
The correct order of atomic radii is:
(i) $ Y{b^{3 + }} < P{m^{3 + }} < C{e^{3 + }} < L{a^{3 + }} $
(ii) $ C{e^{3 + }} < Y{b^{3 + }} < P{m^{3 + }} < L{a^{3 + }} $
(iii) $ Y{b^{3 + }} < P{m^{3 + }} < L{a^{3 + }} < C{e^{3 + }} $
(iv) $ P{m^{3 + }} < L{a^{3 + }} < C{e^{3 + }} < Y{b^{3 + }} $
Answer
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Hint :In the given question all the ions mentioned are $ 4f $ block elements or the lanthanide series. Lanthanide series shows a unique property known as lanthanide contraction which can be defined as the greater than expected decrease in ionic radii of the elements as the atomic number of the element increases. Use this concept to find out the order.
Complete Step By Step Answer:
The elements given in the question are: $ Yb,\,Pm,\,Ce,\,La $ which are elements belonging to the $ 4f $ block or the lanthanide series. Now, the electronic configurations are as follows:
$ Yb:\,\,\left[ {Xe} \right]4{f^{14}}6{s^2} $
$ Pm:\,\,\left[ {Xe} \right]4{f^5}6{s^2} $
$ Ce:\,\,\left[ {Xe} \right]4{f^1}5{d^1}6{s^2} $
$ La:\,\,\left[ {Xe} \right]5{d^1}6{s^2} $ .
We can see the atomic number increases as $ La \to Ce \to Pm \to Yb $ along a period.
Now we know that the lanthanide series shows a unique property of lanthanide contraction where there is a decrease in atomic radii with increase in atomic number. There the increasing order of atomic radii follows the order: $ Yb < Pm < Ce < La $ . The ionic radii also follows the same order and the increasing order is given by: $ Y{b^{3 + }} < P{m^{3 + }} < C{e^{3 + }} < L{a^{3 + }} $ .
Hence the correct answer is (i) $ Y{b^{3 + }} < P{m^{3 + }} < C{e^{3 + }} < L{a^{3 + }} $ .
Additional Information:
As the atomic number increases the positive charge on the nucleus increases whereas the electrons are added to the $ 4f $ subshell. As the shielding effect of the $ 4f $ subshell is less than the $ d $ subshell, the valence shell is pulled more tightly via the nucleus which results in the decrease of the size of the elements with an increase in atomic number and is known as the lanthanide contraction.
Note :
For doing this question you must have a proper idea about the periodic table and the different elements present in the $ s,\,p,\,d,\,f $ blocks of the periodic table. You must also have an idea about lanthanide contraction as you may think that size will increase with increasing atomic number due to decrease in effective nuclear charge.
Complete Step By Step Answer:
The elements given in the question are: $ Yb,\,Pm,\,Ce,\,La $ which are elements belonging to the $ 4f $ block or the lanthanide series. Now, the electronic configurations are as follows:
$ Yb:\,\,\left[ {Xe} \right]4{f^{14}}6{s^2} $
$ Pm:\,\,\left[ {Xe} \right]4{f^5}6{s^2} $
$ Ce:\,\,\left[ {Xe} \right]4{f^1}5{d^1}6{s^2} $
$ La:\,\,\left[ {Xe} \right]5{d^1}6{s^2} $ .
We can see the atomic number increases as $ La \to Ce \to Pm \to Yb $ along a period.
Now we know that the lanthanide series shows a unique property of lanthanide contraction where there is a decrease in atomic radii with increase in atomic number. There the increasing order of atomic radii follows the order: $ Yb < Pm < Ce < La $ . The ionic radii also follows the same order and the increasing order is given by: $ Y{b^{3 + }} < P{m^{3 + }} < C{e^{3 + }} < L{a^{3 + }} $ .
Hence the correct answer is (i) $ Y{b^{3 + }} < P{m^{3 + }} < C{e^{3 + }} < L{a^{3 + }} $ .
Additional Information:
As the atomic number increases the positive charge on the nucleus increases whereas the electrons are added to the $ 4f $ subshell. As the shielding effect of the $ 4f $ subshell is less than the $ d $ subshell, the valence shell is pulled more tightly via the nucleus which results in the decrease of the size of the elements with an increase in atomic number and is known as the lanthanide contraction.
Note :
For doing this question you must have a proper idea about the periodic table and the different elements present in the $ s,\,p,\,d,\,f $ blocks of the periodic table. You must also have an idea about lanthanide contraction as you may think that size will increase with increasing atomic number due to decrease in effective nuclear charge.
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