Chapter 28, Problem 14Q

To determine

Part (a) To Determine:

The check and explain the validity of electronic configuration in the following ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}{\text{3s}}^{\text{2}}{\text{4p}}^{\text{4}}$

Answer to Problem 14Q

Solution:

This is invalid configuration because it violates the Aufbau Principle at 2p, 4p orbitals.

Explanation of Solution

The Aufbau Principle states that electrons will fill the lower energy orbitals completely before moving on to a higher energy state. The series is given as:$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}5d^{10}4f^{14}6p^6\mathrm{....}$

We see that the 2p orbital can accommodate two more electrons. Also, instead of 4p, 3p should be used. The correct configuration is $1s^{2}2s^{2}2p^{6}3s^2$, Calcium.

To determine

Part (b) To Determine:

The check and explain the validity of electronic configuration in the following ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}{\text{2p}}^8{\text{3s}}^1$

Answer to Problem 14Q

Solution:

This is invalid configuration because the 2porbital cannot accommodate more than 6 electrons.

Explanation of Solution

The Aufbau Principle states that electrons will fill the lower energy orbitals completely before moving on to a higher energy state. The series is given as:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}5d^{10}4f^{14}6p^6\mathrm{....}$

Here, the 2p orbital has two extra orbitals. Here the correct configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$, Aluminium.

To determine

Part (c) To Determine:

The check and explain the validity of electronic configuration in the following ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^5{\text{4s}}^2{\text{4d}}^6{\text{4f}}^1$

Answer to Problem 14Q

Solution:

This is invalid configuration because it violates the Aufbau Principle at 3p, 4f, 4dorbitals.

Explanation of Solution

The Aufbau Principle states that electrons will fill the lower energy orbitals completely before moving on to a higher energy state. The series is given as:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}5d^{10}4f^{14}6p^6\mathrm{....}$

Here, the 3p orbital can accommodate another electron. Also, instead of 4d, 3d subshell is used, and then 4p instead of 4f. The correct configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^5$