Chapter 2: Structure of Atom

Ans: From Heisenberg’s uncertainty principle,

\[\Delta x\times \Delta p=\frac{h}{4\pi }\Rightarrow \Delta p=\frac{1}{\Delta x}\frac{h}{4\pi }\] 

Where, $\Delta x$ = uncertainty in position of the electron

$\Delta p$ = uncertainty in momentum of the electron

Substituting the values in the expression of$\Delta p$,

$\Delta p=\frac{1}{0.002nm}\frac{6.626\times {{10}^{-34}}Js}{4\times 3.14}$ 

$\Rightarrow \Delta p=2.637\times {{10}^{-23}}kgm{{s}^{-1}}$ 

Actual momentum = $\frac{h}{4\pi m\times 0.05nm}$= $\frac{6.626\times {{10}^{-34}}Js}{4\times 3.14\times 5.0\times {{10}^{-11}}m}=1.055\times {{10}^{-24}}kgm{{s}^{-1}}$ 

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

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

For n = 3 and l = 2, the orbital occupied is 3d. 

For n = 4 and l = 1, the orbital occupied is 4p. 

Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively. 

Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).

Ans: Nuclear charge experienced by an electron (present in a multi-electron atom) is dependent upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases. 

Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atoms with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.

Ans: Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron(s) in it.

The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.

Ans: 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.

Ans: 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

Ans: Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom. 

The higher the atomic number, the higher is the nuclear charge. 

Silicon has 14 protons while aluminium has 13 protons. 

Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). 

Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

Ans: Atomic number = 15

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

The orbital picture of P can be represented as:

From the orbital picture, phosphorus has three unpaired electrons.

Ans: Atomic number = 14

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

The orbital picture of Si can be represented as:

From the orbital picture, phosphorus has two unpaired electrons.

Ans: Atomic number = 24

The electronic configuration of Cr is: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{1}}3{{d}^{5}}$ 

The orbital picture of chromium is:

From the orbital picture, chromium has six unpaired electrons.

Ans: Atomic number = 26

The electronic configuration is: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{6}}$ 

The orbital picture of iron is:

From the orbital picture, iron has four unpaired electrons.

e)  Krypton (Kr)

Ans: Atomic number = 36

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

The orbital picture of krypton is:

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

Ans: n = 4 (Given) 

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

Hence, l = 0, 1, 2, 3

Thus, four subshells are associated with n = 4, which are s, p, d and f.

Ans: Number of orbitals in the ${{n}^{th}}$ shell =${{n}^{2}}$ 

for n = 4

Number of orbitals = 16

If each orbital is taken fully, then it will have 1 electron with  ${{m}_{s}}$  value of $-\frac{1}{2}$

$\therefore $ Number of electrons with ${{m}_{s}}$  value of $-\frac{1}{2}$ = 16