Unit 2: Unit 2: Atomic Structure — Short Questions

Ans. The p-subshell has three values of magnetic quantum number that can be calculated by m = 2 l +1 formula.
l = 1 for p subshell
So, m = 2(1) + 1 = 2+1 = 3 orbitals and they have values −1, 0, +1.

Ans. Mg has two valance electrons. And it is easy to remove two valance electrons, but for 3rd electron a much bigger amount of energy required due to disturbance in octet rule. That's why I3 is much bigger than I2.

Ans. The least first ionization energy is of alkali metals. And among alkali metals it decreases downward. So, among the given elements lowest ionization energy is of K and highest value of Kr which is a noble gas that has complete octet, as ionization energy increase from left to night in period.

Ans. (i) 19K = 1s2 2s2 2p6 3s2 3p6 4s1

(ii) Electrons are filled according to n + l rule. The values of 4s is n + l = 4+0 = 4, while that of 3d, the values of n + l = 3+2 = 5. So, because the values of 4s of n + l is smaller than 3d that's why electrons are filled in 4s first then in 3d.

(iii) As per ground state configuration: 4pz electrons = 2
Because maximum electrons that can occupy an orbital are 2.

(iv) According to ground state configuration:
Total electrons are 80, the total number of valance electron are 12. The total spin "up" valance electrons (-1/2) are 06.

Ans. (i) An element X has atomic number 17 means it has 17 protons and 17 electrons because X has no charge. While neutrons are 18 (n = A−Z)
Z = 17 p = 17
X has no charge e = 17

(ii) 35X1−
p = 17 Z = 17
e = 18 X1− = 17+1 = 18
n = 18 n = A−Z = 35−17 = 18
Because of -1 charge, it have total of 18 electrons.

Ans. (i) 80 Hg 1s2 2s2 2p6 3s2 3p6 3d10 4p6 4s2 3d10 4p6 5s2
4d10 5p6 4f14 5d10 6s2
Atomic orbitals with n = 3
3s2 3p6 3d10
Electrons = 2+6+10 = 18

(ii) As per ground state configuration: 4d electrons = 10

(iii) As per ground state configuration: 4pz electrons = 2
Because maximum electrons that can occupy an orbital are 2.

(iv) According to ground state configuration:
Total electrons are 80, the total number of valance electron are 12. The total spin "up" valance electrons (-1/2) are 06.

Ans. The massive difference between 2nd and 3rd ionization energies show that 3rd electron is removed from inner shell. It means atom has two electrons in its valence shell. And has total four electrons so, it should be 4 Be and it is of alkaline earth metals s-block (IIA).

Ans. (i) The trend in ionization energies as given is in increasing order as the electrons are removed, because of increasing attraction of nucleus and valance electrons.
Al < Al+ < Al2+ < Al3+

(ii) The large increase between I3 and I4 is due to the difference in their shells. I3 is of valence shell while I4 is of inner shell, that's why it has large difference and massive jump can be seen. Al' has 3 electrons in its valance shell.

Ans. (i) The general order of filling orbitals up to 4p is:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

(ii) According to Aufbau principle the energy of 4s is less than the energy of 3d orbital so, it is filled first.
The values for 4s of n+ l rule is 4+0 = 4 while for 3d is 3+2 = 5, that's why it is filled before 3d.

Ans. 15P = 1s2 2s2 2p6 3s2 3p1 3p1 3p1
Valence shell configuration:
[1s] [1 1 1 1]
3s 3px 3py 3pz
Orbital box diagram of valence electrons of phosphorus.

Ans. It is a relationship between frequency of a particular line in the x-rays and atomic number of element √ν = a (Z - b) a → Proportionality constant b → screening constant. Moseley stated that it is the atomic number and not the atomic mass which describes the properties of an atom. He formulated the modern periodic table.

Ans. 16O2−
e = 8+2 = 10
p = 8
n = 16-8 = 8

Ans. Atoms having same atomic masses but different atomic numbers are called isobars.
Examples: 14N & 16O, 12C & 14N

Ans. Atoms having same atomic numbers but different mass numbers are called isotopes.
Examples: 1H 2H 1H 12C 13C

Ans. Atoms having same number of neutrons are called isotones.
Example: 1H & 4He, 12C & 16O
n = 2, n = 8

Ans. The atoms or ions having same number of electrons are called isoelectronic elements or systems.
Examples: F-, Ne, Na+, Mg2+, All have 10 electrons

Ans. Positive rays consist of protons and cathode rays consist of electrons. Mass of proton is 1836 times greater than that of electron. So e/m value for positive rays will be 1836 times less as compared to cathode rays.

Ans. When the spectral lines are split up into a number of thin lines in the presence of electric field then that is called as Stark effect.

Ans. When emission spectrum is studied in the presence of magnetic field. The spectral lines are split up into thinner lines it is called as Zeeman effect.

Ans. Neutrons have following properties.
(i) They have high penetrating power.
(ii) They cannot ionize gases.
(iii) They can expel high speed protons from paraffin water, paper and cellulose.
(iv) They have no effect of electric and magnetic field.

