Explain why there is a phenomenal decrease in ionisation enthalpy from carbon to silicon.
(N/A) The ionisation enthalpy of carbon (the first element of group $14$) is very high $(1086 \ kJ/mol)$. This is expected due to its small atomic size.
However,on moving down the group to silicon,there is a sharp decrease in the ionisation enthalpy $(786 \ kJ/mol)$.
This significant decrease is primarily due to the appreciable increase in the atomic size of elements as we move down the group,which reduces the effective nuclear attraction on the valence electrons.
However,on moving down the group to silicon,there is a sharp decrease in the ionisation enthalpy $(786 \ kJ/mol)$.
This significant decrease is primarily due to the appreciable increase in the atomic size of elements as we move down the group,which reduces the effective nuclear attraction on the valence electrons.
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Similar Questions
The $CORRECT$ order of first ionisation enthalpy is:
Electronic configurations of four elements $A$,$B$,$C$,$D$ are given below:
$A$) $1s^2 2s^2 2p^6 3s^1$
$B$) $1s^2 2s^2 2p^6 3s^2 3p^1$
$C$) $1s^2 2s^2 2p^6 3s^2$
$D$) $1s^2 2s^2 2p^6 3s^2 3p^2$
The correct order of first ionization enthalpy of these elements is:
$A$) $1s^2 2s^2 2p^6 3s^1$
$B$) $1s^2 2s^2 2p^6 3s^2 3p^1$
$C$) $1s^2 2s^2 2p^6 3s^2$
$D$) $1s^2 2s^2 2p^6 3s^2 3p^2$
The correct order of first ionization enthalpy of these elements is:
Which of the following electronic configurations represents the highest second ionization energy for an element?
Easy
View SolutionIn the $2^{nd}$ period,which element has the highest value of the sum of the $1^{st}$ and $2^{nd}$ ionization energy $(I.E.)$?
Medium
View SolutionThe electronic configurations of elements $A, B$ and $C$ are $[He] 2s^1$,$[Ne] 3s^1$ and $[Ar] 4s^1$ respectively. Which one of the following orders is correct for the first ionization potentials (in $kJ \ mol^{-1}$) of $A, B$ and $C$?
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