Answer
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Hint: Transition elements are those which have unpaired electrons either in their ground state or their most stable oxidation state. All d-block elements are not transition elements but all transition elements are d block elements.
Complete step by step answer:
Here we are discussing the first transition series, i.e. from scandium to zinc containing 10 elements.
Oxidation state is defined as the charge present on any atom in combined form.
All the transition elements except the first and last element of the series, exhibit variable oxidation state.
Now we will find the number of elements having 4 electrons in their +2 oxidation state
Since scandium has atomic number 21. So its electronic configuration is \[3{d^1}4{s^2}\].
Scandium in +2 oxidation state have electronic configuration as
\[S{c^{ + 2}} = 3{d^1}\], i.e. only 1 unpaired electrons
Now Ti has atomic number 22.
So its electronic configuration is \[3{d^2}4{s^2}\]
After losing 2 electrons it gets the electronic configuration as
\[T{i^{ + 2}} = 3{d^2}\], i.e. 2 unpaired electrons
In case of V, the atomic number is 23.
So electronic configuration is \[3{d^3}4{s^2}\].
In +2 oxidation state \[{V^{ + 2}} = 3{d^3}\], i.e. only 3 unpaired electrons
Similarly in case of \[Cr\] due to exceptional case electronic configuration is \[3{d^5}4{s^1}\].
In case of +2 oxidation state i.e. \[C{r^{ + 2}} = 3{d^4}\].
That means there are 4 unpaired electrons
In the case of \[M{n^{ + 2}} = 3{d^5}\], i.e. 5 unpaired electrons.
Then in \[F{e^{ + 2}} = 3{d^6}\]. Here pairing gets started hence 4 unpaired electrons are present.
In \[C{o^{ + 2}} = 3{d^7}\] here also pairing occurs hence only 3 unpaired electrons are present.
In \[N{i^{ + 2}} = 3{d^8}\]. Here 2 unpaired electrons are present.
In the case of \[C{u^{ + 2}}=3{d^9}\]. Only 1 electron due to exceptional configuration.
In the case of \[Z{n^{ + 2}} = {\rm{ }}no {\rm{ }} unpaired {\rm{ }}electrons\].
Hence the correct answer is option (A)
Note:
Since zinc has no unpaired electrons in their ground state as well as in oxidation state so it is not considered to be a transition element. It is the only d block element.
Complete step by step answer:
Here we are discussing the first transition series, i.e. from scandium to zinc containing 10 elements.
Oxidation state is defined as the charge present on any atom in combined form.
All the transition elements except the first and last element of the series, exhibit variable oxidation state.
Now we will find the number of elements having 4 electrons in their +2 oxidation state
Since scandium has atomic number 21. So its electronic configuration is \[3{d^1}4{s^2}\].
Scandium in +2 oxidation state have electronic configuration as
\[S{c^{ + 2}} = 3{d^1}\], i.e. only 1 unpaired electrons
Now Ti has atomic number 22.
So its electronic configuration is \[3{d^2}4{s^2}\]
After losing 2 electrons it gets the electronic configuration as
\[T{i^{ + 2}} = 3{d^2}\], i.e. 2 unpaired electrons
In case of V, the atomic number is 23.
So electronic configuration is \[3{d^3}4{s^2}\].
In +2 oxidation state \[{V^{ + 2}} = 3{d^3}\], i.e. only 3 unpaired electrons
Similarly in case of \[Cr\] due to exceptional case electronic configuration is \[3{d^5}4{s^1}\].
In case of +2 oxidation state i.e. \[C{r^{ + 2}} = 3{d^4}\].
That means there are 4 unpaired electrons
In the case of \[M{n^{ + 2}} = 3{d^5}\], i.e. 5 unpaired electrons.
Then in \[F{e^{ + 2}} = 3{d^6}\]. Here pairing gets started hence 4 unpaired electrons are present.
In \[C{o^{ + 2}} = 3{d^7}\] here also pairing occurs hence only 3 unpaired electrons are present.
In \[N{i^{ + 2}} = 3{d^8}\]. Here 2 unpaired electrons are present.
In the case of \[C{u^{ + 2}}=3{d^9}\]. Only 1 electron due to exceptional configuration.
In the case of \[Z{n^{ + 2}} = {\rm{ }}no {\rm{ }} unpaired {\rm{ }}electrons\].
Hence the correct answer is option (A)
Note:
Since zinc has no unpaired electrons in their ground state as well as in oxidation state so it is not considered to be a transition element. It is the only d block element.
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