Q. 6.1 - An Introduction to General, Organic, and Biological Chemistry

Question

State the number of electrons that must be lost by atoms of each of the following to achieve a stable electron arrangement:

a. $\mathrm{Li}$      b. $\mathrm{Ca}$      c. $\text{Ga}$      d. $\text{Cs}$      e. $\mathrm{Ba}$       

Step-by-Step Solution

Answer

a. One valence electron is lost by Lithium.

b. Two valence electrons are lost by Calcium.

c. Three valence electrons are lost by Gallium.

d. One valence electrons is lost by Caesium.

e. Two valence electrons are lost by Barium.

Step 1: Introduction(part a)

The given is $\mathrm{Li}$

We have to find the number of electrons lost by $\mathrm{Li}$

Step 2: Explanation(part a)

Lithium(Li) has an atomic number of$3$ and belongs to the IA group. As a result, Li's electron configuration is as follows:

$1s^{2 }2s^1$

The configuration of $L{i}^{+}$ is$1s^2$, which is the stable electron configuration of the noble gas helium if it loses one electron to form an ion with$+1$ charge.

As a result, one valence electron is lost by lithium in order to achieve a stable electron configuration.

Step 3: Introduction(part b)

The given is Ca.

We have to find the number of electrons lost by Ca.

Step 4: Explanation(part b)

Calcium(Ca) has an atomic number of $20$ and belongs to the IIA group. As a result, ca's electron configuration is as follows:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^2$

The configuration of $C{a}^{2+}$is$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^2$, which is the stable electron configuration of the noble gas argon (Ar) if it loses two electrons to form an ion with $+2$ charge.

As a result, two valence electron is lost by calcium in order to achieve a stable electron configuration.

Step 5: Introduction(part c)

The given is Ga.

We have to find the number of electrons lost by Ga.

Step 6: Explanation(part c)

Gallium (Ga) has the atomic number $31$ and belongs to the IIIA group. As a result, Ga's electron configuration is as follows:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^1$

The configuration of $G{a}^{3+}$is$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$, which is the stable electron configuration if it loses three electrons to form an ion with $+3$ charge.

As a result, three valence electron is lost by gallium in order to achieve a stable electron configuration.

Step 7: Introduction(part d)

The given is Cs.

We have to find the number of electrons lost by Cs.

Step 8: Explanation(part d)

Cesium has the atomic number $55$ and belongs to the IA group. As a result, Ga's electron configuration is as follows:

$1s^{2}2s^{2}2p^{6}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^1$

The configuration of $C{s}^{+}$ is $1s^{2}2s^{2}2p^{6}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^1$, which is the stable electron configuration of the noble gas Xenon (Xe) if it loses one electron to form an ion with +1 charge.

As a result, one valence electron is lost by cesium in order to achieve a stable electron configuration.

Step 9: Introduction(part e)

The given is Ba.

We have to find the number of electrons lost by Ba.

Step 10: Explanation(part e)

Barium has an atomic number of 56 and belongs to the IIA group. As a result, Ba's electron configuration is as follows:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$

The configuration of $B{a}^{2+}$is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$, which is the stable electron configuration of the noble gas Xenon (Xe) if it loses two electron to form an ion with $+2$ charge.

As a result, two valence electron is lost by Barium in order to achieve a stable electron configuration.