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 2px1)
ii) 1s2, 2px2
(Aufbau principle; 1s2, 2s2)
iii) 1s2, 2s2, 2px2 2py1
(Hund’s rule; 1s2, 2s2 2px1 2py1 2py1)
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 3
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 ms)
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 px, px and pz. 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 2px, 2py and 2pz orbitals, we will fill single electron in each orbital 2px↿, 2py↿,2pz↿ rather than double electrons 2px↿⇂ 2py↿⇂ 2pz. 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↿⇂ 2px↿ 2px↿ 2pz (Two unpaired electrons)
N = 7 = 1s↿⇂ 2s↿⇂ 2px↿ 2px↿ 2pz↿ (Three unpaired electrons)
O = 8 = 1s↿⇂ 2s↿⇂ 2px↿⇂ 2px↿ 2pz↿ (Two unpaired electrons)
F = 9 = 1s↿⇂ 2s↿⇂ 2px↿⇂ 2px↿⇂ 2pz↿ (One unpaired electron)
Ne = 10 = 1s↿⇂ 2s↿⇂ 2px↿⇂ 2px↿⇂ 2pz↿⇂ (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↿⇂ 2px↿ 2px 2pz (One unpaired electrons)
Excited State Electronic configuration of 5B =1s↿⇂ 2s↿ 2px↿ 2px↿ 2pz (Three unpaired electrons)
Ground State Electronic configuration of 6C =1s↿⇂ 2s↿⇂ 2px↿ 2px↿ 2pz (Two unpaired electrons)
Excited State Electronic configuration of 6C =1s↿⇂ 2s↿ 2px↿ 2px↿ 2pz (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 2px1)
ii) 1s2, 2px2
iii) 1s2, 2s2, 2px2 2py1
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.
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