IX Chemistry Model Test Questions Chemistry Test # 6 for Chapter # 2 (Atomic Structure)

 

Model Test Questions Chemistry Test # 6 for Chapter # 2 (Atomic Structure)

Short Questions Answers

 

Q1. What are Limitations of Bohr's Atomic Model?

Q2. Differentiate between shell and sub shell with examples?

Q3.What is maximum number of electrons that can be accommodate in's' subshell?

Q4. How many electrons will be in L shell of an atom having atomic number 11?

Q5. In the distribution of electrons of an atom, which shell filled first and why?

Q6. If both K and L shells of an atom are completely filled, what is the total number of electrons are present  in them?

Q7. An atom has 5 electrons in M shell than:

(a) Find out its atomic number?

(b) Write Electronic configuration of atom?

(c) Name the element of atom?

Q8. Describe wave particle duality of electron of De Broglie Hypothesis?

             

Long Questions Answers

Q9. Prove that modern theory of De Broglie is related with Einstein and Plank's equations.

Q10. Describe the schrodinger atomic model.

Q11.Explain how Bohr's atomic model is different from Rutherford atomic model.

OR

State postulates of Bohr’s atomic model.

Q12. Write down electronic configuration of B, F, N, Na, P, Cl, Ca, K⁺, O^2-, S^2-, Mg²⁺, Cl^-.

 

 

جو ہم  نہ ہوں تو زمانے کی سانس  رک  جائے

قتیلؔ  وقت  کے  سینے  میں  ہم  دھڑکتے  ہیں

Solutions of Questions 

Q1. What are Limitations of Bohr's Atomic Model?

Answer

1.  It failed to explain the Zeeman Effect (effect of magnetic field on the spectra of atoms).

2.   It also failed to explain the Stark effect (effect of electric field on the spectra of atoms).

3.  It is against the Heisenberg Uncertainty Principle.

4.   It could not explain the spectra obtained from larger atoms.

5.   It explained the mono-electronic species like H⁺, Li²⁺, B³⁺.

Q2. Differentiate between shell and sub shell with examples?

Q8. Describe wave particle duality of electron of De Broglie Hypothesis?

Answer 

In 1923 Lois De Broglie extend the wave particle duality to electron, and propose a hypothesis that all matter has particle as well as wave nature at the submicroscopic level.

 

De Broglie combined the Einstein and Planck equations to derive its de-Broglei’ equation. The wave nature of a particle is quantified by De Broglie wavelength defined as l = h/p where p is the momentum of the particle.

All matter (material) particles in motion have a dual character exhibiting both particle and wave nature i.e. electrons, protons, neutrons, atoms, and molecules possess the characteristics of both the material particle and a wave. This is called wave-particle Duality in matter.

 

de-Broglie derived a mathematical equation known as de-Broglie’s equation which relates the wavelength (l) of the material particle (electron) of mass m moving with velocity v to its momentum (mv = p):

Q9. Prove that modern theory of De Broglie is related with Einstein and Plank's equations.

Answer 

Q10. Describe the schrodinger atomic model.

Answer

In 1926 Erwin Schrödinger, an Austrian physicist, took the Bohr’s atomic model one step forward. Schrödinger used mathematical equations to describe the likelihood of finding an electron in a certain position. This atomic model is known as the quantum mechanical model of the atom.

The quantum mechanical model determines that electron can be find in various location around the nucleus. He found electrons are in orbit as an electron cloud.

Basic Postulates

1. The quantum mechanical model determines that electron can be found in various location around the  nucleus as an electron cloud (Electrons in orbits are found as an electron cloud).

2. Each energy subshell in an orbit has different shapes which determine the presence of electron.

3. Different subshells or orbitals are named as s, p, d and f with different shapes. e.g. s-orbital is spherical and p-orbital is dumbbell shaped.

4. The numbers and kind of atomic orbitals depends on the energy subshell.

 

According to quantum mechanical model probability of finding an electron within certain volume of space surrounding the nucleus can be represented as a fuzzy cloud. The cloud is denser the probability of finding

electron is high which are called atomic orbitals. 

Q11. Explain how Bohr's atomic model is different from Rutherford atomic model.

OR

State postulates of Bohr’s atomic model.

