Consider the following sequence of reaction:

If electronic configuration of element $X$ is $[N e] 3 s^{1}$ , then which of the following order is incorrect regarding given enthalpies?

| Δ H_{4} | = | Δ H_{5} |

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| Δ H_{2} | > | Δ H_{1} |

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| Δ H_{2} | > | Δ H_{3} |

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| Δ H_{1} | = | Δ H_{6} |

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Solution

The correct option is D $| Δ H_{1} | = | Δ H_{6} |$
A) When a reaction is reversed, magnitude of enthalpy change remains same.
B) Second ionization energy will be more than the first
C) Third ionization energy will be more than the second
D) When a reaction is reversed, magnitude of enthalpy change remains same.

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Similar questions

[N e] 3 s^{1}

4

= [N e] 3 s^{2} 3 p^{3}

= [A r] 3 d^{10} 4 s^{2} 4 p^{3}

= [H e] 2 s^{2} 2 p^{5}

= [N e] 3 s^{1}

[H e] 2 s^{1}, [N e] 3 s^{1}

[A r] 4 s^{1}

(i n K J m o l^{1})

[N e] 3 s^{2} 3 p^{2}

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

1 s^{1} 2 s^{2} 2 p^{3}

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

1 s^{2} 2 s^{2} 2 p^{6} {3 s}^{1}

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