Among the following electronic configurations which one will have low electron affinity value?

1 s^{2}

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1 s^{2} 2 s^{2}

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1 s^{2} 2 s^{2} 2 p^{4}

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1 s^{2} 2 s^{2} 2 p^{5}

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Solution

The correct option is A $1 s^{2}$
Among the given electronic configurations $1 s^{2}$ belongs to helium. As it is a noble gas, it will have low electron affinity value.

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Similar questions

Element	Subshell electronic configuration	No. of electrons present in the last subshell	Group number
$L i$	$1 s^{2} 2 s^{2}$	$1$	$1$
$N a$	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$	-	$1$
$M g$	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$	-	$2$
$C a$	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 s^{2} 3 p^{6} 4 s^{2}$	-	-

1 s^{2}

2 s^{2}

2 p^{6}

3 s^{2}

3 p^{6}

3 d^{5}

4 s^{2}

Element	Subshell electronic configuration	No. of the outer most shell	Period number
$_{4} B e$	$1 s^{2} 2 s^{2}$	$2$	$2$
$_{6} C$	$1 s^{2} 2 s^{2} 2 p^{2}$	$2$	$2$
$_{11} N a$	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$	$3$	-
$_{19} K$	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}$		-

B

1 s^{2} 2 s^{2}

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