Answer

Introduction

After Planck and Einstein’s discoveries, a Danish Physicist Neil Bohr in 1913 offered a theoretical explanation of line spectra of hydrogen atom and proposed a new model for structure of atom based on Planck’s Quantum theory of Max Planck.

 

Basic Postulates

Postulate # 1

The atom has a number of fixed energy orbits or energy levels called stationary orbit’s or permissible orbits in which electron revolves around the nucleus. Bohr adopted Planck’s idea that energies are quantized. These orbits or shells have certain fixed amount of energy and are named as K, L, M, N. Electrons in atoms move only in certain allowed energy levels in which they are completely stable and will not emit or absorb energy continuously and therefore will not spiral (fall) into the nucleus.

Postulate # 2

When an electron jumps from lower energy level (E₁) to higher energy level (E₂), it absorbs a definite discrete quantity of energy called quantum of energy. When electron jumps from higher energy level (E₂) to lower energy level (E₁), it emits energy (as a light in a definite discrete quantity called quantum of energy). A quantum of energy is directly proportional to the frequency of the radiation. Since the process of emission of energy is not continuous, so a line spectrum is produced by atoms.     

     

Postulate # 3

an electron cannot revolve in an arbitrary orbits but only those orbits are permissible (which are called allowed orbits or stationary states) in which angular momentum of electrons is an integral (whole number) multiple of h/2p.

mvr = nh/2p. 

[Where; n = Principal quantum number = number of orbits = 1, 2, 3 …., h = Planck’s constant]

Answer

Boron     (₅B)    = 5 eˉ     = 1s², 2s² 2p¹

Carbon   (₆C)    = 6 eˉ     = 1s², 2s² 2p²

Nitrogen (₇N)   = 7 eˉ     = 1s², 2s² 2p³

Fluorine  (₉F)    = 9 eˉ     = 1s², 2s² 2p⁵

                                                                                               

Sodium   (₁₁Na) = 11 eˉ  = 1s², 2s² 2p⁶, 3s¹

Phosphorus (₁₅P) = 15 eˉ= 1s², 2s² 2p⁶, 3s² 3p³

Chlorine   (₁₇Cl) = 17 eˉ = 1s², 2s² 2p⁶, 3s² 3p⁵

Calcium   (₂₀Ca) = 20 eˉ = 1s², 2s² 2p⁶, 3s² 3p⁶, 4s²

                                                                                               

K⁺ = 19 –1 = 18 eˉ  = 1s², 2s² 2p⁶, 3s² 3p⁶

                                                                                               

O²⁻ = 8 + 2 =10 eˉ   = 1s², 2s² 2p⁶

S²⁻ = 16 + 2 = 18 eˉ = 1s², 2s² 2p⁶, 3s² 3p⁶

                                                                                               

Mg²⁺= 12 – 2 = 10 eˉ = 1s², 2s² 2p⁶

                                                                                               

Cl⁻  = 17 + 1= 18 eˉ  = 1s², 2s² 2p⁶, 3s² 3p⁶

 

 

XI Chemistry Test Model Questions for Chapter # 2 (Atomic structure)

 

Q1. Differentiate between Orbit and Orbital

Q2. What are quantum numbers? Give a brief account of 4 quantum numbers. Write all possible value of l, m and s for n=2   and n=3

Q3. Write down the four Quantum numbers of both electrons of Helium atom.

Q4. State and illustrate the following rules of electronic configuration.

 (i) Pauli Exclusion principle 

(ii) Hund’s rule of maximum multiplicity

Q5. Write down the Electronic Configuration of Boron and Carbon atom in ground state and excited state.

Q6. Draw shapes of orbitals for third energy level (l=0, l=1, l=2) (s, p and d-orbitals).

