The electronic configuration of titanium ion $(\text{T}{{\text{i}}^{2+}})$ is:
A. $(\text{Ar})3{{d}^{2}}\text{ 4}{{\text{s}}^{2}}$
B. $(\text{Ar})3{{d}^{2}}$
C. $(\text{Ar})3{{d}^{2}}\text{ 4}{{\text{s}}^{1}}$
D. $(\text{Ar})\text{4}{{\text{d}}^{1}}$
Answer
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Hint: For this problem, firstly we should know the atomic number of titanium after which when two electrons are lost and the formation of titanium ions take place. Then we can write the electronic configuration of the titanium ion.
Complete Step-by-step answer:
- In the given question, we have to choose the correct electronic configuration of titanium ions in the form of a noble gas.
- As we know that titanium is placed in the periodic in period 4 as well as in group 4.
- So, as we know that the parent or neutral atom has the atomic number of 22 and when we will remove 2 electrons we will get titanium ion that is $\text{T}{{\text{i}}^{2+}}$.
- So, the number of electrons in the titanium ion will be 20.
- So, the electronic configuration of titanium will be:
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{2}}$ or $\left( \text{Ar} \right)\text{4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{2}}$
- Now, the loss of electrons will take place through that shell which will have the maximum energy.
- According to the Aufbau principle, the energy of the 3d is slightly more than the 4s orbital so, electrons will first fill the 4s shell.
- Now, after the 4s shell completes, the energy of the 4s orbital becomes more than the 3d orbital so the loss of electrons will take place from the 4s orbital.
- That's why the electronic configuration of the titanium ion will be:
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 3}{{\text{d}}^{2}}$ or $\left( \text{Ar} \right)\text{3}{{\text{d}}^{2}}$
Therefore, option B is the correct answer.
Note: In the electronic configuration of titanium, the noble gas argon was used because before titanium, argon is an only noble gas which comes. Argon has the atomic number of 18 and has the electronic configuration of $\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}$.
Complete Step-by-step answer:
- In the given question, we have to choose the correct electronic configuration of titanium ions in the form of a noble gas.
- As we know that titanium is placed in the periodic in period 4 as well as in group 4.
- So, as we know that the parent or neutral atom has the atomic number of 22 and when we will remove 2 electrons we will get titanium ion that is $\text{T}{{\text{i}}^{2+}}$.
- So, the number of electrons in the titanium ion will be 20.
- So, the electronic configuration of titanium will be:
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{2}}$ or $\left( \text{Ar} \right)\text{4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{2}}$
- Now, the loss of electrons will take place through that shell which will have the maximum energy.
- According to the Aufbau principle, the energy of the 3d is slightly more than the 4s orbital so, electrons will first fill the 4s shell.
- Now, after the 4s shell completes, the energy of the 4s orbital becomes more than the 3d orbital so the loss of electrons will take place from the 4s orbital.
- That's why the electronic configuration of the titanium ion will be:
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 3}{{\text{d}}^{2}}$ or $\left( \text{Ar} \right)\text{3}{{\text{d}}^{2}}$
Therefore, option B is the correct answer.
Note: In the electronic configuration of titanium, the noble gas argon was used because before titanium, argon is an only noble gas which comes. Argon has the atomic number of 18 and has the electronic configuration of $\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}$.
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