
The ground state electronic configuration of neutral titanium atoms is _____.
Answer
556.5k+ views
Hint:Just keep in mind that for a neutral atom, the atomic number is just equal to the number of electrons. As we start to write the electron configuration for titanium, we should be aware of how many electrons it has. Atomic number informs us about the number of electrons a neutral atom has.
Complete step by step answer:
To know the atomic number we simply need to have a look in the periodic table.
For a neutral atom, the atomic number is just equal to the number of electrons. This is due to the atomic number depending on the number of protons it has. As that neutral means a charge of zero, so to acquire a zero charge, the positively charged particles that are protons must be equal to the negatively charged particle which is electrons.
With the help of a periodic table, we get to know that Titanium has an atomic number of $22$ so which means we have $22$ electrons as well.
Now, we write the no. of electrons that each orbital has in superscript. Remember that the maximum no. of electrons that s, p, d, and f have is $2$, $6$ and $10$ respectively. Then we continue writing through the pattern until the target number of electrons which is $22$ reached.
$ \Rightarrow $$3{d^2}4{s^2}$
$ \Rightarrow $$2 + 2 + 6 + 2 + 6 + 2 = 20$
As we need $22$ so $2$ electrons are still left.
As the next in the pattern is 3d. Note that from what we have noted earlier that d can have a maximum of ten electrons but it doesn't really need to have ten, we can actually have less. So with that, the electron configuration should be like
$ \Rightarrow $$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}4{s^2}$
Now as the final answer is concerned, we should write this as the condensed form using noble gases. This actually means that we change a part of the electron configuration above with a symbol of a noble gas. We decide that noble gas by looking at the last noble gas that just comes before our element. In this case, the last noble gas before Titanium is $Ar$. $Ar$ contains $18$ electrons hence it would replace the part of $18$ electrons in our ground state electron configuration:
$ \Rightarrow $ $\left[ {Ar} \right]4{s^2}3{d^2}$
Ti: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}4{s^2}$
Hence, the answer \[\left[ {Ar} \right]3{d^2}4{s^2}\].
Note:
Firstly we need to check the periodic table for checking the electrons titanium has and then need to write electronic configuration based on that. Electronic configuration should be written by following all the rules.
Complete step by step answer:
To know the atomic number we simply need to have a look in the periodic table.
For a neutral atom, the atomic number is just equal to the number of electrons. This is due to the atomic number depending on the number of protons it has. As that neutral means a charge of zero, so to acquire a zero charge, the positively charged particles that are protons must be equal to the negatively charged particle which is electrons.
With the help of a periodic table, we get to know that Titanium has an atomic number of $22$ so which means we have $22$ electrons as well.
Now, we write the no. of electrons that each orbital has in superscript. Remember that the maximum no. of electrons that s, p, d, and f have is $2$, $6$ and $10$ respectively. Then we continue writing through the pattern until the target number of electrons which is $22$ reached.
$ \Rightarrow $$3{d^2}4{s^2}$
$ \Rightarrow $$2 + 2 + 6 + 2 + 6 + 2 = 20$
As we need $22$ so $2$ electrons are still left.
As the next in the pattern is 3d. Note that from what we have noted earlier that d can have a maximum of ten electrons but it doesn't really need to have ten, we can actually have less. So with that, the electron configuration should be like
$ \Rightarrow $$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}4{s^2}$
Now as the final answer is concerned, we should write this as the condensed form using noble gases. This actually means that we change a part of the electron configuration above with a symbol of a noble gas. We decide that noble gas by looking at the last noble gas that just comes before our element. In this case, the last noble gas before Titanium is $Ar$. $Ar$ contains $18$ electrons hence it would replace the part of $18$ electrons in our ground state electron configuration:
$ \Rightarrow $ $\left[ {Ar} \right]4{s^2}3{d^2}$
Ti: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}4{s^2}$
Hence, the answer \[\left[ {Ar} \right]3{d^2}4{s^2}\].
Note:
Firstly we need to check the periodic table for checking the electrons titanium has and then need to write electronic configuration based on that. Electronic configuration should be written by following all the rules.
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