Q7. Arrange the following energy levels in ascending order using (n+l) rule:  

5d, 3s, 4f, 7s, 6p, 2p

Q8. Write down the E.C. of  S, Na+, Cl⁻, Cr, Fe, Cu, Ag, Mo, Br⁻, I, P³⁻, S²⁻, C⁴⁻, Cu⁺, Sr²⁺, Ca²⁺, Mg²⁺, Al³⁺, Fe²⁺, Fe³⁺

(i) S     (Z = 16) = 1s², 2s², 2p⁶, 3s², 3p⁴                       

(No. of electrons in S (atom) = Z =16)

(ii) Na⁺ (Z =11) = 1s², 2s², 2p⁶    

 (No. of electrons in Na⁺ (cation) = Z – charge = 11 – 1 = 10) 

(iii)  Cl⁻  (Z = 17) = 1s², 2s², 2p⁶, 3s², 3p⁶                     

(No. of electrons in Cl⁻ (anion) = Z + charge = 17+ 1 = 18)

Q9. Which rule and principle is violated in writing the following E.C.

i)  1s², 2s³                                             

(Pauli’s exclusion principle; 1s², 2s²  2p_x¹)  

 ii)  1s², 2p_x²                                                                

(Aufbau principle; 1s², 2s²)                 

 iii) 1s², 2s², 2p_x² 2p_y¹                        

(Hund’s rule; 1s², 2s² 2p_x¹ 2p_y¹ 2p_y¹)           

iv) 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁴ 4s³         

(Pauli’s exclusion principle and Hund’s rule; 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁵ 4s¹ )

Q10.  Identify the orbital of higher energy in the following pairs

(i) 4s and 3d (3d >4s)       

(ii) 4f and 6p (6p>4f)        

(iii) 5p and 6s (6s>5p)     

(iv) 4d and 4f (4f>4d)  

Q11.  Explain why the filling of electron is 4s orbital takes place prior to 3d?

Answers of Model Test Questions 2 

Q1. Differentiate between Orbit and Orbital

Answer

Difference between Orbit and Orbital

Q2. What are quantum numbers? Give a brief account of 4 quantum numbers. Write all possible value of l, m and s for n=2 and n=3

Answer

Definition

The solution of Schrodinger’s Wave Equation gives a set of mathematical integers or constant numbers called Quantum Numbers which describe energy levels, sub-levels and orbitals available for electrons

A set of constant integral numbers which describe the complete behaviour (position and energy) of an electron in an orbital showing energy of an electron, shape of orbital, orientation of orbital in space around the nucleus and the direction of movement (spin) of an electron in an orbital obtained by solving Schrodinger’s Wave Equation are called Quantum Numbers.

 

Types of Quantum Numbers

There are four Quantum Numbers, the first three are the solutions of Schrodinger Wave Equation:

 

Principle Quantum Number (n)

1.  It represents orbits or shell.

     e.g.  n  =  1 then electrons are in k-shell.

2. It gives size of the orbit.  As value of n increases, the size of orbit increases.

3.  It also gives energy of electron.

4. Its value gives maximum number of electrons in an orbit by 2n² formula.

 

Azimuthal / Subsidiary Quantum No. (l)

1. Its values show shape of orbitals. It has values l = 0 to (n – 1). e.g.

l = 0, used for s-orbital, spherical in shape

 l = 1, used for p-orbital, dumb-bell in shape

l = 2, used for d-orbital, double dumb-bell in shape

l = 3, used for f-orbital, complicated shape

2.‘l’ values show different sub-shells which are represented by small letters s, p, d, f e.g.

l  =  0 stands for s-orbital

l  =  1 stands for p-orbital

l  =  2 stands for d-orbital

3.  The maximum number of orbitals in an orbit are determined by the formula (2l + 1). 

4.  Its values show capacity of electrons in a sub-shell by 2(2l + 1) formula

 

Magnetic Quantum Number (m)

1. It gives different orientations or directions of an orbital in space in a magnetic field.

2. It shows further breaking up s, p, d, f orbitals.

3.The value of m depends upon the values of l.

4.Orbitals of same sub-shell having different orientations but same energy shapes are called degenerated orbitals.

Spin Quantum Number (s or m_s)

1. It specifies the spin of electron in an orbital.

2.It shows that electrons are in a pair, spin in anti-clock and clockwise direction. Its values are +½ for anticlockwise spin (↿) and –½ for clockwise spin (⇂).

 

all possible value of l, m and s for n=2 and n=3

Q3.  Write down the four Quantum numbers of both electrons of Helium atom.

Answer

In He (Z = 2), there are two electrons in K-shell. It can be seen that these two electrons have same values for n, l and m but due to opposite electron spins, they have different value of spin quantum number.

