Electronic configurations of four elements A, | vedclass

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(D) The electronic configurations are:$A$: $1s^2 2s^2 2p^6 3s^1$ (Sodium,$Na$)$B$: $1s^2 2s^2 2p^6 3s^2 3p^1$ (Aluminium,$Al$)$C$: $1s^2 2s^2 2p^6 3s^

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$D > C > B > A$

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(D) The electronic configurations are:$A$: $1s^2 2s^2 2p^6 3s^1$ (Sodium,$Na$)$B$: $1s^2 2s^2 2p^6 3s^2 3p^1$ (Aluminium,$Al$)$C$: $1s^2 2s^2 2p^6 3s^2$ (Magnesium,$Mg$)$D$: $1s^2 2s^2 2p^6 3s^2 3p^2$ (Silicon,$Si$)Ionization enthalpy generally increases across a period from left to right.The elements belong to the $3^{rd}$ period in the order $Na However,$Mg$ $(3s^2)$ has a fully filled orbital,making it more stable than $Al$ $(3s^2 3p^1)$. Thus,$IE_1$ of $Mg > Al$.The order is $Na Therefore,the correct order of first ionization enthalpy is $D > C > B > A$.

$(IP)_1$	$7.1 \ eV$
$(IP)_2$	$10.3 \ eV$
$(IP)_3$	$36.2 \ eV$
$(IP)_4$	$39.1 \ eV$

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:

Similar Questions

An atom with high electronegativity has

The successive ionization energies of an element $(A)$ are given as $IE_1 = 20 \ eV, IE_2 = 45 \ eV, IE_3 = 150 \ eV, IE_4 = 900 \ eV, IE_5 = 1800 \ eV$. What is the formula of the halide of $(A)$?

The ionization potential $(IP)$ values for an element are given below. Identify the element:
$(IP)_1$ $7.1 \ eV$
$(IP)_2$ $10.3 \ eV$
$(IP)_3$ $36.2 \ eV$
$(IP)_4$ $39.1 \ eV$

The first ionization potential of $Na, Mg, Al$ and $Si$ are in the order:

$A$ neutral atom will have the lowest ionization potential when its electronic configuration is

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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: