Page No 69: - Chapter 2 Structure Of Atom class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 2.63:

The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?

Answer:

Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant 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 atom 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.

Question 2.64:

Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p

Answer:

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.

(i) 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.

(ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.

(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

Question 2.65:

The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Answer:

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.

Question 2.66:

Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Answer:

(a) Phosphorus (P):

Atomic number = 15

The electronic configuration of P is:

1s² 2s² 2p⁶ 3s² 3p³

The orbital picture of P can be represented as:

From the orbital picture, phosphorus has three unpaired electrons.

(b) Silicon (Si):

Atomic number = 14

The electronic configuration of Si is:

1s² 2s² 2p⁶ 3s2 3p²

The orbital picture of Si can be represented as:

From the orbital picture, silicon has two unpaired electrons.

(c) Chromium (Cr):

Atomic number = 24

The electronic configuration of Cr is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵

The orbital picture of chromium is:

From the orbital picture, chromium has six unpaired electrons.

(d) Iron (Fe):

Atomic number = 26

The electronic configuration is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

The orbital picture of chromium is:

From the orbital picture, iron has four unpaired electrons.

(e) Krypton (Kr):

Atomic number = 36

The electronic configuration is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

The orbital picture of krypton is:

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

Question 2.67:

(a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having ms value of –1/2 for n = 4?

Answer:

(a) n = 4 (Given)

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

∴ l = 0, 1, 2, 3

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

(b) Number of orbitals in the n^th shell = n²

For n = 4

Number of orbitals = 16

If each orbital is taken fully, then it will have 1 electron with ms value of .

∴ Number of electrons with ms value of   is 16.