Q4. State and illustrate the following rules of electronic configuration

(i) Pauli Exclusion principle 

(ii) Hund’s rule of maximum multiplicity

Answer

Pauli’s Exclusion Principle

Introduction

It is an empirical rule but agrees fully with experimental observations. It was put forward by Wolfgang Pauli in 1925 A.D.  It is used to assign the values of four quantum numbers to an electron of an atom.

Statement

In an orbital of an atom, no two electrons can have the same set of four quantum numbers, at least one quantum number must be different. Thus an orbital can contain a maximum of two electrons with opposite spins.

 

Applications or Significance

1.From the Pauli’s principle, it is concluded that an orbital can accommodate only two electrons and these two electrons must have opposite spins (↿⇂).

2. According to this principle, two electrons in an orbital may have same values of three quantum numbers (n, l, m) but the value of fourth quantum number (s) must be different. It means that if one electron of same orbital has clockwise spin then second electron must have anti-clockwise spin.

 

Examples

In He (Z = 2), there are two electrons in s-orbital of first shell (K-shell i.e. He  =  1s²  or 1s^↿⇂

The set of four quantum numbers will be written as:

 

It shows that, these two electrons have same values for n, l and m but due to opposite electron spin, they have different value of spin quantum number.

 

Hund’s Rule of Maximum Multiplicity

Degenerated Orbitals

Orbitals of same sub-shell possess same energy and are known as degenerated orbitals. The orbitals given by a particular value of ‘l’ if n is same, have the same energy and such orbitals of equal energy are called Degenerate Orbitals. For example; p-sub shell consists of three same energy degenerate orbitals p_x, p_x and p_z. Similarly five orbitals of d-sub shell are also degenerated orbitals

 

Introduction

The filling of degenerate orbitals with electrons takes place according to Hund’s Rule of Maximum Multiplicity given by German Physicist Friedrich Hund in 1927.

 

Statement

In available degenerated orbitals of p, d and f-subshells, electrons are distributed in such a way that maximum number of half-filled orbitals (single electron in orbital) are obtained.

 

For example

if we have three electron to fill the 2p_x, 2p_y and 2p_z orbitals, we will fill single electron in each orbital 2p_x^↿, 2p_y^↿,2p_z^↿  rather than double electrons 2p_x^↿⇂ 2p_y^↿⇂ 2p_z. Unpaired electrons are more stable than paired electrons because electron create repulsion.

Examples

Electronic configuration of some elements considering Hund’s rule are given below:

C = 6   = 1s^↿⇂  2s^↿⇂  2p_x^↿  2p_x^↿  2p_z    (Two unpaired electrons)

N = 7      = 1s^↿⇂  2s^↿⇂  2p_x^↿  2p_x^↿  2p_z^↿   (Three unpaired electrons)

O = 8      = 1s^↿⇂  2s^↿⇂  2p_x^↿⇂  2p_x^↿  2p_z^↿      (Two unpaired electrons)

F = 9      = 1s^↿⇂  2s^↿⇂  2p_x^↿⇂  2p_x^↿⇂  2p_z^↿      (One unpaired electron)

Ne = 10 = 1s^↿⇂  2s^↿⇂  2p_x^↿⇂  2p_x^↿⇂  2p_z^↿⇂ (No unpaired electron)

 

Q5. Write down the Electronic Configuration of Boron and Carbon atom in ground state and excited state

Answer

Ground State Electronic configuration of ₅B =1s^↿⇂  2s^↿⇂  2p_x^↿  2p_x 2p_z  (One unpaired electrons)

excited State Electronic configuration of ₅B =1s^↿⇂  2s^↿  2p_x^↿  2p_x^↿  2p_z      (Three unpaired electrons)

Ground State Electronic configuration of ₆C =1s^↿⇂  2s^↿⇂  2p_x^↿  2p_x^↿  2p_z    (Two unpaired electrons)

excited State Electronic configuration of ₆C =1s^↿⇂  2s^↿  2p_x^↿  2p_x^↿  2p_z      (Four unpaired electrons)

 

Q9.  Which rule and principle is violated in writing the following E.C.

 i)  1s², 2s³                                            

(Pauli’s exclusion principle; 1s², 2s²  2p_x¹)                

ii)  1s², 2p_x²                                                            

iii) 1s², 2s², 2p_x² 2p_y¹                      

 iv) 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁴ 4s³   

Answer 

Q6. Draw shapes of orbitals for third energy level (l=0, l=1, l=2) (s, p and d-orbitals).

