How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
(N/A) The first ionization enthalpy of $Na$ is lower than that of $Mg$ due to two main reasons:
$1.$ The atomic size of $Na$ $([Ne] 3s^1)$ is larger than that of $Mg$ $([Ne] 3s^2)$.
$2.$ The effective nuclear charge of $Mg$ is higher than that of $Na$.
Consequently,the energy required to remove the first electron from $Mg$ is higher than that for $Na$.
However,the second ionization enthalpy of $Na$ is higher than that of $Mg$. After the loss of the first electron,$Na^+$ attains the stable noble gas configuration $(1s^2 2s^2 2p^6)$. Removing a second electron from this stable configuration requires a very high amount of energy.
In contrast,$Mg^+$ $([Ne] 3s^1)$ still has one electron in the $3s$-orbital. Removing this electron is relatively easier compared to removing an electron from the stable noble gas core of $Na^+$. Therefore,the second ionization enthalpy of $Na$ is significantly higher than that of $Mg$.
$1.$ The atomic size of $Na$ $([Ne] 3s^1)$ is larger than that of $Mg$ $([Ne] 3s^2)$.
$2.$ The effective nuclear charge of $Mg$ is higher than that of $Na$.
Consequently,the energy required to remove the first electron from $Mg$ is higher than that for $Na$.
However,the second ionization enthalpy of $Na$ is higher than that of $Mg$. After the loss of the first electron,$Na^+$ attains the stable noble gas configuration $(1s^2 2s^2 2p^6)$. Removing a second electron from this stable configuration requires a very high amount of energy.
In contrast,$Mg^+$ $([Ne] 3s^1)$ still has one electron in the $3s$-orbital. Removing this electron is relatively easier compared to removing an electron from the stable noble gas core of $Na^+$. Therefore,the second ionization enthalpy of $Na$ is significantly higher than that of $Mg$.
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Similar Questions
Among the second period elements,the actual ionization enthalpies are in the order $Li < B < Be < C < O < N < F < Ne$. Explain why:
$(i)$ $Be$ has higher $\Delta_{i}H$ than $B$
$(ii)$ $O$ has lower $\Delta_{i}H$ than $N$ and $F$?
$(i)$ $Be$ has higher $\Delta_{i}H$ than $B$
$(ii)$ $O$ has lower $\Delta_{i}H$ than $N$ and $F$?
Difficult
View Solution$A$ sudden large jump between the values of second and third ionisation energies of an element would be associated with the electronic configuration:
Consider the following changes:
$M_{(s)} \to M_{(g)}$ ........$(1)$
$M_{(s)} \to M^{2+}_{(g)} + 2e^-$ .......$(2)$
$M_{(g)} \to M^{+}_{(g)} + e^-$ .........$(3)$
$M^{+}_{(g)} \to M^{2+}_{(g)} + e^-$ .........$(4)$
$M_{(g)} \to M^{2+}_{(g)} + 2e^-$ ..........$(5)$
The second ionization energy of $M$ could be calculated from the energy values associated with:
$M_{(s)} \to M_{(g)}$ ........$(1)$
$M_{(s)} \to M^{2+}_{(g)} + 2e^-$ .......$(2)$
$M_{(g)} \to M^{+}_{(g)} + e^-$ .........$(3)$
$M^{+}_{(g)} \to M^{2+}_{(g)} + e^-$ .........$(4)$
$M_{(g)} \to M^{2+}_{(g)} + 2e^-$ ..........$(5)$
The second ionization energy of $M$ could be calculated from the energy values associated with:
As one moves along a given row in the periodic table,ionization energy
Easy
View SolutionGiven below are two statements:
Statement $I$: The decrease in first ionization enthalpy from $B$ to $Al$ is much larger than that from $Al$ to $Ga$.
Statement $II$: The $d$ orbitals in $Ga$ are completely filled.
In the light of the above statements,choose the most appropriate answer from the options given below:
Statement $I$: The decrease in first ionization enthalpy from $B$ to $Al$ is much larger than that from $Al$ to $Ga$.
Statement $II$: The $d$ orbitals in $Ga$ are completely filled.
In the light of the above statements,choose the most appropriate answer from the options given below:
DifficultJEE MAIN 2023
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