What is the correct order of ionisation energy:
$$(1)K < Cu < C{u^ + } < {K^ + }$$
$$(2)K < C{u^ + } < Cu < {K^ + }$$
$$(3)C{u^ + } < K < Cu < {K^ + }$$
$$(4){K^ + } < C{u^ + } < Cu < K$$
Answer
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Hint: Ionisation energy can be defined as the energy required to remove a loosely bound valence electron from an isolated gaseous atom. Ionisation energy is always positive as energy is always given to the atom to remove an electron. It can vary on the basis of some parameters like size, electronegativity, charge and so on.
Complete answer:
Ionisation energy can be defined as the energy required to remove a loosely bound valence electron from an isolated gaseous atom. So, we know that, when we give energy to a system (in this case an atom), the value of energy is always positive.
Ionisation energy can be arranged in correct order by looking at the electronic configuration of given atoms and ions. So, now we will write down the electronic configuration for each given atom or ion.
$K = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}$
${K^ + } = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
$Cu = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^{10}}$
$C{u^ + } = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}$
So we know that the cations have high ionisation energy as compared to atoms because of high nuclear charge and the electrons are tightly bound.
Now, there are ten d electrons in the Cu atom which have poor shielding effect and the nuclear charge will be holding the valence electron tighter than in the case of a potassium atom as there are no d electrons. So, Cu will have greater ionisation energy than K.
${K^ + }$ has a noble gas configuration so it is very stable and has extremely high ionisation energy than the $C{u^ + }$ .
Therefore, from the above discussion we can say that the correct order of ionisation energy is:
$$K < Cu < C{u^ + } < {K^ + }$$
The correct option is: $$(1)K < Cu < C{u^ + } < {K^ + }$$.
Note:
The ionisation energy of any atom or ion changes with many parameters. If the size of the atom is small, then it will have greater ionisation energy. If the atom is electronegative in nature, it will not lose any electron and hence, it will have greater ionisation energy. More the positive charge on an ion, more will be its ionisation energy.
Complete answer:
Ionisation energy can be defined as the energy required to remove a loosely bound valence electron from an isolated gaseous atom. So, we know that, when we give energy to a system (in this case an atom), the value of energy is always positive.
Ionisation energy can be arranged in correct order by looking at the electronic configuration of given atoms and ions. So, now we will write down the electronic configuration for each given atom or ion.
$K = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}$
${K^ + } = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
$Cu = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^{10}}$
$C{u^ + } = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}$
So we know that the cations have high ionisation energy as compared to atoms because of high nuclear charge and the electrons are tightly bound.
Now, there are ten d electrons in the Cu atom which have poor shielding effect and the nuclear charge will be holding the valence electron tighter than in the case of a potassium atom as there are no d electrons. So, Cu will have greater ionisation energy than K.
${K^ + }$ has a noble gas configuration so it is very stable and has extremely high ionisation energy than the $C{u^ + }$ .
Therefore, from the above discussion we can say that the correct order of ionisation energy is:
$$K < Cu < C{u^ + } < {K^ + }$$
The correct option is: $$(1)K < Cu < C{u^ + } < {K^ + }$$.
Note:
The ionisation energy of any atom or ion changes with many parameters. If the size of the atom is small, then it will have greater ionisation energy. If the atom is electronegative in nature, it will not lose any electron and hence, it will have greater ionisation energy. More the positive charge on an ion, more will be its ionisation energy.
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