From the ground state electronic configuratio | vedclass

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(B) The second ionization energy is the energy required to remove an electron from a unipositive ion $(M^+)$.$1$. For option $A$ $(Mg)$: $Mg^+ (1s^2,

Vedclass · Accepted Answer

$1s^2, 2s^2, 2p^6, 3s^1$

Vedclass · Answer

(B) The second ionization energy is the energy required to remove an electron from a unipositive ion $(M^+)$.$1$. For option $A$ $(Mg)$: $Mg^+ (1s^2, 2s^2, 2p^6, 3s^1) \rightarrow Mg^{2+} + e^-$. The electron is removed from the $3s$ orbital.$2$. For option $B$ $(Na)$: $Na^+ (1s^2, 2s^2, 2p^6) \rightarrow Na^{2+} + e^-$. The electron is removed from the stable noble gas configuration $(2p^6)$,which requires very high energy.$3$. For option $C$ $(Ne)$: $Ne^+ (1s^2, 2s^2, 2p^5) \rightarrow Ne^{2+} + e^-$. The electron is removed from the $2p$ orbital.$4$. For option $D$ $(F)$: $F^+ (1s^2, 2s^2, 2p^4) \rightarrow F^{2+} + e^-$. The electron is removed from the $2p$ orbital.Comparing these,removing an electron from the stable octet configuration of $Na^+$ requires the highest energy.

Element	$IE_1 + IE_2$ $(kJ/mol)$	$IE_3 + IE_4$ $(kJ/mol)$
$P$	$2.45$	$8.82$
$Q$	$2.85$	$6.11$

From the ground state electronic configuration of the elements given below,pick up the one with the highest value of second ionization energy.

Similar Questions

The first ionization enthalpies of $Na, Mg$ and $Si$ are respectively $496, 737$ and $786 \ kJ \ mol^{-1}$. What would be the first ionization enthalpy of $Al$ in $kJ \ mol^{-1}$?

Which of the following statements is incorrect?

Which of the following orders is correct for the property mentioned in brackets?

The sum of $IE_1 + IE_2$ and $IE_3 + IE_4$ for elements $P$ and $Q$ are given below:

Element $IE_1 + IE_2$ $(kJ/mol)$ $IE_3 + IE_4$ $(kJ/mol)$

$P$ $2.45$ $8.82$

$Q$ $2.85$ $6.11$

Then,according to the given information,the incorrect statement$(s)$ is/are:

The correct order of first ionisation enthalpy of group-$13$ elements is

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