg)(811.579m/s)}$

$=8.9625\times {10}^{-7}m$

Question 2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons?

${\text{Na}}^{+},{\text{K}}^{+},{\text{Mg}}^{2+},{\text{Ca}}^{2+},{\text{S}}^{2-},\text{Ar}$

Answer :

Calculating the number of electrons for each species.

$Na$ has 11 electrons then, $N{a}^{+}$ will have $(11-1)=10$ electrons.

$K$ has 19 electrons then, $K^{+}$ will have $(19-1)=18$ electrons.

$Mg$ has 12 electrons then, $M{g}^{2+}$ will have $(12-2)=10$ electrons.

$Ca$ has 20 electrons then, $C{a}^{2+}$ will have $(20-2)=18$ electrons.

$S$ has 16 electrons then, $S^{2-}$ will have $(16+2)=18$ electrons.

$Ar$ has 18 electrons.

Hence, the following are isoelectronic species:

$N{a}^{+}andM{g}^{2+}$ having 10 electrons each.

$K^{+},C{a}^{2+},S^{2-},andAr$ having 18 electrons each.

Question : 2.23 (i) Write the electronic configurations of the following ion:

(a) ${\text{H}}^{-}$

Answer :

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

Now, the electronic configuration of ${\text{H}}^{-}$ will be $1s^2$ .

Question 2.23 (i) Write the electronic configurations of the following ion :

(b) ${\text{Na}}^{+}$

Answer :

The electronic configuration of $Na$ having $Z=11$ is : $1s^{2}2s^{2}2p^{6}3s^1$

Now, the electronic configuration of ${\text{Na}}^{+}$ will be $1s^{2}2s^{2}2p^6$ .

Question 2.23 (i) Write the electronic configurations of the following ion :

(c) ${\text{O}}^{2-}$

Answer :

The electronic configuration of $O$ having $Z=8$ is : $1s^{2}2s^{2}2p^4$

Now, the electronic configuration of ${\text{O}}^{2-}$ will be $1s^{2}2s^{2}2p^6$ .

Question 2.23 (ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a) ${\text{3s}}^1$

Answer :

With given outermost electrons ${\text{3s}}^1$ ,

The complete electronic configuration is $1s^{2}2s^{2}2p^{6}3s^1$

Hence the number of electrons present in the atom of the element is:

$=2+2+6+1=11$ .

Therefore, the atomic number of the element is $11$ which is Sodium (Na) .

Question 2.23 (ii) What are the atomic numbers of elements whose outermost electrons are represented by

(b) $2p^3$

Answer :

With given outermost electrons $2p^3$ ,

The complete electronic configuration is $1s^{2}2s^{2}2p^3$

Hence the number of electrons present in the atom of the element is:

$=2+2+3=7$ .

Therefore, the atomic number of the element is 7 which is NItrogen (N) .

Question 2.23 (ii) What are the atomic numbers of elements whose outermost electrons are represented by

(c) $3p^5$

Answer :

With given outermost electrons $3p^5$ ,

The complete electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$

Hence the number of electrons present in the atom of the element is:

$=2+2+6+2+5=17$ .

Therefore, the atomic number of the element is 17 which is Chlorine (Cl) .

Question 2.23 (iii) Which atoms are indicated by the following configurations?

(a) $\left[\text{He}\right]2s^1$

Answer :

The electronic configuration of the element is $\left[\text{He}\right]2s^1$ or $1s^{2}2s^1.$

Therefore, the atomic number of the element is 3 which is Lithium , a p-block element.

Question 2.23 (iii) Which atoms are indicated by the following configurations?

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

Answer :

The electronic configuration of the element is $[\text{Ne}]3s^{2}3p^3$ or $1s^{2}2s^{2}2p^{6}3s^{2}3p^3.$

Therefore, the atomic number of the element is 15 which is Phosphorus , a p-block element.

Question 2.23 (iii) Which atoms are indicated by the following configurations?

(c) $[Ar]4s^{2}3d^1$

Answer :

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

Therefore, the atomic number of the element is 21 which is Scandium , a d-block element.

Question 2.24 What is the lowest value of n that allows g orbitals to exist?

Answer :

For g-orbital, the value of Azimuthal quantum number (l) will be 4.

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

Therefore, for l =4, the minimum value of n should be 5.

Question 2.25 An electron is in one of the 3 d orbitals. Give the possible values of n , l and m _l for this electron.

Answer :

For d-orbital, the value of Azimuthal quantum number (l) = 2.

When l =2, the values of m are: $-2,-1,0,+1,+2$

Now, for the 3d orbital:

The value of Principal quantum number, $n=3$

Azimuthal quantum number, $l=2$

Magnetic quantum number, $m_l$ = $-2,-1,0,1,2$

Question 2.26(i) An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons

Answer :

Given an atom of an element contains 29 electrons and 35 neutrons.

Now, for an atom to be neutral, the number of protons is equal to the number of electrons.

Therefore, the number of protons in the atom of the given element will be 29 .

Question 2.26(ii) An atom of an element contains 29 electrons and 35 neutrons. Deduce (ii) the electronic configuration of the element.

Answer :

Given an atom of an element contains 29 electrons and 35 neutrons.

The electronic configuration of the atom will be:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}$ which is the electronic configuration of copper .

Question 2.27 Give the number of electrons in the species ${\text{H}}_2^{+}$ , ${\text{H}}_2$ and ${\text{O}}_2^{+}$

Answer :

The number of electrons in ${\text{H}}_2$ molecule is $(1+1)=2$ .

The number of electrons in ${\text{H}}_2^{+}$ molecule will be one less than the number of electrons in $H_2$ molecule. i.e, $(2-1)=1$ .

The number of electrons present in ${\text{O}}_2^{+}$ molecule will be one less than the number of electrons present in $O_2$ molecule. i.e., $(16-1)=15$

Question 2.28 (i) An atomic orbital has n = 3 . What are the possible values of l and m _l ?

Answer :

For a given value of the principal quantum number $(n)$ , the azimuthal quantum number $(l)$ can have values from $0$ to $(n-1)$ .

Therefore, for given atomic orbital $n=3$ ,

The value of $l$ can take values from $0$ to $(3-1)=2$ , i.e., $l=0,1,2$ .

And for a given value of $l$ , the Magnetic quantum number $m_l$ can have $(2l+1)$ values.

When value of $l=0$ then, $m=0$ ,

or $l=1$ then, $m=-1,0,+1$

or $l=2$ then, $m=-2,-1,0,+1,+2$

or $l=3$ then, $m=-3,-2,-1,0,+1,+2,+3$

Question 2.28 (ii) List the quantum numbers ( m _l and l ) of electrons for 3d orbital.

Answer :

For 3d-orbital, the values of Principal quantum number is $(n)=3$ and Azimuthal quantum number $(l)=2$ .

Therefore, for $(l)=2$ ,