Ans. K = 19
1s2 2s2 2p6 3s2 3p6 4s1
Cu = 29
1s2 2s2 2p6 3s2 3p6 4s1 3d10

Ans.
(i) Fe (26)
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6
(ii) Br (35)
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p5
(iii)Cr (24)
1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5

Ans. The formula is 2(2ℓ+1)
For s-orbital 2(2×0+1) = 2 electrons
For p-orbital 2(2×1+1) = 6 electrons
For d-orbital 2(2×2+1) = 10 electrons
For f-orbital 2(2×3+1) = 14 electrons

Ans. The orbitals are arranged in the increasing order of (n+ℓ) if any two orbitals have the same (n+ℓ) values, then that orbital should be placed first whose 'n' value is smaller.
For example: (n+ℓ)
1s (1+0) = 1
2s (2+0) = 2 and so on.

Ans. A spectrum formed by the radiation after being absorbed by the absorbing substance is called as atomic absorption spectrum.
It is represented by dark lines separated by bright spaces. It appears as dark lines against the bright background.
The spectrum of light when passed through an excited hydrogen atom is the atomic absorption spectrum.

Ans. "A spectrum formed by the radiations emitted by the heated substance is called as atomic emission spectrum".
It is represented by bright lines separated by dark spaces. It appears as bright lines on a dark background.

Ans. We can calculate the energy of an electron experimentally by measuring the ionization energy of an electron.
Example: The energy of electron of hydrogen atom is actually its 1st ionization energy.
H(g) ----→ H(g)+e- ΔH1I = +1313.315kJ/mol
Na(g) ----→ Na(g)+e- ΔH1I = +496kJ/mol

Ans. Amount of energies required to remove all the electrons one by one until the nucleus is left. For example: An element "X" has three electrons, its successive ionization energies are:
X ----→ X+ + e- ΔH11
X+ ----→ X2+ + e- ΔH12
X2+ ----→ X3+ + e- ΔH13

Ans. All the ionization energies are strongly endothermic because they required energy to separate the electrons from an atom. Due to attractive forces of nucleus.

Ans.
(a) n = 1, ℓ = 0, m = 0 (b) n = 2, ℓ = 0, m = 0

Ans.
n = 2, ℓ = 1, m = 0, −1, +1

Ans. The total number of electrons in a shell can be calculated by "2n2" (where n is orbital)
K shell (n=1); 2 electrons (1s orbital)
L shell (n=2); 8 electrons (2s and 2p orbitals)
M shell (n=3); 18 electrons (3s, 3p and 3d orbitals)
N shell (n=4); 32 electrons (4s, 4p, 4d and 4f orbitals)

Ans. We can identify the position of an element based on its valance shell electron ended up in the particular subshell. If the valance electron presents in s-subshell, then the element present in s-block. Similarly, p, d and f block elements have their valance electrons in their respective subshells.
Examples: Na is s-block element & O is p-block element.

Ans. An element X-19 has electronic configuration as: X = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 as per valance shell configuration, the valance electron present in 4s. So, it is s-block element. And 4s1 shows it is of 1st group of s-block, means it is an alkali metal present in 4th period as of 4s end of configuration and the element is K−19.

Ans. The number of electrons in a subshell can be calculated by 2(2ℓ+1) formula.

For s, ℓ = 0
s: 2(2ℓ+1) = 2(2(0)+1) = 2(1) = 2 electrons

For p, ℓ = 1
p: 2(2ℓ + 1) = 2(2(1)+1) = 2(3) = 6 electrons

For d, ℓ = 2
d: 2(2ℓ + 1) = 2(2(2)+1) = 2(5) = 10 electrons

For f, ℓ = 3
f: 2(2ℓ + 1) = 2(2(3) +1) = 2(7) = 14 electrons

Ans. Mg has two electrons in its valance s-subshell, so due to the full filled rule, Mg requires more energy than Al' because it has one electron in its valance p subshell that makes Al unstable as compared to Mg, although Mg has bigger size than Aℓ.

Ans. The electrons in an atom in the outermost shell are called valance electrons. These are such electron that are primarily involved in chemical reactions. The similarities among the configurations of valance electrons account for similarities of chemical properties among groups of elements.

Ans. Semi-conductors are the materials that can conduct electricity under some conditions. They conduct electricity more than insulator and less than conductors. Some examples are Si, Ge, As, Sb, etc.

Ans. Impurity atoms with three valence electrons (like Al) are added to the pure semiconductor. Some of the trivalent atoms take place of the Si atoms in the crystals. The silicon atoms cannot make four bonds due to the lack of electrons. For this reason, there are created holes in the crystal lattice, which act as positive charge carriers. This process creates a positive-type semiconductor or P-type semiconductor.

Ans. When impurity atoms with five valence electrons (like phosphorus) are added to the pure semiconductor, some of Si atoms are replaced with the pentavalent phosphorus atoms. The Si atoms in the vicinity of these atoms can make four bonds and the fifth electron is an extra electron. These impurity atoms contribute extra electrons to the crystal lattice, which become free to move and act as negative charge carriers. This process creates N-type semiconductor.

Ans. See Q5. from theory.

Ans. See Q6. from theory.

Ans. See Q4. from theory.