Answer

 

Q7. Arrange the following energy levels in ascending order using (n+l) rule:  5d, 3s, 4f, 7s, 6p, 2p

Answer 

Q8. Write down the E.C. of  S, Cr, Fe, Cu, Ag, Mo, Br⁻, I, P³⁻, S²⁻, C⁴⁻, Cu⁺, Sr²⁺, Ca²⁺, Mg²⁺, Al³⁺, Fe²⁺, Fe³⁺, Na⁺, Cl⁻

Answer

Sulphur; ₁₆S (Z = 16)  = 16 eˉ  = 1s², 2s², 2p⁶, 3s², 3p⁴                        

(No. of electrons in S atom = Z =16)

Chromium ₂₄Cr (Z = 24) = 24 eˉ = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁵, 4s¹

Copper; ₂₉Cu (Z = 29) = 29 eˉ  = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰, 4s¹

Molybdenum; ₄₂Mo (Z = 42) = 42 eˉ   = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d⁵, 5s¹

Silver; ₄₇Ag (Z = 47)  = 47 eˉ   = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d¹⁰, 5s¹

Bromine; ₃₅Br (Z = 35) = 35 eˉ   = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰, 4s², 4p⁵

Iodine; ₅₃I (Z = 53)  = 53 eˉ   = 1s², 2s² 2p⁶, 3s² 3p⁶, 4s², 3d¹⁰, 4p⁶, 4d¹⁰, 5s² 5p⁵

₃₈Sr²⁺    (Z = 38)   = 38 –2    =  36 eˉ  = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰, 4s², 4p⁶   or   [Kr]

₁₁Na⁺(Z =11) = 11–1 = 10 eˉ   = 1s², 2s², 2p⁶    or   [Ne]               

(No. of electrons in Na⁺ (cation) = Z – charge = 11 – 1 = 10)       

₁₂Mg²⁺ (Z =12)    = 12– 2    = 10 eˉ   = 1s², 2s² 2p⁶     or   [Ne]

₁₃Al³⁺    (Z =13)    = 13– 2    = 10 eˉ   = 1s², 2s² 2p⁶        or   [Ne]

₆C⁴ˉ       (Z = 6)     = 06 + 4   = 10 eˉ  = 1s², 2s² 2p⁶        or   [Ne]

₁₇Cl⁻ (Z = 17)= 17+1 = 18 eˉ   = 1s², 2s², 2p⁶, 3s², 3p⁶   or   [Ar] 

(No. of electrons in Cl⁻ (anion) = Z + charge = 17+1=18)

₁₇Clˉ  (Z = 17) = 17 + 1   = 18 eˉ  = 1s², 2s² 2p⁶, 3s² 3p⁶  or   [Ar]

₁₆S²ˉ; (Z = 16)   = 16 + 2   = 18 eˉ  = 1s², 2s² 2p⁶, 3s² 3p⁶   or  [Ar]

₂₆Fe²⁺ (Z = 26) = 26–2 = 24 eˉ = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁶, 4s⁰  

or   

[Ar] 3d⁶

₂₆Fe³⁺ (Z = 26) = 26 –3 = 23 eˉ   = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁵, 4s⁰  or   

[Ar] 3d⁵

 

 

Q10. Identify the orbital of higher energy in the following pairs

(i) 4s and 3d (3d >4s)    

(ii) 4f and 6p (6p>4f)     

(iii) 5p and 6s (6s>5p) 

(iv) 4d and 4f (4f>4d)  

Q11.  Explain why the filling of electron is 4s orbital takes place prior to 3d?

Answer

According to the Aufbau principle, atomic orbitals are filled in order of increasing energies. The energy of the orbital can be determined by the (n+l) rule:

For 3d orbital; n = 3 and l=2; so its n + l = 3 + 2 = 5

For 4s orbital; n = 4 and l=0, so its n + l = 4 + 0 = 4

Thus the energy of 4s orbital is less than 3d orbital, on the basis of (n+l) rule, 4s orbital is filled prior to 3d orbital.