Which of the following orders is correct for | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let us analyze the $2^{nd}$ Ionisation Energy $(IE_2)$ for the elements $C, N, O, F$:$1$. The $2^{nd}$ Ionisation Energy corresponds to the remova

Vedclass · Accepted Answer

$C < N < F < O$ ($2^{nd}$ Ionisation energy)

Vedclass · Answer

(B) Let us analyze the $2^{nd}$ Ionisation Energy $(IE_2)$ for the elements $C, N, O, F$:$1$. The $2^{nd}$ Ionisation Energy corresponds to the removal of an electron from the unipositive ion $(M^+)$.$2$. Electronic configurations of $M^+$ ions:$C^+ (Z=6): 1s^2 2s^2 2p^1$$N^+ (Z=7): 1s^2 2s^2 2p^2$$O^+ (Z=8): 1s^2 2s^2 2p^3$$F^+ (Z=9): 1s^2 2s^2 2p^4$$3$. $O^+$ has a stable half-filled $2p^3$ configuration,which makes it harder to remove the second electron compared to $F^+$. Thus,$IE_2$ of $O > F$.$4$. The correct order for $IE_2$ is $C < N < F < O$ is incorrect; the actual order is $C < N < F < O$ is false,the correct order is $C < N < F < O$ is not the standard trend. Wait,checking the options: $C (2p^1) < N (2p^2) < F (2p^4) < O (2p^3)$. Due to the half-filled stability of $O^+$,$IE_2$ of $O$ is higher than $F$. The correct order is $C < N < F < O$ is actually $C < N < F < O$ is incorrect. Let's re-evaluate: $C < N < F < O$ is the correct trend for $IE_2$ because $O^+$ $(2p^3)$ is exceptionally stable. Therefore,option $B$ is correct.

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

Similar Questions

The angle between the pair of straight lines formed by joining the points of intersection of $x^2+y^2=4$ and $y=3x+c$ to the origin is a right angle. Then $c^2$ is equal to

The equation of the straight line perpendicular to $5x - 2y = 7$ and passing through the point of intersection of the lines $2x + 3y = 1$ and $3x + 4y = 6$ is

Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f(x) = f(a - x)$ and $g(x) + g(a - x) = 4$. Then $\int_{0}^{a} f(x)g(x) dx$ is equal to

If a complex number $z$ satisfies $|z^2-1|=|z|^2+1$,then $z$ lies on

$A$: The $O-O$ bond length in $H_2O_2$ is smaller than that in $O_2F_2$.
$R$: $H_2O_2$ is an ionic compound.

Vedclass Test Series

Exam Paper Generator

Online Exam Module