Which of the following pairs have identical v | vedclass

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

Question

(A) The bond order is calculated using the formula: $Bond \ order = \frac{1}{2} 	imes (N_b - N_a)$, where $N_b$ is the number of bonding electrons an

Vedclass · Accepted Answer

$N_2^+$ and $O_2^+$

Vedclass · Answer

(A) The bond order is calculated using the formula: $Bond \ order = \frac{1}{2} 	imes (N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.(1) For $N_2^+$ ($13$ electrons): $Bond \ order = \frac{1}{2} 	imes (9 - 4) = 2.5$.For $O_2^+$ ($15$ electrons): $Bond \ order = \frac{1}{2} 	imes (10 - 5) = 2.5$.Since both have a bond order of $2.5$, this pair is correct.(2) For $F_2$ ($18$ electrons): $Bond \ order = 1$.For $Ne_2$ ($20$ electrons): $Bond \ order = 0$.(3) For $O_2$ ($16$ electrons): $Bond \ order = 2$.For $B_2$ ($10$ electrons): $Bond \ order = 1$.(4) For $C_2$ ($12$ electrons): $Bond \ order = 2$.For $N_2$ ($14$ electrons): $Bond \ order = 3$.

Which of the following pairs have identical values of bond order?

Similar Questions

Among the following species,the diamagnetic molecule is

Among the following,the molecule expected to be stabilized by anion formation is

Which of the following species is not paramagnetic?

If $P$ to $T$ are second period $p-$block elements,then which of the following graphs shows the correct relation between the number of valence electrons in $P_2$ to $T_2$ (corresponding molecules) and their bond order?

Which of the following statements is correct for the peroxide ion $(O_2^{2-})$?
$A.$ It has completely filled $\pi^*$ antibonding molecular orbitals.
$B.$ It is diamagnetic.
$C.$ It has a bond order of one.
$D.$ It is isoelectronic with Neon.

Vedclass Test Series

Exam Paper Generator

Online Exam Module